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CHAPTER 3
80 mm
PROBLEM 3.1
A foot valve for a pneumatic system is hinged at B. Knowingthat a - 28°, determine the moment of the 1 6-N force about
Point B by resolving the force into horizontal and vertical
components.
SOLUTION
Note that
and
=a- 20° = 28° -20° = 8°
Fx = (1 6 N)cos 8° = 1 5.8443 NFv =(16N)sin8° = 2.2268N
^£*. \©* — tL> Kl
* U^r^-^T^P,,/|t^^^u^d
^r^^^^-,Also x = (0. 1 7 m) cos 20° = 0. 1 59748 m
y = (0.17 m)sin 20° = 0.058143 m.
>n^y c
Noting that the direction of the moment of each force component about B is counterclockwise,
MB =xFy +yFx= (0.1 59748 m)(2.2268N)
+(0.058143 m)(l 5.8443 N)
= 1.277 N-m or MB =1.277N-m^)4
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153
PROBLEM 3.2
A foot valve for a pneumatic system is hinged at B. Knowingthat a = 28°, determine the moment of the 1 6-N force about
Point B by resolving the force into components along ABC and
in a direction perpendicular to ABC.
SOLUTION
First resolve the 4-lb force into components P and Q, where
g = (16 N) sin 28°
= 7.5115 N
Then MB = rmQ°^7^
= (0.17m)(7.5115N)
= 1.277N-m or MB = 1.277 N-m^^
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154
200 in..,
25"
-100 mm- 200 mm
i'"> mm
PROBLEM 3.3
A 300-N force is applied at A as shown. Determine (a) the momentof the 300-N force about D, (b) the smallest force applied at B that
creates the same moment about D.
SOLUTION
(a)
O.2.
PC
0>T-**>
Fv =(300N)cos25°
= 27.1.89 N
Fy=(300 N) sin 25°
= 126.785 NF = (27 1 .89 N)i + (1 26.785 N)
j
r = ZM = -(0.1m)i-(0.2m)j
MD =rxF
MD = HO. 1 m)i - (0.2 m)j] x [(271 .89 N)i + (1.26.785 N)j]
= -(12.6785 N • m)k + (54.378 N • m)k
= (41.700 N-m)k
M =41.7N-nO^
(b) The smallest force Q at B must be perpendicular to
DB at 45°^£L
MD =Q(DB)
41 .700 N m = (2(0.28284 m) Q = 147.4 N ^L 45° <
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155
200 mn.
25°
•-lOOnim-* -200 mm *.
125 il
•C
H
PROBLEM3.4
A 300-N force is applied at A as shown. Determine (a) the
moment of the 300-N force about D, (b) the magnitude and
sense of the horizontal force applied at C that creates the
same moment about D, (c) the smallest force applied at C that
creates the same moment about D.
SOLUTION
(a) See Problem 3.3 for the figure and analysis leading to the determination ofMd
M =41.7N-m^HCl^n
0>\7.Siy\
c = at.
(b) Since C is horizontal C = Ci
r = DC = (0.2 m)i - (0. 125 m)j
M D =rxCi = C(0.l25m)k
4l.7N-m = (0.l25m)(C)
C = 333.60 N
(c) The smallest force C must be perpendicular to DC; thus, it forms a with the vertical
C = 334N <
tan6^0.125 m0.2 m
a = 32.0°
M D = C(£>C); DC = V( -2 m)
2+ (°- 125 m)'
= 0.23585 m
41.70 Nm = C(0.23585m) C = 176.8 N^L 5HX)°<
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156
PROBLEM 3.5
An 8-1b force P is applied to a shift lever. Determine the moment of V about Bwhen a is equal to. 259.
SOLUTION
First note Px = (8 lb) cos 25°
= 7.2505 lb
/^ =(8 lb) sin 25°
= 3.3809 lb
Noting that the direction of the moment of each force component about B is
clockwise, have
= -(8in.)(3.3809 1b)
- (22 in.)(7.2505 lb)
= -186.6 lb -in.
i*22. •**.
or Mj =186.6 lb -in. J) ^
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157
PROBLEM 3.6
For the shift lever shown, determine the magnitude and the direction of the smallest
force P that has a 21 0-lb • in. clockwise moment about B.
22 in.
SOLUTION
For P to be minimum it must be perpendicular to the line joining Points A. and B. Thus,
a =e
, 8 in ^T^s*. K
22 in. /= 19.98° i
and MB =dP^n p ZZ i«.
Where d = rAIB fl$ u=^m.y+(22m.y JB w= 23.409 in.
Then_210Ib-in.
' min " 23.409 in.
-8.97 lb Pmin =8.97 lb^ 19.98° <
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158
PROBLEM 3.7
An 1 1 -lb force P is applied to a shift lever. The moment of P about B is clockwise
and has a magnitude of250 lb • in. Determine the value ofa.
22 in.
SOLUTION
By definition
where
and
also
Then
or
or
and
MB =rmPsm.6
= a + (9Q°-tf>)
_i 8 in.<p - tan" 19.9831'
22 in.
rf/fl =V(8in.)
2+(22in.)
2
= 23.409 in.
250lb-in = (23.409in.)(lllb)
xsin(tf + 90° -19.9831°)
sin (« + 70.01 69°) = 0.97088
a + 70.0 169° = 76. 1391°
a+ 70.0169° = 103.861° a = 6.12° 33.8° <
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159
PROBLEM 3.8
It is known that a vertical force of 200 lb is required to remove the nail
at C from the board. As the nail first starts moving, determine (a) the
moment about B of the force exerted on the nail, (b) the magnitude of
the force P that creates the same moment about B if a ~ .1 0°, (c) the
smallest force P that creates the same moment about B.
V—Ar 4 in.n
SOLUTION
(a) We have MB =raBFN(4 in.)(200 lb)
800 lb -in.
or MB =H00\b-m.)<
A- m.(/;) By definition MB ~rA/BPsin
6» = 10° + (180°~70°)
= 120°
Then 800 lb • in. = (18 in.) x Psin 1 20°
or P = 51.3 lb <
(c) For P to be minimum, it must be perpendicular to the line joining
Points A and B. Thus, P must be directed as shown.
Thus
or
or
A*W™.cl = fA IB
800 lb • in. = (18 in.)Pm
^i„=44.4 1b Pmitl
=44.41b^l20 ^
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160
PROBLEM 3.9
A winch puller AB is used to straighten a fence post. Knowing that the tension in cable BC is 1040 N andlength d is 1 ,90 m, determine the moment about D of the force exerted by the cable at C by resolving that
force into horizontal and vertical components applied (a) at Point C, (b) at Point E.
0.2 m
0.875 m
SOLUTION
(a) Slope of line
Then
EC =0.875 m
1.90 m + 0.2 m 12
*abx - ,~ (Tab)
12
13
960 N
(1040 N) 0,1«y
and *W=-0040N)
= 400 N
Then
(b) We have
MD - TABx (0.875 m)~TABy (0.2 m)
= (960 N)(0.875 m) - (400 N)(0.2 m)
= 760 N •
m
MD ~rm (y) + TABx (x)
= (960 N)(0) + (400 N)(I .90 m)
= 760 N •
m
or M./J=760N-m
>
)^
or M7>= 760N-m
v)<«
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161
PROBLEM 3.10
It is known that a force with a moment of 960 N • m about D is required to straighten the fence post CD. If
d- 2,80 m, determine the tension that must be developed in the cable of winch puller AB to create the
required moment about Point D.
0.875 m
0.2 j.i»
SOLUTION
\*.ȣ^
*AB
'My
o. a? 5^i
z-acw OiZ^
Slope of line
Then
and
We have
EC =0.875 m 7
7'
2.80 m + 0.2 m 24
24My
r/ffl)>
25
125
T/)B
rAH
24 7960N-m=—7^(0)+—7^(2.80*)
7^= 1224 N or r^=1224N ^
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162
PROBLEM 3.11
It is known that a force with a moment of 960 N • m aboutD is required to straighten the fence post CD. Ifthe
capacity of winch puller AB is 2400 N, determine the minimum value of distance d to create the specified
moment about Point D.
fe
0.2 in
0.875 in
SOLUTION
o.mtn
czom
The minimum value ofd can be found based on the equation relating the moment of the force TAB about D:
MD ={TABmK ),(d)
where
Now
MD =960N-m
(^flmax )y = TAIHmx sin & = (2400 N)sfo
. . 0.875msin #
960 N • m = 2400 N
^(t/ + 0.20)2+(0.875)
2m
0.875(d)
+ 0.20)2+(0.875)
2
or ^ + 0.20)2 + (0.875)
2 = 2. ! 875d
or (J + 0.20)2 + (0.875)
2 = 4.7852rf2
or 3 .7852</2 - 0.40c/- .8056 =
Using the quadratic equation, the minimum values ofd are 0.51719 m and -.41151 m.
Since only the positive value applies here, d — 0.5 1 7 1 9 mor d ~ 5 1 7 mm ^
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163
5.3 in. PROBLEM 3.12
12.0 in.
2.33 in.
mmM1 >iii. i
The tailgate of a car is supported by the hydraulic lift BC. If the
lift exerts a 125-lb force directed along its centerline on the ball
and socket at B, determine the moment of the force about A.
SOLUTION
First note
Then
and
Now
where
Then
dcB == 7(12.0 in.)2
= 12.224 lin.
+ (2.33 in.)2
cos 9 =12.0 in.
12.2241 in.
sin 9-2.33 in.
12.2241 in.
*cb = FCB cos 9\ -FCB $m9l1251b
12.2241 in.
[(12.0 in.)i- (2.33 in.) j]
M.A= VB/A X ^CB
XBIA = (15.3in.)i-- (12.0 in.+ 2.33 in.) j
M
= (15.3 in.) i--(14.33 in.) j
ru-nimn* 1251b
\«5»,a> im.
\Z.& >N.
2.33 ttvi.
— (12.01 — 2.331)12.2241 in.
(1393.87 lb in.)k
(116.156 lb -ft)k or M„ = 116.2 lb- ft *)<
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164
20.5 in.
h—
-
4.38 in
i
*Ti
7.62
1
t§|1.7.2 hi.
PROBLEM 3.13
The tailgate of a car is supported by the hydraulic lift EC. If the lift
exerts a 125-lb force directed along its centerline on the ball andsocket at B, determine the moment of the force about A.
SOLUTION
First note
Then
dCB := 7(17.2 in.)2
= 18.8123 in.
+ (7.62 in.)2
cos -17.2 in.
18.8123 in.
sin 6 -7.62 in.
18.8123 in.
Z.O.'S «w.
and
Now
where
Then
fcd = (Ft:v*cos 0)\ - (FCB sin 6>)j
=S(,7 '2i"-)i + (7 '62i^
r^ = (20.5 in.)i- (4.38 in.)j
M, = [(20.5 in.)i - (4.38 in.)j] xt
.
1251b(1 7.21 - 7.62j)
18.8 123 in.
n.z .«»».
4.?.e. >Ni.
XKvZ >KJ.
(1538.53 lb -in.)k
(128.2 lb -ft)k or M^ =128.2 lb- ft^H
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165
120mm
65 mm
PROBLEM 3.14
A mechanic uses a piece of pipe AB as a lever when tightening an
alternator belt. When he pushes down at A, a force of 485 N is
exerted on the alternator at B. Determine the moment of that force
about bolt C if its line of action passes through O.
SOLUTION
We have M c =rwc xFj,
Noting the direction of the moment of each force component about C is
clockwise.
Where
and
Mc =xFBy +yFBx
x - 120 mm - 65 mm = 55 mm
y - 72 mm + 90 mm - 1 62 mm
4 j,
F65
lix
Fa
7(65)2+(72)
2
72
V(65)2 + (72)
3
-(485N) = 325N
.(485 N)- 360 N
iWc = (55 mm)(360 N)+ (1 62)(325 N)
= 72450 N- mm= 72.450 N-m or Mc =72.5 N-m J) <
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166
By definition:
Now
and
PROBLEM 3.15
Form the vector products B x C and B' * C, where B = B\ and use the results
obtained to prove the identity
sin a cos/? = -sin (a + P) + -sin (a - p).
SOLUTION
N(>te: B = £(cos/?i + sin/?j)
B'= 5(cos/?i-sin/?j)
C - C(cosa i + sin a j)
|BxC| = flCsin(a-jff)
|B'xC| = 5Csin(flf + y?)
B xC = Z?(cos /?i + sin y9j)x C(cos tfi + sin orj)
= BC(cos /?sin « - sin /?cos «)k
B'xC - /?(cos /?i - sin /?j)x C(cos ai -f sin #j)
- £C(cos yffsin ar+ sin /?cos ar)k
Equating the magnitudes of BxC from Equations (I ) and (3) yields:
BCs'm(a-p)~ BC(cos ps'm a- sin pcos a)
Similarly, equating the magnitudes of B'xC from Equations (2) and (4) yields;
BCsm(a + p) = BC(cos ps'm a + sln pcos a)
Adding Equations (5) and (6) gives:
sin(a- p) + s'm(a+ p) - 2cos /?sin a
0,)
(2)
(3)
(4)
(5)
(6)
or sin « cos /? --sin(ar + /?) + -sin(flf - /?) ^
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167
PROBLEM 3.16
A line passes through the Points (20 m, 16 m) and (-1 m, ~4 m). Determine the perpendicular distance rffrom
the tine to the origin O of the system of coordinates.
SOLUTION
dAB = V[20m- (-1 m)]2 + [1 6 m - (-4 m)f
- 29 m
Assume that a force F, or magnitude F(N), acts at Point A and is
directed ixomA to B.
Then,,
Where
By definition
Where
Then
¥~FX m
'All*B~ rA
d.41!
= -—<21i + 20j)29
V
MQ = \rA xF\ = dF
r,=-(lm)i-(4m)j
F
E> (Zom, K-rti^
M =[-(-1 m)i-(4 m)j]x—-[(21 m)i + (20 m)j]29 m
a -(20)k + (84)k]
~F|k N-m29
-*x
Finally64
29F = </(F)
. 64rf-— m
29</ = 2.21m ^
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168
PROBLEM 3.17
The vectors P and Q are two adjacent sides of a parallelogram. Determine the area of the parallelogram when(tf)P--7i + 3j-3kandQ = 2i + 2j + 5k,(6)P = 6i-5j-2kandQ = -2i + 5j~k.
SOLUTION
(a) We have
where
Then
(b) We have
where
v4 = |PxQf
P = -7i + 3j-3k
Q = 2i + 2j + 5k
PxQk
-3
> J
-7 3
2 2 5
= [(15 + 6)i + (-6 + 35)j + (-14-6)k]
= (21)1 + (29)j(-20)k
^ = V(20)2+(29)
2+(-20)
2
A = \PxQ\
P = 6i~5j-2k
Q = ~2i + 5in.j~lk
or 4 = 41.0 <4
Then PxQi fci
6 -5
-2 5
= [(5 + 1 0)i + (4 + 6)j + (30-1 0)k]
«(15)i + (10)j + (20)k
A=j(i5)-+(\oy + (2oy or 4 = 26.9 ^
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169
PROBLEM 3.18
A plane contains the vectors A and B. Determine the unit vector normal to the plane when A and B are equal
to, respectively, (a) i + 2j - 5k and 4i - 7j - 5k, (b) 3i - 3j + 2k and -2i + 6j - 4k.
SOLUTION
{a) We have
where
Then
and
(b) We have
where
Then
and
AxB
A
B
AxB
|AxB|
:li + 2j-5k
= 4i - 7 j- 5k
i J *
1 +2 -5
4 -7 -5
(-1 ~ 35)i + (20 + 5)j + (-7 - 8)k
15(31 -1J -Ik)
|AxB|
X--
X-
A
B
AxB
15V(-3)2 +H)2
+(-l)2-I5>/ri
15(-3f-lJ-lk)or X
15VH
AxB
(-3i-j~k) 4
|AxB|
3i-3j + 2k
:-2i + 6j-4k
I J k
3-3 2
-2 6 -4
(12-12)i + (-4+ l2)j + (18-6)k
(8j + 12k)
|AxB| 4^(2)2 +(37 = 4^13
4(2j + 3k)
4^13or X,
V^(2j + 3k) <
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170
PROBLEM 3.19
Determine the moment about the origin O of the force F = 4i + 5j - 3k that acts at a Point A. Assume that the
positioji vector ofA is {a) r - 2i - 3j + 4k, (£) r - 2i + 2.5j - 1 .5k, (c) r - 2i + 5j + 6k.
SOLUTION
(a) M,
(h) M,
' J k
2-3 4
4 5 -3
(9-20)i + (16 + 6)j + (I0+ 12)k
i j k
2 2.5 -1.5
4 5 -:s
Ma = -lli + 22j + 22k ^
(-7.5 + 7.5)i + (-6+ 6)j + (10-1 0)k M, «
(<0 M,
' J
2 5
4 5
(-1 5 - 30)i + (24 + 6)j + (10- 20)k M, -45i + 30j-10k A
Note: The answer to Part b could have been anticipated since the elements ofthe last, two rows of the
determinant are proportional.
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171
PROBLEM 3.20
Determine the moment about the origin of the force F == -2i + 3j + 5k that acts at a Point A. Assume that the
position vector ofA is (a) r == i + j + k,(6)r == 2i + 3j- 5k, (c)r == -4i + 6j + 10k.
SOLUTION
i j k
(a) M(}= 1 1 1
-2 3 5
= (5-3)i + (-2-5)j + (3 + 2)k M = 2i-7j + 5k A
i j k
(h) M - 2 3 -f
-2 3 5
= (15 + 15)i + (10-10)j + (6+ 6)k Mo =30i + I2k A
1 J k
(c) M a - -4 6 10
-2 3 5
= (30 - 30)i + (-20 + 20)j + (-.12 + 1 2)k M = <
Note: The answer to Part c could have been anticipated since the elements of the last two rows of the
determinant are proportional.
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172
-VI
200 N
r
PROBLEM 3.21
A 200-N force is applied as shown to the bracket ABC. Determinethe moment of the force about A.
SOLUTION
We have
where
Then
M, =%.i x *c
rCIA = (0.06 m)i + (0.075 m)j
*c = ~(200 N)cos 30°j + (200 N)sin 30°k
i j &
MA-.= 200 0.06 0.075
-cos 30° sin 30°
= 200[(0.075sin 30°)i - (0.06sin 30°)j ~ (0.06 cos 30°)k]
or M,, = (7.50 N m)i- (6.00 N-m)j-(l 0.39 N-m)k <4
PROPRIETARY MATERIAL <Q 201 The McGraw-Hill Companies, inc. All rights reserved. No part of this Manual may be displayed,reproduced or distributed in anyform or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educatorspermittedby McGraw-Hillfor their individual course preparation. Ifyou are a student using this Manual,you are using it without permission.
173
'/ PROBLEM 3.22
£^-^
Before the trunk of a large tree is felled, cables AB and BC are
attached as shown. Knowing that the tensions in cables AB and BCare 555 N and 660 N, respectively, determine the moment about O
. .'vjB 4.25 in
of the resultant force exerted on the tree by the cables at B.
/ ,.-"r
^J""
0.75 in/\-^1 "I
.T
SOLUTION
We have
where
and
M,
1
ff/O
rl3/Q
X F#
(7m)j
Tdff + Tec
'yf/f ~ ,VBA' AB
l J!C
-(0.75m)i-(7m)J+ (6m)k
(.75)2+(7)
2+(6)
2 m
*-BC*BC
(555 N)
(4.25m)i-(7m)j + (lm)k
^(4.25)2+(7)
2+(l)
2 m-(660 N)
Fi?= [-(45.00 N)i - (420.0 N)j -f- (360.0 N)k]
+[(340.0 N)i - (560.0 N)j + (80.00 M)k]
= (295.0 N)i - (980.0 N)j + (440.0 N)k
M,
i J k
7
295 980 440
Nm
(3080 N • m)i - (2070 N • m)k or M,}= (3080 N-m)i- (2070 N-m)k <
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174
fi in
PROBLEM 3.23
The 6-m boom ^5 has a fixed end /*. A steel cable is stretched fromthe free end B of the boom to a Point C located on the vertical wall.
If the tension in the cable is 2.5 kN, determine the moment about A ofthe force exerted by the cable at B.
SOLUTION
First note
Then
We have
where
Then
^c-VK>)2+(2.4)
2+(-4)
:
= 7.6 m
V=-?4^(-« + 2.4j-4k)7.6
^A ~ r/i//l
XTBC
lB/A (6 m)i
M;4 ^(6m.)ix^4r^-(~6» + 2.4j-4k)
7.6
or M //== (7.89 kN-m)j + (4.74 kN-m)k <
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced or distributed in anyform or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permittedby McGraw-Hillfor their individual coursepreparation. Ifyou are a student using this Manual,you are using it without permission.
175
3{ji)i.. C- 48itii
:'6-iftl
PROBLEM 3.24
A wooden board AB, which is used as a temporary prop to support
a small roof, exerts at Point A of the roof a 57-lb force directed
along BA. Determine the moment about C of that force.
SOLUTION
We have
where
and
rA/c = (48 in.)i - (6 in.)j + (36 in.)k
\<BA— xM pM
-(5in.)i + (90in.)j-(30in.)k
V(5)2+(90)
2 + (30)2in.
= -(31b)i + (541b)j-(181b)k
(57 lb)
M c lb -in.
i J k
48 6 36
3 54 18
-(1 836 lb • in.)i + (756 lb • in.)j + (2574 lb • in.)
or Mc = -(1 53 .0 lb • ft)i + (63.0 lb • ft)j + (2 1 5 lb • ft)k A
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176
/.)«'
0.6 id
0.6 m "X
PROBLEM 3.25
The ramp ABCD is supported by cables at corners C and D.
The tension in each of the cables is 810 N. Determine the
moment about A of the force exerted by (a) the cable at D,
(b) the cable at C.
SOLUTION
(a) We have M^riy,xTM
where **/•/,(= (2-3 m)j
Tw: =*"DE^f.)E
- (0.6m)l + (3.3in)J-(3m)k
V(0.6)2+(3.3)
2+(3)
2m= (1 08 N)i + (594 N)j - (540 N)k
i J k
M,,= 2.3
108 594 -540
Nm
= -(1 242 N - m)i - (248.4 N • m)k
or M/i=-(1242N-m)i-(248N-m)k <
(b) We have ^A=rG//lxTCG
where r(,,=(2.7m)i + (2.3m)j
T — 1 T*CO ~~ *"CG'CG
= -(.6m)i + (3.3m)j-(3m)k(810N)
V(.6)2+(3.3)
2+(3)
2m= -(108 N)i + (594 N)j - (540 N)k
i J k
M A= 2.7 2.3
-108 594 -540
N-m
= -(1 242 N • m)i + (1458 N • m)j + (1 852 N • m)k
or MA = -(1242 N • m)i + (1 458 N • m)j + (1 852 N • m)k A
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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177
PROBLEM 3.26
A smaJI boat hangs from two davits, one of which is shown in the
figure. The tension in line ABAD is 82 lb. Determine the momentabout C of the resultant force R,, exerted, on the davit at A.
SOLUTION
We have
where
and
Thus
Also
Using Eq. (3.21):
R.=2R(/)
. + F,D
,„~-(82 1b)j
ADAD I AD— = (82 lb)
AD6i~7.75j-3k
10.25
F,/3=(48lb)i-(62 1b)j-(24 1b)k
RA = 2¥AB +FAD = (48 lb)i - (226 lb)j - (24 lb)k
l A/C
Mc
(7.75ft)j + (3ft)k
1 J
7.75
48 -226
k
3
-24
(492 lb • ft)i + (1 44 lb • ft)j - (372 lb • ft)k
Mc = (492 lb • ft)i + (1 44.0 lb ft)j - (372 lb • ft)k <
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178
PROBLEM 3.27
In Problem 3.22, determine the perpendicular distance fromPoint O to cable AB.
PROBLEM 3.22 Before the trunk of a large tree is felled,
cables AB and BC are attached as shown. Knowing that the
tensions in cables AB and BC are 555 N and 660 N,
respectively, determine the moment about O of the resultant
force exerted on the tree by the cables at B.
SOLUTION
We have
where
Now
and
IM TBAd
d = perpendicular distance from O to line AB.
M, rB/0 X *BA
rBIO ={lm)\
BA' AB
(0.75m)i-(7m)j + (6m)k
*BA ~~ '"BA 1 AB
M,
and
or
(555 N)'(0.75)
2+(7)
2+(6)
2m-(45.0 N)i - (420 N)j + (360 N)k
i j k
7 N-m-45 -420 360
(2520.0 N • m)i + (315.00 N • m)k
|M1= V(252o-°)
2 + (31 5.00);
= 2539.6 N-m
2539.6 N-m = (555 N)rf
d = 4.5759 m or d = 4.58 m <
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179
PROBLEM 3,28
In Problem 3.22, determine the perpendicular distance from
Point O to cable BC.
PROBLEM 3.22 Before the trunk of a large tree is felled,
cables AB and BC are attached as shown. Knowing that the
tensions in cables AB and BC are 555 N and 660 N,
respectively, determine the moment about O of the resultant
force exerted on the tree by the cables at.#.
SOLUTION
We have
where
[M, Tscd
d — perpendicular distance from O to line BC.
M,
lB/0
rB/0 X ^liC
7mj
P — 1 T'BC ™ *"BC* HC
(4.25m)i-(7m)j + (lm)k(660 N)
M,
and
V(4.25)2+(7)
2+(l)
2m
(340 N)i - (560 N)j + (80 N)k
I J k
7
340 -560 80
(560 N • m)i - (2380 N • m)k
|M!= 7(560)
2+(2380)
= 2445.0 N-m
2445.0 N-m = (660 N)rf
d = 3.7045 m or d~ 3.70 m
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180
A."(i in.
•IS in.
Cj-iii.
.90 in,:
()(> in.
~~*S B ~~^~£
PROBLEM 3.29
In Problem 3.24, determine the perpendicular distance from Point Dto a line drawn through Points A and B.
PROBLEM 3.24 A wooden board AB, which is used as a temporary
prop to support a small roof, exerts at Point A of the roof a 57-lb
force directed along BA. Determine the moment about C of that
force.
SOLUTION
We have
where
and
IMJ^rf
d — perpendicular distance from D to line AB.
M.l}^rAlii x¥]iA
r,/D =-(6in.)j + (36in.)k
^liA ~ ^BA^BA
(-(5in.)i + (90in.)j-(30 in.)k)
V(5)2+(90)
2 + (30)2in.
-(31b)i + (541b)j-(181b)k
(57 lb)
M,;)
I J k
-6 36 lb -in.
-3 54 -18
-(1 836.00 lb • in.)i - (1 08.000 lb • in.)j- (1 8.0000 lb • in.)k
IM 836.00)2+(108.000)
2 + (18.0000)'
= 1839.26 lb -in.
1839.26 lb -in =(57 lb)e/
</ = 32.268 in. or </ = 32.3in. <
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181
(if) in.
B^-<
PROBLEM 3.30
In Problem 3.24, determine the perpendicular distance from Point C to a
line drawn through Points A and B.
PROBLEM 3.24 A wooden board AB, which is used as a temporary
prop to support a small roof, exerts at Point A of the roof a 57-3b force
directed along BA. Determine the moment about C of that force.
SOLUTION4
We have
where
\Mc \~FBAd
d - perpendicular distance from C to line AB.
3fc.. .
M.c ^rA/c x¥BA ^%f M / >lD ^-.
rAIC - (48 in.)i - (6 in,)j + (36 in.)k ^^\ \ //S^fylA ~ ^BA^IIA f*
_ (~(5 in.)i + (90 in.)j - (30 in.)k) fe*
7(5)2+(90)
2+(30)
2in.
= ™(3 lb)i + (54 lb)j - (1 8 lb)k
i J k
Mc = 48 -6 36
-3 54 -18
lb -in.
and
= -(1 836.001b -i
|M
n.)i - (756.00 lb • in.)j + (2574.0 lb • in.)k
2
c |= V(l 836.00)
2 + (756.00)2 + (2574.0)
-3250.8 lb -in.
3250.8 lb in. = 57 lb
d- 57.032 in. or d = 57.0 in. ^
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182
0.6 m
PROBLEM 3.31
In Problem 3.25, determine the perpendicular distance fromPoint A to portion DE of cable DEF.
PROBLEM 3.25 The ramp ABCD is supported by cables at
comers C and D, The tension in each of the cables is 810 N.
Determine the moment about A of the force exerted by (a) the
cable at D, (b) the cable at C. .
SOLUTION
We have
where
IM TDEd
and
d
—
perpendicular distance from A to line DE,
Mr„HIA (2.3 m)j
'DE ~ A"DE*-DE
(0.6m)i + (3.3m)j-(3m)k(8I0N)
M
V(0.6)2 + (3.3)
2+(3)
2 m(108 N)i + (594 N)j - (540 N)k
i J b
2.3 N-m108 594 540
~(1 242.00 N • m)i - (248.00 N • m)k
|MJ =-N/(1242.00)
2+(248.00)
:
= 1266.52 N-m
1266.52Nm = (8 10 N)d
d = 1 .56360 m
O'fcn,
or d = 1.564 m
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educatorspermittedby McGraw-Hillfor their individual coursepreparation. Ifyou are a student using this Manual,you are using it withoutpermission.
183
PROBLEM 3.32
In Problem 3.25, determine the perpendicular distance from
Point A to a line drawn through Points C and G.
PROBLEM 3.25 The ramp ABCD is supported by cables at
corners C and D. The tension in each of the cables is 810 N.
Determine the moment about A of the force exerted by (a) the
cable at D, (h) the cable at C.
0.6 m
SOLUTION
We have
where d - perpendicular distance from A to line CG.
rG/A X IceM
IVu/A
lCG
M
(810 N)
(2.7m)i + (2.3m)j
-(0.6.m)i + (3.3m)j-(3 m)k
A/(0.6)2+(3.3)
2 + (3)2m
-(108N)i + (594N)j-(540N)k
i J k
2.7 2.3 N-m-108 594 -540
-(1242.00 N • m)i + (1458.00 N • m)j +(1 852.00 N m)k
and
= 2664.3 N-m
2664.3 N-m = (8 ION)d
d = 3.2893 m or d = 3.29 m 4
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184
PROBLEM 3.33
.In Problem 3.26, determine the perpendicular distance fromPoint C to portion AD of the line ABAD.
PROBLEM 3.26 A small boat hangs from two davits, one ofwhich is shown in the figure. The tension in line ABAD is
82 lb. Determine the moment about C of the resultant force
R4 exerted on the davit at A.
SOLUTION
First compute the moment about C of the force FDA exerted by the line on D:
From Problem 3.26:
= -(48 lb)i+ (62 lb)j + (24 lb)k
Mc =rm; xF^= +(6 ft)i x[-(48 lb)i + (62 lb)j + (24 lb)k]
= -(1441b-ft)j + (372 1b-ft)k
Mc =V044>2+(372)
2
= 398.90 lb • ft
Then Mc = VMd
Since F,w = 82 1b
398.90 lb -ft
82 1bd = 4.86 ft <
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced or distributed in anyform or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educatorspermittedby McGraw-Hillfor their individual course preparation. Ifyou are a student using this Manual,yon are using it without permission.
185
*-* PROBLEM 3.34
Determine the value of a that minimizes the
i
16 ft perpendicular distance from Point C to a section
of pipeline that passes through Points A and B.
SOLUTION
Assuming a force F acts along AB,
\M.cMrA/c xF\^F(d)
Where d - perpendicular distance from C to line AB
%ABF
(24ft)i + (24ft)j-(28)k
F =W7(24)
2+(24)
2+(18)
2ft
•F
(6)i + (6)j-(7)k
AK; = (3 ft)i - (1 ft)j- {a - 1 ft)k
i J k
3 -10 10a
6 6-7Mc
.[(10+ 6fl)i + (81-6.ai)j + 78k]11
Since
12!
^/c xr or r^xF2
|=W
(1 + 6a)2 +(81- 6a)
1 + (78)2 = d*
d / j2Setting -j-{d )- to find a to minimize c/
1
[2(6)(l + 6a) + 2(-6)(8 1 - 6a)] =
Solving
121
a = 5.92 ft or o - 5.92 ft ^
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186
PROBLEM 3.35
Given the vectors P =3i -
P S, and Q • S.
-J + 2k,Q == 4i + 5j-3k, and S = -2i + 3j - k, compute the scalar products P • Q,
SOLUTION
P-Q = (31-1j + 2k)-(4i-5J-3k)
= (3)(4) + (-l)(~5) + (2)(-3)
= 1 or P Q = I AP • S = (3i - lj + 2k) • (-2i + 3j - Ik)
= (3)(-2) + H)(3) + (2)H)= -U. or P-S = -U <
Q-S-(4i-5j~3k)-(-2i + 3j-l.k)
= (4K-2) + (5)(3)+ (-3X-l)
= 10 or Q-S = 10 <
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced or distributed in anyform or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers andeducatorspermittedby McGraw-Hillfor their individual coursepreparation, ifyou are a student using this Manual,you are using it without permission.
187
PROBLEM 3.36
Form the scalar products B • C and B' • C, where B = B', and use the
results obtained to prove the identity
cos a cosP ~— cos (a + /?)+— cos (a - ft).
SOLUTION
y By definition
where
(1)
B-C = £Ccos(a~/?)
B = /?[(cos/?)i + (sin/?)j]
C = C[(cos a)\ + (sin a)j]
(B cos /?)(Ccos a) + (B sin /?)(Csin a) - BCcos(a~ (3)
or cos /?cos a+ sin /?sin a = cos(a - j3)
By definition B' • C - BC cos (or + /?)
where B' = [(cos fi)i ~ (sin /?)j]
(B cos p)(C cos or) + (~B sin /?)(C sin or) = BCcos (a + /?)
or cos /?cos « - sin ft sin a = cos{a + /?)
Adding Equations (1) and (2),
2 cos /?cos a = cos (a-fi)+ cos (a+ /?)
or cosacosfi = ~-cos(a + fi) + ~cos(a~ ft) A
(2)
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188
PROBLEM 3,37
Section AB of a pipeline lies in the yz plane and forms an angle
of 37° with the z axis. Branch lines CD and EF join AB as
shown. Determine the angle formed by pipes AB and CD.
SOLUTION
First note AB = AB{$m 37°j - cos 37°k)
CD = CD(- cos 40° cos 55°j + sin 40°j- cos 40° sin 55°k)
fc.b\
Now
or
or
AB- CD = (AB)(CD) cos
AB(sm 37°j - cos 37°k) • CD(-cos 40° cos 55°i + sin 40°j - cos 40°sin 55°k)
~(AB)(CD) cos
cose? = (sin 37°)(sin 40°) + (~cos 37°)(--cos 4()°sin 55°)
= 0.88799
or (9 = 27.4° A
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies., Inc. All rights reserved. No part of this Manual may be displayed,reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educatorspermittedby McGraw-Hillfor their individual coursepreparation. Ifyou are a student using this Manual,you are using it without permission.
189
PROBLEM 3.38
Section AB of a pipeline lies in theyz plane and forms an angle
of 37° with the z axis. Branch lines CD and EF join AB as
shown. Determine the angle formed by pipes AB and EF.
SOLUTION
First note AB = /f£?(sm37 j-cos37 k)__.
= £F(cos 32° cos 45°i + sin. 32°j - cos 32° sin 45°k)
1
£
F
l/^ (€t\ *W A
Now AB-EF--= (AB)(EF)cosO
or AB(sm
:
17°j - cos 37°k) - £F(cos 32° cos 45°j + sin 32°j - cos 32° sin 45°k)
= (AB)(EF)cos0
or cos = (sin 37°)(s.in 32°)+ (-cos 37°)(-cos 32° sin 45°)
= 0.79782
or (9 = 37.1° ^
PROPRIETARY MATERIAL © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No parr of this Manual may be displayed,
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1 90
PROBLEM 3.39
Consider the volleyball net shown.
Determine the angle formed by guywires AB and AC,
SOLUTION
First note
and
By definition
or
or
AB - V(-6.5)2 + (-8)
2 + (2)2 = 1 0.5 ft
AC^yj(0f+(~S)2 + (6)2 =]0 ft
AB = -(6.5 ft)i - (8 ft)j + (2 ft)k
7c = -(8ft)j + (6ft)k
AB-AC = (ABXAC)cos0
(-6,51 - 8j + 2k) • (-8j + 6k) = (1 0.5)0 0) cos -
(-6.5)(0) + (-8)(-8)+ (2)(6) - 1 05 cos 6
cos = 0.72381 or = 43.6° <
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced or distributed in anyform or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permittedby McGraw-Hillfor their individual coursepreparation. Ifyou are a student using this Manual,you are using it withoutpermission.
191
sit
PROBLEM 3.40
Consider the volleyball net shown.
Determine the angle formed by guy
wiresAC and AD.
SOLUTION
First note
and
By definition
or
or
^C = V(0)2+(~8)
2+(6)
2
-10 ft
AD = J(4)2+(-&)
2 + (l?
= 9 ft
IC = ~(8ft)j + (6ft)k
75 = (4ft)j-(8ft)j + (lft)k
AC AD = (AC)(AD)cos0
(-8j+ 6k) • (4i - 8j + k) = (1 0)(9) cos t
(0)(4) + (-8)(-8) + (6)(1) = 90cos 6
cos0 = 0.77778 or # = 38.9° <
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192
1.2 in
2.-1
PROBLEM 3.41
Knowing that the tension in cable AC is 1260 N, determine
(a) the angle between cable AC and the boom AB, (b) the
projection on AB ofthe force exerted by cableAC at Points.
-f
2.6 in
2.4 m
SOLUTION
(a) First note
and
By definition
or
or
or
(b) We have
AC~-=^j(~2A)2 + (0.8)
2+(1.2)
2
= 2.8m
AB = yj(-2A)2+(-].&? +(Q)
2
= 3.0m
^C = -(2.4 m)i + (0.8 m)j + (1 .2 m)k
I/? = -(2.4m)i-(K8m)j
AC -AB = (AC){AB) cos
(-2.41 + 0.8j + 1 .2k) • (-2.4i - 1 .8j) = (2.8)(30) x cos
(-24X-2.4) + (0.8X-1 .8) + (1 .2X0) - 8.4cos
cos# = 0.51429
= 7^c cos<9
= (1260N)(0.51429)
or = 59.OC
or C^cU=648N
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193
2.4 m
PROBLEM 3.42
Knowing that the tension in cable AD is 405 N, determine (a) the
angle between cable AD and the boom AB, (b) the projection on
AB of the force exerted by cable AD at Point A.
SOLUTION
{a) First note AD = >/(-2.4)2 + (i ,2)
2 + (-2.4)2
= 3.6m
AB = V(-2.4)2 + (-1 .8)
2 + (0)2
-3.0m
and AD = -(2.4 m)i + (1 .2 m)j - (2.4 m)k
AB = -(2.4m)i~(1.8m)j
By definition, ADAB = {AD)(AB)cos0
(-2.4i + 1 .2j - 2.4k) (-2.4i - .1 .8j) = (3.6)(3.0)cos#
(-2.4)(-2.4) 4- (1 .2)(-I.8) + (-2.4X0) = 10.8 cos
cos $ — —3
^^70.5° <
(b) (*AD'AB ~ *AD ' ^Ali
= TAD cos&
=(405N)^ Pad)ab = 135.0 N <
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194
PROBLEM 3.43
Slider P can move along rod OA. An elastic cord PC is
attached to the slider and to the vertical member BC. Knowingthat the distance from O to P is 6 in. and that the tension in thecord is 3 lb, determine (a) the angle between the elastic cordand the rod OA, (b) the projection on OA of the force exertedby cord PC at Point P.
SOLUTION
First note
Then
^ = >/(12)2+(12)
2+(-6)
2=18in.
» OA 1 ,
= i(2i + 2j~k)
Now OP = 6 in, => OP = -(04)
The coordinates ofPoint P are (4 in., 4 in., -2 in.)
PC = (5 in.)i + (1 .1 in..)j + (14 in.)k
PC = 7(5)2+ (1 1)
2 + (14)2 = V342 in.
~PCXOA ^(PC)cosO
so that
and
(a) We have
or
or
(5i + llj + 14k)--(2i + 2(i-k) = 7342cos^
cos]
3V342
0.32444
[(5)(2) + (ll)(2) + (14)(~l)j
(b) We have
= {Tpc'K]>c)-XOA
PCftpc OA
— Tpr COS&
= (3 1b)(0.32444)
or = 71.1° ^
or (Tpc)oa =0-973 lb ^
=^^S~S5£=SS£=f£S195
PROBLEM 3.44
Slider P can move along rod OA. An elastic cord PC is
attached to the slider and to the vertical member BC.
Determine the distance from O to P for which cord PC and
rod OA are perpendicular.
SOLUTION
First note
Then
(M^/(12)2+(12)
2 +(-6)- =18 in.
OA 1
XCH ~ — (12i + 12j-6k)OA OA 18
(2i + 2j-k)
Let the coordinates of Point P be (x in., j> in., z in.). Then
PC = [(9 - x)in.]i + (15- y)in.]j + [(12- z)in.]k
Also,
and
OP = rfo^o, = -^(21 + 2J-k)
OP ~ (x in.)i + (>' in.)j + (z in.)k
2 2dOP
The requirement that CM and PC" be perpendicular implies that
^PC-0
or -(2j + 2j-k)-[(9-.v)i + (15-y)j + (12-2)k] =
or (2)|9--^>| + (2) 'l5-|rfw |+ H) 12 -rf0P
or Jnp = 12.00 in. ^W
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196
PROBLEM 3.45
Determine the volume of the parallelepiped of Fig. 3.25 when(a) P = 4i - 3j + 2k, Q = -2i - 5j + k, and S = 7i + j - k,
(/;) P = 5i - j + 6k, Q = 2i + 3j + k5 and S = -3i - 2j + 4k.
SOLUTION
Volume of a parallelepiped is found using the mixed triple product.
O) VoI = P-(QxS)
4-3 2
-2 -5 1 in.3
7 1 -1
(20-21.-4 + 70 + 6-4)
67
or Volume=67.0 4(b) Vol=P-(QxS)
5 -1 6
2 3 I in.3
-3 -2 4
(60 + 3-24 + 54 + 8+10)
111
or Volunie= 111.0 A
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197
PROBLEM 3.46
Given the vectors P - 4i - 2j + 3k,Q = 2i + 4j - 5k, and S = SJ - j + 2k, determine the value of Sx for which
the three vectors are coplanar.
SOLUTION
If P, Q, and S are coplanar, then P must be perpendicular to (Qx S).
P-(QxS) = ()
(or, the volume of a parallelepiped defined by P, Q, and S is zero).
4-2 3
Then 2 4 -5-0
Sx -1 2
or 32 + lO.S: -6-20 + 8-125;. =0 S =7 <
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198
0.1 1 in...
PROBLEM 3.47
The 0.61x].00-m lid ABCD of a storage 'bin is hingedalong side AB and is held open by looping cord DEC over a
frictionless hook at E. If the tension in the cord is 66 N,determine the moment about each of the coordinate axes ofthe force exerted by the cord at D.
SOLUTION
First note
Then
and
Now
where
Then
z = Vi°-61)2-(0.Jl)
2
0.60 m0.11m
dm = V(0.3)2+(0.6)
2
+(-0.6r
= 0.9 m
1 l)E
66 N
0.9(0.3i + 0.6j-0.6k)
= 22[(lN)i + (2N)j-(2N)k]
MA = rD/A xTDf:
r/)//(
= (0.ll m)j + (0.60m)k
i J k
M/(= 220 0.11 0.60
1 2 -2
= 22[(-0.22 ~ 1 .20)i + 0.60j - 0. 1 1k
]
= - (3 1 .24 N • m)i + (1 3.20 N • m)j - (2.42 N • m)k
-3 1 .2 N • m, Mv= 1 3.20 N • m, A/
2= -2.42 N - m A
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//™ Manual may be displayed,reproduced or distributed in anyform or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permittedby McGraw-Hillfor their individualcoursepreparation. Ifyou are. a student using this Manual,you are using it without permission.
199
PROBLEM 3.48
The 0.61xl.00-m lid ABCD of a storage bin is hinged
along side AB and is held open by looping cord DEC over a
frictionless hook at E. If the tension in the cord is 66 N,
determine the moment about each of the coordinate axes of
the force exerted by the cord at C.
SOLUTION
First note
Then
and
Now
where
Then
x6\y-(o.uy
0.60 m
= l.lm
lCE66 N1.1
(-0.7i + 0.6j-0.6k)
= 6[-(7N)i + (6N)j-(6N)k]
M A ^vm x\E
r£M = (0.3m)I + (0.71m)J
i J k
MA -6 0.3 0.71
-7 6 -6
= 6[-4.26i + 1 .8j + (1 .8 + 4.97)k]
= - (25.56 N • m)i + (1 0.80 N • m)j + (40.62 N • m)k
Mx- -25.6 N • m, M
y= 1 0.80 N • m, Afz = 40.6 N •m -4
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200
PROBLEM 3,49
To lift a heavy crate, a man uses a block and tackle attached to thebottom of an I-beam at hook B. Knowing that the moments about the yand the z axes of the force exerted at B by portion AB of the rope are,
respectively, 120 N • m and -460 N • m, determine the distance a.
SOLUTION
First note
Now
where
Then
Thus
^ = (2.2m)i-(3.2 m)j-(am)k
™*D ~ VAID X '«/j
'AID (2.2m}i + (1.6m)j
TT,BA
I!Ad
ISA
M,T,BA
dBA
(2.2i-3.2j-ak)(N)
i J k
2.2 1.6
2.2 -3.2 -a
T,BA
dt .
{- 1 .6a \ + 22a\ + [(2.2)(~3 .2) - (1 .6)(2.2)]k}
Mv =2.2-^-a
BA
Then forming the ratio
M,
itM.
-I0.56-7*4
d
(N • m)
(N • m)HA
120 N-m 2 -27Z"(N-m)dB,t
-460 N-m -10.56-^- (N-m)or a = 1 .252 m ^j
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201
PROBLEM 3.50
To lift a heavy crate, a man uses a block and tackle attached to the
bottom of an I-beam at hook B. Knowing that the man applies a 195-N
force to end A of the rope and that the moment of that force about they
axis is 132 N • m, determine the distance a.
SOLUTION
First note
and
Now
where
Then
Substituting for My and dtHA
dBA - J(22f+(-3.2)2 + (-af
= Vl5.08 + «2 m
195 NTs,=^^-!-(2.2i-3.2j-ok)
My~\ -(rA/D xTw )
Vf/0
M,
(2.2m)i + (1.6m)j
195
195
d
1
2.2 1.6
2.2 -3.2 -a
(2.2a) (N • m)HA
132 N-m195
J\Jm+a*(2.2a)
or 0.30769^1 5.08 + a* - a
Squaring both sides of the equation
0.094675(15.08 + a2 ) = a2
or a = 1.256 m A
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202
PROBLEM 3.51
A small boat hangs from two davits, one of which is shown in
the figure. It is known that the moment about the z axis of the
resultant force Rtexerted on the davit at A must not exceed
279 lb • ft in absolute value. Determine the largest allowable
tension in line ABAD when x ~ 6 ft.
SOLUTION
First note R 21^ +TADAlso note that only TAD will contribute to the moment about the z axis.
Now
Then,
Now
where
Then for 71.
= 10.25 ft
'AD
M..
yAIC
279
TADADr
(6i-7.75j-3k)10.25
(7.75ft)j + (3ft)k
7\
10.25
1
7,75 3
6 -7.75 -3
~^H-(1)(7.75)(6)|10.25'
A ;I
or 7" =61.5 lb <
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203
PROBLEM 3.52
For the davit of Problem 3.51, determine the largest allowable
distance x when the tension in line ABAD is 60 lb.
SOLUTION
From the solution ofProblem 3.51, TAD is now
l AI)
ADAD
60 lb
yjX2 +(-lJS)
2+(-lf
(xi-7.75j-3k)
Then M = k • (rA/c xT,(0) becomes
279
279.
60
V*2+(-7.75)
2 +(-3)'
60
1
7.75 3
jc -7.75 -3
six2 +69.0625
(D(7.75)(x)
279>/x2 + 69^0625 = 465.x
0.6Vjc2 + 69.0625 =x
Squaring both sides: 0.36x2 + 24.8625 = r
x1 =38.848 x = 6.23 ft 4
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204
PROBLEM 3,53
To loosen a frozen valve, a force F of magnitude 70 lb is
applied to the handle of the valve. Knowing that 6 = 25°,
Mx - -6.1 lb • ft, and Mz~ -43 lb • ft, determine (j) and d.
SOLUTION
We have
where
Pol-
and
From Equation (3)
From Equation (1)
rivi(
VAIO'
F
F
ML:
M.
My
M.
^ = cos
-(4in.)i + (llin.)j-(rf)k
F(cos #eos fi- sin $\ + cos #sin <J)k)
70 1b, 9 = 25°
- (70 lb)[(0.9063 1 cos 0)i - 0.42262j + (0.9063 1 sin 0)k]
ij k
(701b) -4 11 -J in.
-0.90631 cos -0.42262 0.90631 sin
(70 lb)[(9.9694sin - 0.42262^)1 + (-0.9063 Wcos + 3.6252sin <p) j
+ (1.69048 -9.9694 cos 0)k] in.
(70 lb)(9.9694sin - 0.42262d)m. = -(61 lb • ft)(I2 in./ft) (1)
(70 lb)(-0.9063 Irfcos (j) + 3.6252 sin<f>)
in. (2)
(70 lb)(l .69048 - 9.9694cos^ in. - -43 lb • ft(l 2 in./ft) (3)
U 634.33
697.8624.636'
d1022.90
29.583= 34.577 in.
or (p = 24.6° <
or c/ = 34.6in. ^
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205
i I in.
PROBLEM 3.54
When a force F is applied to the handle of the valve shown, its
moments about the x and z axes are, respectively, Mx ~ -11 lb • ft
and M.z- -81 lb • ft. For d~21 in., determine the moment My of
F about they axis.
SOLUTION
We have
Where
and
ZM : rw xF =M(
r^=-(4in.)i + (lim.)j-(27in.)k
F = F(cos 0cos <p\ - sin. $\ + cos sin flk)
i J k
-4 1 .1 -27
cos 0cos (p -sin 6 cos #sin
F [(11 cos sin 0-27 sin 0)i
+ (-27 cos Bcos <p+ 4 cos #sin 0)j
+ (4sin - 1 1 cos 6*cos ^)k](lb • in.)
Ma =F lb in.
Mx= F(l. 1 cos (9sin0-27 sin 0)(lb • in.)
MJ?
= F(-21 cos 0cos0+ 4cos 0sin 0) (lb • in.)
Mz= F(4 sin <9 - 1 .1 cos cos (j)) (lb • in.)
Now, Equation (1) cos #sin <f>-11
MF^+ 27sin#
cos 0cos 0-— 4sin —
—
-
n{ Fand Equation (3)
Substituting Equations (4) and (5) into Equation (2),
0)
(2)
(3)
(4)
(5)
M, =/N-27 4sin0-F
+ 41 (M111 F
*- + 27sin0
or M.11
(27MZ+4.M
V )
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206
PROBLEM 3.54 (Continued)
Noting that the ratiosyp and± are the ratios of lengths, have
Mv=— (-81 lb -ft)+—(-77 lb -ft)
- 11 11
= 226.82 lb • ft or Mv-= -227 lb - ft A
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207
0.35 in
0.75 m
0.75 in*
PROBLEM 3.55
The frame ACD is hinged at A and D and is supported by a
cable that passes through a ring at B and is attached to hooks
at G and H. Knowing that the tension in the cable is 450 N,
determine the moment about the diagonal AD of the force
exerted on the frame by portion BH ofthe cable.
SOLUTION
MAD ~ "'AD ' \VBIA X *Bll)
Where X„,D ---(4i-3k)
'a/a (0.5 m)i
and
Then
dm - vv0.375)2+(0.75)
2+(-0.75)
2
= 1.125 m
T.SH
450 N—(0.375i + 0.751 - 0,75k)
1.1.25J
(1 50 N)i + (300 N)j - (300 N)k
Finally Hu> =
4 -3|
0.5!
150 300 -300;
[(-3X0.5X300)]
or MAD 90.0 N-m
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208
0.35 m
0.75 m
PROBLEM 3.56
In Problem 3.55, determine the moment about the diagonal ADof the force exerted on the frame by portion BG of the cable.
SOLUTION
Where
and
Then
Finally
mad=^ad-(*biaXTbg)
M/) =-(4i-3k)
r»MA (0.5 m)j
BG - V(-°-5) + (0.925)2+ (-0.4)
2
= 1.125 ra
450 NT liC!=- -(-0.5i + 0.925j-0.4k)
1.125J '
= -(200 N)i + (370 N)j - (1 60 N)k
-3
u»'i
4
0.5
-200 370 -160
[(-3X0.5X370)] MAD -Ul.ON-m
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of (his Manual may be displayed,reproduced or distributed in anyform or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permittedby McGraw-Hillfor their individual course preparation. Ifyou are a student using this Manual,you are using it without permission.
209
0.7 m
.O.G.Ti]
PROBLEM 3.57
The triangular plate ABC is supported by ball-and-socket joints
at B and D and is held in the position shown by cables AE and
CF. If the force exerted by cable AE at A is 55 N, determine the
moment of that force about the line joining Points D and B.
SOLUTION
First note
Then
Also
Then
Now
where
Then
4E = V(°-9) + (~0-6) + (0-2)2=1.1 m
«=-yp(0.9i-0.6J + 0.2k)
= 5[(9N)i-(6N)j + (2N)k]
DB - y[(\ 2f + (-0.35)2 + (0)
2
X
1.25m
DBOB DB
1.25
1
(1.2I-0.35J)
25(241 - 7j)
MDB ~ '"DB"*DB '\VAID X T Ui)
TO/}=-(0.1m)j + (0.2m)k
!M»-^(5)24 -7
-0.1 0.2
9 -6 2
H1.8- 12.6 + 28.8)
or MDfl=2.28N-m ^
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210
PROBLEM 3.58
The triangular plate ABC is supported by ball-and-socket joints
at B and D and is held in the position shown by cables AE and
CF. If the force exerted by cable CF at C is 33 N, determine the
moment of that force about the line joining Points D and B.
SOLUTION
First note dCF -- 0.2)2 = 1.1 m= V(0.6)
2+(-0.9)
2 + (-
Then \:f'-
33 N-_ (0.6i-0.9j + 0.2k)
Also DB :
= 3[(6N)i~(9N)j-(2N)k]
= V0-2)2+(-0.35)
2+(0)
2
~ 1 .25 m
Then ^DB~~
_ DB' DB
=— (I.2i- 0.35j)1.25
J
= _L(24i-7j)25
V J;
Now Mm -A'OB '(rC/D X *cp)
where yan :-(0.2m)j-(0.4m)k
24 -7
Then Mm = -(3)25
0.2 -0.4
6 -9 -2
= —(-9.6 + 16.8-86.4)25
or Mm ^~9.50N-m <
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211
l>
PROBLEM 3.59
A regular tetrahedron has six edges of length a. A force P is
directed as shown along edge BC. Determine the moment of Pabout edge OA.
>-
SOLUTION
We have
where
From triangle OBC
Since
or
Then
and
MoA=*OA<*aoX*)
{OA)x2
f i \
(OA)z=(OA)x tanW
n/3, 2>/3vv-1 /
(OA)2=(OA)l +(OA)l +{OAzf
2 [a
2+ {OA)l +
( Ya
a , 12 . a4,0
2 h 2>/3
J. 2. 1 ,
x- =l,+Vl
J+ivr
k
p = = (a sin30°)i-(a coS 30°)k(/)) =
P(|_^
« 2
rc/0 =oi
M(A!
2 V3
I
1
2V3
Kf)]
-73"
2
' Tv - 3
y
0X->/3) =aP
Moa =aP_
72"
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212
PROBLEM 3.60
A regular tetrahedron has six edges of length a. (a) Show that two
opposite edges, such as OA and BC, are perpendicular to each
other, (b) Use this property and the result obtained in Problem 3.59
to determine the perpendicular distance between edges OA and BC.
SOLUTION
(a) For edge OA to be perpendicular to edge BC\
OA-BC = Q
where
From triangle OBC (OA)x
(OA)z =(OA)x tm\30° = -
04 = I-|I + (CM)J +
( i>
\
V3 J 2^3
ka
\2Sj
and BC = (asm 30°) i - (a cos 30°)k
Then
or
so that
a .
2
2
i+(.oa)vi+
2
c \a
^+ (O^)v(0)-~ =
4 } 4
OA-BC^0
(i-V3k)~ =
<9/i is perpendicular to #C.
(6) Have A^fW = Pt/, with P acting along BC and d the perpendicular distance from OA to 5C.
From the results ofProblem 3.57
ft?M0A
Pa
4i
4i
~Pd or d =—T A72
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213
y PROBLEM 3.61
y^ji \ 45 in. A sign erected on uneven ground is
4 V guyed by cables EF and EG. If the
11 _^#g& force exerted by cable £7** at E is 46 lb,
si /,: .^-^.,-f?;"" .--.. determine the moment of that force
96 ii r. 8»v jm^M^^^W% about the line joining Points A and D.;|*:
=
i|
r\ .
1 fPIMl
'/
L#j >' M^:
^^i^Ibr>>V - 47 in .
l:-;iy';!t'^ " ':>-^ :'::
:
'"'
' <*><. -
.--•'-•
. •• -'• -/••- ,"-~-J
" "^v "^
\ .--^ &%8 ift^o^
i 1^Xl7in. >
• v-ii:—;T*V ->''' T. •
-'*'"- I'-i i"»?»*" ^' fT /^ "''
i
"x*'' '\
SOLUTION
First note that EC ~-V(48)
2+ (36)
2 - 60 in. and that J§ = -g = |. The coordinates of Point E are
then(fx48,96, J> 36) or (36 in., 96 in,, 27 in,). Then
^,=>/H5)2 + H10)
2+(30)
2
= 115 in.
Then T^==^(-l5I-110J + 30k)
Also
= 2[-(3 lb)i - (22 lb)j + (6 lb)k]
AD = 7C48)2 + (-12)
2 + (36)2
= 3.2726 in.
Then
_
=—!—(48i-12j + 36k)12^26
=ir(4M+3k)
Now MAD =-Kiiy(-rEI,4 XTEF)
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214
PROBLEM 3.61 (Continued)
where
Then M
rm = (36 in.)i + (96 in.)j + (27 in.)k
1
2
(2)
4 -I 3
36 96 27
-3 -22 6
(2304 + 81-2376 + 864 + 216 + 2376)
or M^, =1359 lb -in. <«
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215
>J PROBLEM 3.62
% in.
\-1 > in. A sign erected on uneven ground is
guyed by cables EF and EG. If the
^0^ force exerted by cable EG at E is 54 lb,
,, ,,^'y^^ determine the moment of that force
Is. ^M^M^M:-}-^:-^ about the line joining Points A and D.
"&ftK?'?^^T?^i*K;
: V^lS ]ff ;1 - :i' p":
>^' *
*-"' "\ ^^^x in.XJ7in-
?|V ; 36iri7 \^
\| 12 1 ®^,
SOLUTION
First note that BC = - 60 in. and that ~ ~~ — ~. The coordinates of Point E areBC 60 4=V(48>
2+ (36)
2=
then (Jx 48, 96, f <36) or (36 in , 96 in., 27 in.). Then
4*;=Vai)2+(-88)
2+(-44)
2
= 99 in.
Then T*;= ^(lli-88j -44k)
Also
= 6[(llb)i-(8 1b)j-(4 1.b)k]
^/> - s/(48)2 + (~1 2)
2 + (36)2
= 12>/26 in.
Then AD AD
=—!=(48i~I2j + 36k)12^26
J
= ~=-(4i-j + 3k)V26
Now MAD ^kAD <rm xTEG )
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21fi
where
Then
PROBLEM 3.62 (Continued)
rBA = (36 in.)i + (96 in.)j + (27 in.)k
(6)
(-1536 -27 »864 -.288 -144 + 864)
MAD26
_6_
V26
4 -I 3
36 96 27
I -8 -4
or M,n =~23501b-in. <
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217
PROBLEM 3.63
Two forces F| and F, in space have the same magnitude F, Prove that the moment of F, about the line of
action of F2 is equal to the moment of F2about the line of action of F,
,
SOLUTION
First note that ¥x~f\Ay and F2 ~F2
a\2
Let Mj = moment of F2 about the line of action of M, and M2= moment of F, about the line of
action of M,
Now, by definition
Since
Using Equation (3.39)
so that
itfi=4-(rBM xF2 )
= A\-( i:wa x A2)f2
Fi=F2 =F and xmMi=^r(»*^xi2
)F
M2 =X2-{-*
BIA xAx)F
h i?BIA XX2 ) = ^2 HfiM X ^1
)
M2= A
}-(rm xZ
2 )F
-r»BM
M[2 =M2i A
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218
0.35 m
^m
0.925 m
0.875 m
//
0.75 in
L _y_
0.75 m0.5 n>*K-^^ 1C,<
0.5 m "A,
PROBLEM 3.64
In Problem 3.55, determine the perpendicular distance betweenportion BH of the cable and the diagonal AD.
PROBLEM 3.55 The frame ACD is hinged at A and D and is
supported by a cable that passes through a ring at B and is
attached to hooks at G and H. Knowing that the tension in the
cable is 450 N, determine the moment about the diagonal ADofthe force exerted on the frame by portionBH of the cable.
SOLUTION
From the solution to Problem 3.55: 7)W =450N
TBH = (1 50 N)i + (300 N)j - (300 N)k
|A/^! = 90,0N-m
X,D =-(4i.-3k)
Based on the discussion of Section 3.1 1, it follows that only the perpendicular component of TBH will
contribute to the moment ofTbh about line AD.
Now\'lill )para!lei
~~ *IIH '"'AD
= (1 501 + 300j - 300k) • I(4i - 3k)
= ^l'050)(4) + (-300)(-3)]
= 300 N
Also
so that
T*liH
v BH 'perpendicular
=V */>•// /parallel
+ ( '«// ^perpendicular
= V(450)2 - (300)
2 = 335.41 N
Since %AD and (TM )pcrpcndicuiQr are perpendicular, it follows that
MAD ~ "('BH Jpcipendiculftr
or 90.0 N-m
d
= rf(335.41N)
= 0.26833 m d = 0.268 m A
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219
0.35 in
(;:v.l 0.875 m
PROBLEM 3.65
In Problem 3.56, determine the perpendicular distance between
portionBG of the cable and the diagonal AD,
PROBLEM 3.56 In Problem 3.55, determine the moment
about the diagonal AD of the force exerted on the frame by
portion BG of the cable.
SOLUTION
From the solution to Problem 3.56: T}iG- 450 N
Tbg =-(200M)i + (370N)j-(160N)k
\MAD \= \WH-m
^D--(4i-3k)
Based on the discussion of Section 3.11, it follows that only the perpendicular component of TSo will
contribute to the moment ofTbg about line AD.
Now (Tbg )parallel ~ * BO ' ^ 41)
Also
so that
- (-2001 + 370j - 1 60k) - ~(4i - 3k)
= |[(-200X4) + (-I60X-3)]
= -64 N
*BG ~ '*/J6" /parallel
"*~V *BG )perpendicular
(TBG )pcrpenc.icuiar= V(450)
2 - (~64)2= 445.43 N
Since XAD and (Tso )pC1.
p,ndicularare perpendicular, it follows that
™AD ~ "('ag)perpendicular
llJN-m = J(445.43N)
</ = 0.24920 m
or
d = 0.249 m A
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220
PROBLEM 3.66
In Problem 3.57, determine the perpendicular distance betweencable AE and the line joining Points D and B.
PROBLEM 3,57 The triangular plate ABC is supported by
ball-and-socket joints at B and D and is held in the position
shown by cables AE and CF. If the force exerted by cable AEat A is 55 N, determine the moment of that force about the line
joining Points D and B,
SOLUTION
From the solution to Problem 3.57 T„? =55N
T^=5[(9N)i-(6N)j + (2N)k]
|MD/i |= 2.28N-m
X/«-'—(24i-7j)
Based on the discussion of Section 3.11, it follows that only the perpendicular component of TAE will
contribute to the moment ofTAe about line DB.
Now (*AE )parallel ~ AE * ^ Dli
5(9i-6j + 2k)-—(24i~7j)
[(9)(24) + (-6)(-7)]
Also
so that
5
= 51.6N
*AE ~ ( *Ae)parallel+ ( *AEJperpendicular
(''./: ),,, neodfcul* - J(5S)2+(51.6)
2 - 1 9.0379 N' perpendicuiar
Since A,ra and (T(/i )pcrpendicular are perpendicular, it follows that
MDB ^"(T^Operpendieiilar
or 2.28 N-m = </(!. 9.0379 N)
d~ 0.1 19761 J = 0.1198 m ^
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Ill
PROBLEM 3.67
In Problem 3.58, determine the perpendicular distance
between cable CF and the line joining Points D and B.
PROBLEM 3.58 The triangular plate ABC is supported by
ball-and-socket joints at B and D and is held in the position,
shown by cables AE and CF. If the force exerted by cable CF at
C is 33 N, determine the moment of that force about the line
joining Points D and B.
:0.3iJi
SOLUTION
From the solution to Problem 3.58 Tcr = 33N
Tc/,=3[(6N)i-(9N)j-(2N)k]
jA*D«| = 9.50N-m
Xm =25
(241 ~ 7,)
Based on the discussion of Section 3.11, it follows that only the perpendicular component of TCp will
contribute to the moment ofTcr about line DB,
Now ( i-CF )parallcl~" *CF " * o«
= 3(61-9j-2k)~ (241 - 7 j)
= ^[(6)(24) + (-9)(-7)]
- 24.84 N
AlSO Tc y;- — (TcF )jwral le!
+ ( *-C/' )perpendicular
s° that (Ta.Wendicuiar = V(33)2 - (24.84)
2
= 2.1 .725 N
Since XDB and (TCF )pcrpendicillar are perpendicular, it follows that
\MDB I "V'CY'Vperpcndicperpendicular
or 9,50N-m = </x21.725N
or d~ 0.437 m A
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Ill
% ill
PROBLEM 3.68
In Problem 3.61, determine the perpendicular
distance between cable EF and the line joining
Points A and D.
PROBLEM 3.61 A sign erected on uneven
ground is guyed by cables EF and EG. If the
force exerted by cable EF at E is 46 lb,
determine the moment of that force about
the line joining Points A. and D.
SOLUTION
From the solution to Problem 3.61 7V„=46 1bEl-
\M
X
T^-2H3 1b)i-(22 1b)j + (6 1b)k]
lb in.
(4i-j + 3k)
1Di -1359 1b-in.
1
AD
Based on the discussion of Section 3.11, it follows that only the perpendicular component of TEF will
contribute to the moment ofTEF about line AD.
Now (TeF /parallel ~ ^EF ' ^, AD
2(-3i-22j + 6k)
2
(4i-j + 3k)
Also
so that
[(~3)(4)+ (-22)H) + (6)(3)j'26
= 10.9825 lb
iEF - ( lEF )para||c]+ (lEF )perpemiicular
(^perpendicular = V(46)2 - (10.9825)
2 = 44.670 lb
Since \AD and OW ^^are perpendicular, it follows that
or
MAD - d{TEl, )pcrpeiidicular
1359 lb • in. = rfx 44.670 lb or t/ = 30.4in. <
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223
PROBLEM 3.69
In Problem 3.62, determine the perpendicular
distance between cable EG and the line joining
Points A and D.
PROBLEM 3.62 A sign erected on uneven
ground is guyed by cables EF and EG. If the
force exerted by cable EG at E is 54 lb,
determine the moment of that force about, the
line joining Points A and D.
SOLUTION
From the solution to Problem 3.62 TB0 = 54 lb
TfiC =6[(llb)i-(81b)i-(41b)k]
|M/JD |= 23501b-in.
^ =4^(41 -j + 3k)V26
Based on the discussion of Section 3.11, it follows that only the perpendicular component of TgG will
contribute to the moment of Teg about line AD.
N»w (TEG )para„e,= TEG XAD
- 6(i - 8j - 4k) •-fL(4i - j + 3k)
V26
= ~™L[(I)(4) + (-8X-D + (-4X3)] -
Thus, (TBG )perpendiciliar=T£G = 54 lb
Since %A0 and (T£6 pe[pcildicutarare perpendicular, it follows that
1™ AD \~ "\*EG /perpendicular
or 2350 lb • in. = dx 54 lb
or rf = 43.5 in. 4
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224
PROBLEM 3.70
Two parallel 60-N forces are applied to a lever as shown.Determine the moment of the couple formed by the two forces
(a) by resolving each force into horizontal and vertical componentsand adding the moments of the two resulting couples, (b) by using
the perpendicular distance between the two forces, (c) by summingthe moments of the two forces about Points.
SOLUTION
(a) We have
where
(b) We have
(c) We have
SM 8 : -t/,Cv +rf2C,,=M
</, = (0.360 m) sin 55°
= 0.29489 md2 =(0.360 m) sin 55°
= 0.20649 mC, =(60 N) cos 20°
- 56.382 NC,, =(60 N) sin 20°
= 20.521 N
M = -(0.29489 m)(56.382 N)k + (0.20649 m)(20.52 1 N)k
--=-(12.3893 N-m)k
M = Fd(-k)
= 60 N [(0.360 m)sin(55° - 20°)](-k)
= -(12.3893 N-m)k
£M A : £(rtx F) = rm x FB + rCIA xFc =M
i J k
M = (0.520 m)(60 N) cos 55° sin 55°
-cos20° -sin 20°
I J k
cos 55° sin 55°
cos 20° sin 20°
(1 7.8956 N •m - 30.285 N • m)k
-(12.3892 N-m)k
or M = 12.39 N-mJ^
or M = 12.39 N-m
+(0.800 m)(60N)
or M = 12.39 N-m
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225
1211./
A"1 ft
21. II >
1
1' lb 16 in.
PROBLEM 3.71
A plate in the shape of a parallelogram is acted upon by two
couples. Determine (a) the moment of the couple formed by the
two 21 -lb forces, (/?) the perpendicular distance between the 12-lb
forces if the resultant of the two couples is zero, (c) the value of aifthe resultant couple is 72 lb- in. clockwise and dfc 42 in.
SOLUTION
\Z»V?
(a) We have Mx=d^
c where dx
~ 1 6 in.
Fj=211b
M, =(16in.)(21 lb)
-336 lb -in.
(/?) We have M,+M.2=0
1 2. Ho
or M, -336 lb -in. H
</, = 28.0 in. <or 336 lb • in. -^(1 2 lb) =
(c) We have Mlolll=M
1+M2
or -72 lb -in. = 336 lb -in. - (42 in.)(sin a){\ 2 lb)
sina~ 0.80952
and a = 54.049° or a = 54.0° ^
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226
100 imn KiOi
140 mm160 mm
2-10 mil
V^
SOLUTION
P D
1 nP A
(a) We have
or
(A)
We have
(c)
,32- .tv,
o
r%^ We have
PROBLEM 3,72
A couple M of magnitude 18 N-m is applied to the
handle of a screwdriver to tighten a screw into a
block of wood. Determine the magnitudes of the
two smallest horizontal forces that are equivalent to
M if they are applied (a) at corners A. and D, (b) -at
corners B and C, (c) anywhere on the block.
M = Pd
18N-m = P(.24m)
P = 75.0N
dBC ^yj(BE)2 +{ECf
= V(-24m)2+(.08m)2
= 0.25298 m
M = Pd
18N-m = P(0.25298m)
P-71J52N
dAC =yl(ADf+(DCf
or Pmin =75.0N <
or P = 71.2N <
).24m)2+ (0.32 m)2
0.4 m
M = PdAC18N-m = />(0.4m)
P = 45.0N or P = 45.0N A
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227
.',:.) I!>
6 in.
25 lii
:25 ih
L Sin.-
PROBLEM 3.73
Four 1 -in.-diameter pegs are attached to a board as shown. Two
strings are passed around the pegs and pulled with the forces
indicated, (a) Determine the resultant couple acting on the board.
(/;) If only one string is used, around which pegs should it pass
and in what directions should it be pulled to create the same
couple with the minimum tension in the string? (c) What is the
value ofthat minimum, tension?
SOLUTION
35"&
sty*
(a) +)M = (35 lb)(7 in.) + (25 lb)(9 in.)
-245 lb -in. + 225 lb -in.
M = 470 lb- in. ^)<
(/;) With only one string, pegs A and D, or B and C should be used. We have
6tan
8
= 36.9< 90°- = 53.1°
Direction of forces:
With pegs /I andD:
With pegs/? and C:
(c) The distance between, the centers ofthe two pegs is
53.1° -4
53.1° <
F A s
<J&+62 =10 in.
Therefore, the perpendicular distance d between the forces is
We must have
c/ = 10in. + 2|-in.
= 1 1 in.
M = Fd 4701b-in. = F(llin.) F = 42.7 lb <
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228
25 lb
PROBLEM 3.74
Four pegs of the same diameter are attached to a board as shown.Two strings are passed around the pegs and pulled with the
forces indicated. Determine the diameter of the pegs knowingthat the resultant couple applied to the board is 485 lb in.
counterclockwise.
SOLUTION
M ^ dADpAD+dBCFBC485 lb • in. = [(6+ rf)in.](35 lb) + [(8 + rf)in.j(25 lb) rf = 1.250 in. <
PROPRIETARY MAIERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of Urn Manual mav be displavedreproduced or distributed in anyjorm or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educatorspermittedby McGraw-Hillfor their individual coursepreparation. Ifwit are a student using this Manualyou are using if withoutpermission.
229
SllH'i
PROBLEM 3.75
The shafts of an angle drive are acted upon by the two couples shown.
Replace the two couples with a single equivalent couple, specifying its
magnitude and the direction of its axis.
filb.fl
SOLUTION
Based on
where
I2M
M == M,+M 2
M,== -(81b-ft)j
M 2== -(61b-ft)k
M == -(8tb-ft)j-(61b-ft)k
|M| == >/(8)2+(6)
2 =10 lb -ft
3l =M
"|m:|
-(8 lb • ft)j - (6 lb • fl)k
or M = 10.00 lb- ft <
or M
10ib-ft
-0.8j-0.6k
jMjl = (101b-ft)(~0.8j-0.6k)
cos0v=O #
v=90°
cos<9v=-0.8 6>
(
.=143.130°
cos0z --O,6 Z=126.870°
or 9X - 90.0° 6 = 143. 1° O
z=\ 26.9° <
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230
170 mm
160 mm
18 N
150 mm
.1.50 mm
!8N
3-3 N
PROBLEM 3.76
If P = 0, replace the two remaining couples with a single
equivalent couple, specifying its magnitude and the direction
of its axis.
SOLUTION
We have
where
Also,
M = M, + M,
M,
lGIC
rt?/c
x *i
-(0.3 m)i
(18N)k
M,=-(0.3m)ix(18N)k
= (5.4N-m)J
M 2 =rD/r xF2
'D/F -(.15m)i + (.08m)j
(.15m)i + (.08m)j + (.17m)k
M
(.15)2+(.08)
2+(.1.7)
2 m
= 141.421 N-m(.15i + .08j + .l7k)
i J k
= 141,421 N-m -.15 .08
-.15 .08 .17
- 141 .42 1(.01 361 + 0.0255j)N • m
(34 N)
PROPRIETARY MATERIAL © 2010 The McGraw-Hill Companies, Inc. Aft rights reserved. No part of this Manual may be displayed,reproduced or distributed in anyform or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educatorspermittedby McGraw-Hillfor their individual coursepreparation. Ifyou are a student using this Manual,you are using it withoutpermission.
231
PROBLEM 3.76 (Continued)
and M = [(5.4 N -m)j] + [14 1.42.1(.01 36i + .0255j) N • m]
= (1 .92333 N • m)i + (9.0062 N • m)j
|m:|-^(m,)2 +(m;v )
2
= V(l-92333)2+(9.0062)
2
= 9.2093 N • m or M = 9,2 1 N • m <
M = (1 .92333 N • m)i + (9.0062 N • m)j
~|Mj~ 9.2093 N-m= 0.20885 + 0.97795
cos 6>v=0.20885
0,= 77.945° or 9^11.9° <
cosy=0.97795
0, =12.054° or 0, =12.05°*
cos Z=0.0
=90° or 6' =90.0° <
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232
PROBLEM 3.77
If P~0, replace the two remaining couples with a single
equivalent couple, specifying its magnitude and the direction
of its axis.
SOLUTION
dOE
E,
M = M, +M 2 ; /'J= 16 lb, F
2= 40 lb
M, = rc x F, - (30 in.)* xH) 6 lb)jj = -(480 lb • in.)k
M 2 = rm x F2 ; rm = (1 5 in.)i - (5 in.)j
>/(0)2+ (5)
2+(10)
2 =5^ in.
= 8>/5[(llb)j-(2ib)k]
i j k
M 2 -8n/5 15 ~5
1 -2
- 8>/5[(10 lb • in.)i + (30 lb • in.)j + (1 5 lb • in.)k]
-(480 lb in.)k + 8V5[(1 lb • in.)i + (30 lb • in.)j + (1 5 lb • in.)k]
(178.885 lb • in.)i + (536.66 lb • in.)j - (21 1 .67 lb • in.)k
^/(178.885)2+(536.66)
2 +(-21 1.67)2
M
M
K
603.99 lb in
MA/ = 604 lb -in. <
cos0v
COS
COS 0..
M0.29617
0.88852
-0.35045
0.29617! + 0.88852j - 0.35045k
72.8° 0=27.3° 0,,=110.5° A
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233
PROBLEM 3.78
If P = 20 lb, replace the three couples with a single
equivalent couple, specifying its magnitude and the
direction of its axis.
SOLUTION
From the solution to Problem. 3.77
M, = -(480 lb • in.)k
M2= 8>/5 [(10 lb • in.)i + (30 lb • in.)j + (1 5 lb • in.)k]
16 lb force:
40 lb force:
F - 20 lb M3= rc x P
= (30in.)ix(20lb)k
= (600 lb - in.)j
M-M, +M 2 +M 3
= ~(480)k + 8V5 (1 Oi + 30j + 1 5k) + 600j
-(178.885 lb -in)i + (1136.66 lb -in.)j -(211.67 lb-in.)k
M = ./{178.885)2 + (1
1
3.66)2+(211 .67)
2
= 1169.96 lb in M = 1170 lb- in. <
X„MM
0. 1 52898i + 0.97 154j - 0. 1 8092 1k
cos ex = 0.152898
cos^ = 0.97 154
cos 0.= -0.1 8092.1 9. =81.2° 0=13.70° 6'=100.4° <
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234
160 nun,- \
18 N
PROBLEM 3.79
If P = 20 N, replace the three couples with a single
equivalent couple, specifying its magnitude and the
direction of its axis.
SOLUTION
We have
where
M-M, +M2 +M 3
M^r^xF,i J k
0.3
18
N-m = (5.4N-m)j
M3=r /fXF2
i J k
,15 .08 141.421 N-m.15 .08 .17
141.421(.0136i + .0255j)N-m
(See Solution to Problem 3.76.)
M3= r(X4 xF3= 0.3 0.17 N-m
20
= -(3.4N-m)i + (6N-m)k
M=[(1.92333-3.4)i + (5.4 + 3.6062)j + (6)k]N.m
= -(1 .47667 N • m)i+ (9.0062 N • m)j + (6 N in)
llVff— 1**2 , nj2 , xj2M; +ML
V+ m;
V(l .47667) + (9.0062) + (6)2
10.9221 N-m or M = 10.92N-m A
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235
PROBLEM 3.79 (Continued)
XM -1.47667 + 9.0062 + 6
jM| 10.9221
= -0. 1 35200i + 0.82459J + 0.54934k
cos(?T= -0.135200 6X =97.770 or <9
V-97.8° <
cosV= 0.82459 #,,=34.453 or
y= 34.5° ^
cos#. = 0.54934Z=56.678 or <9
2= 56.7° ^
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236
z A } 1600 N-m
] 200 N-iii
1 1.20 N«m
PROBLEM 3.80
Shafts A and B connect the gear box to the wheel assemblies
of a tractor, and shaft C connects it to the engine. Shafts A andB lie in the vertical yz plane, while shaft C is directed along
the x axis. Replace the couples applied to the shafts with a
single equivalent couple, specifying its magnitude and the
direction of its axis.
SOLUTION
Represent the given couples by the following couple vectors:
M^ = -1 600sin 20°j + 1 600cos20°k
M:
Mc
-(547.232 N m)j + (1 503.5 1 N • m)k
1200sin 20°j + 1200cos20°k
(41 0.424 N • m)j + (11 27.63 N • m)k
-(1120N-m)i
The single equivalent couple is
M^M,, +M 5 +MC
= -(1 120 N • m)i - (1 36.808 N • m)j + (263 1.1 N- m)k
M = yj(\ 1 20)2 + (1 36.808)
2 + (263 1 . I)
2
= 2862.8 N-m-1120
cos 6,
COS tfy ^
cos 0„
2862.8
-136.808
2862.8
2631.1
2862.8
M = 2860N-m 0=113.0° B' =92.7* 23.2° <
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced or distributed in anyform or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educatorspermittedby McGraw-Hillfor their individualcourse preparation. Ifyou are a student using this Manual,you are using it without permission.
237
Aim W
20°
A*'
PROBLEM 3.81
The tension in the cabJe attached to the end C of an
adjustable boom ABC is 560 lb. Replace the force
exerted by the cable at C with an equivalent force-
couple system (a) at A, (b) at B.
SOLUTION
{a) Based on ZF: F,,=r = 560lb
mK^*^t^^ or ¥A = 560 lb ^ 20° ^
X^^S TscnSo ZMA : M^^sinSO )^,)
= (560lb)sin50°(18ft)^A
(b)
or
Based on
= 7721.7 lb -ft
M/(=7720.1b-ftjH
XF: 7^ = 7* = 560 lb
Ffi= 560 lb ^C 20° <
ZMB : MB =(Tsm50°)(dB )
I & or
«eA = (560 lb) sin 50° (10 ft)
3*6 \
^MB \ c
or
= 4289.8 lb ft
Mg =42901b-ftjH
\£&<!2th c*** F*
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238
PROBLEM 3.82
A 1 60-lb force P is applied at Point A of a structural member.
Replace P with (a) an equivalent force-couple system at C,
(b) an equivalent system consisting of a vertical force at Band a second force at D.
SOLUTION
(a) Based on
where
(b) Based on
2.F: Pc ^P = \60\b
XMC : Mc —Pxdcy + PydCx
Px =(160 lb) cos 60°
= 80 lb
/;=(1601b)sin60°
= 138.564 lb
rftt =4ft
JCl ,= 2.75 ft
Mc = (80 lb)(2.75 ft) + (1 38.564 lb)(4 ft)
= 220 lb -ft + 554.26 lb- ft
= 334.26 lb -ft
EFV : PDx =Pcos 60°
or P, =160 lb ^T.60°^
or Mc =334 lb • ft )<
= (160lb)cos60c
= 80 lb
TM, (Pcos60°)(clDA ) = PB(dDB )
[(160 lb)cos60°](1.5 ft) = PB (fi ft)
PB = 20.0 lb or P„=20.01bH
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239
PROBLEM 3.82 (Continued)
ZFV
: Psin60o = PB + PI)},
(160 lb) sin60° = 20.0 lb + PDy
PDy =118.564 lb
2
^=>/('k)2 +('y
= 7(8°)2 +0 18 -564)
2
= 143.029 lb
# = tan_1(p \
p
J 118.564^- tan] — —I 80 )
= 55.991° or PD ,= 143.0 lb ^56.0°^
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240
-*-r-HS A PROBLEM 3.83
50 nun F,- i
«i
100 mm
The 80-N horizontal force. P acts on a bell crank, as shown.
(a) Replace P with an equivalent force-couple system at B.
(b) Find the two vertical forces at C and D that are equivalent
to the couple found in Part a.
SOLUTION
JvJ:
A(a) Based on IF: F
/?=F = 80N
XM: M n =Fdr
80 N (.05 m)
4.0000 N • m
or FB = 80.0 N --•••- ^
or M«-4,00N-m }^
5«*
^
1
(b) If the two vertical forces are to be equivalent to MB, they must be
a couple. Further, the sense of the moment of this couple must be
counterclockwise.
Then, with Fc and FD acting as shown,
XM: MD =Fcd
4.0000 N-m = Fc (.04m)
Fc =100.000 N or Fc =100.0nH
2Fy
: = FD -Fc
Fft =l 00.000 N or F„ =100.0nH
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241
PROBLEM 3.84
tsiSsif^S&&^-fift^iisKSS ......' A dirigible is tethered by a cable attached to its cabin at B. If
the tension in the cable is 1 040 N, replace the force exerted by
the cable at B with an equivalent system formed by twoA B C parallel forces applied at A and C,
6.7 m 4 m \
60°/\......,lA.^ ,
SOLUTION
Require the equivalent forces acting at A and C be parallel and at an
angle ofa with the vertical.
Then for equivalence,
XFX : (1 040 N) sin 30° = FA sin a +FB sina
ZFy
: -(1 040 N) cos 30° = -FA cosa - FB cosa
Dividing liquation (1) by Equation (2),
(1040 N) sin 30° __{FA +FB)s\na
-(.1040 N) cos 30°~ -{FA + FB )cosa
Simplifying yields a ~ 30°
Based on
XMC : [(1040 N)cos30°](4 m) = {FA cos30°)(l 0.7 m)
FA = 388.79 N
(1)
(2)
or
Based on
or
F,, = 389 N "^ 60° M
XM.A :- [(1 040 N) cos 30°](6.7 m) = (Fc cos 30°)(l 0.7 m)
Fc =651.21 M
FC =651N ^60°^
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242
:,-l
A\ 200 mm
300 mm
PROBLEM 3.85
The force P has a magnitude of 250 N and is applied at the end Cof a 500-mm rod AC attached to a bracket at A and B. Assuming
a ~ 30° and j3 — 60°, replace P with (a) an equivalent force-couple
system at B, (b) an equivalent system formed by two parallel forces
applied at A and B.
SOLUTION
(a) Equivalence requires £F: F = P or F = 250N^60°
1M B : M = -(0.3 m)(250 N) = -75 N m
The equivalent force-couple system at B is
F = 250N "^60° M = 75.0 N • m ) <
(b) Require
%/f/ MO NJ ™ 5^1*
9c
Equivalence then requi :es
£FV
: = F,, cos + FA> cos
F^ —~FR or cos0 = O
2)Fy
: --25 = -F4sin -Fe sin
Now if F.f= -Fn => -250 = reject
cos =
or = 90°
and F, + FB = 250
Also £.A/B : - (0.3 m)(250 N) = (0.2m)F/4
or F.,=-375N
and Fi?=625N
F^=375N ^60° FB = 625 N ^ 60° ^
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243
1'
'A\ 200 mm
300 mm
PROBLEM 3.86
Jk\ Solve Problem 3.85, assuming a-fi~ 25°.
SOLUTION
P= 2SO N\
id) Equivalence requires
£F: ¥B = P or Ffi= 250 N X' 25.0°
XMB : Mfi= ~-(0.3 m)[(250 N)sin 50°] = -57.453 N • m
The equivalent force-couple system at. B is
F« = 250 N ^ 25.0C M„^57.5N-m )<
(b) Require
Z5"o n! 25"0 M0-2. w>
Equivalence requires
Adding the forces at B:
MB = dAEQ (0.3 m)[(250 N) sin 50°]
= [(0.2m)sin50°]£
= 375N
F,, = 375 N ^ 25.0° F»=625N X: 25.0° <
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244
450 mm
rA
PROBLEM 3.87
A force and a couple are applied as shown to the end of a
cantilever beam, (a) Replace this system with a single force Fapplied at Point C, and determine the distance d from C to a
line drawn through Points D and E. (h) Solve Part a if the
directions of the two 360-N forces are reversed.
pi.v
i 150 mm
SOLUTION
A3fcON
A
3(>ON
1
E COON E
—*-J
w
—
p
Tc r
150 win
3&0N
(a) We have IF: F = (360N)j-(360N)j-(600N)k
or F = -(600N)k A
and SMD : (360N)(0.]5m) = (600N)(rf)
d - 0.09 m
or d - 90.0 mm below ED A
(b) We have from Part a F = -(600N)k 4
and 1MD : -<360N)(0.15m) = -(600N)(rf)
d = 0.09 m
or d = 90.0 mm above ED A
-H---'r '""fe
d
D
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245
|*-]2()inm-*)*i+
i '' [—~—-"'-^^i^;-; ; --::;:-: ;j
I
250 N y90 mm M
C
1 900 N90 ram
PROBLEM 3.88
The shearing forces exerted on the cross section of a steel channel can
be represented by a 900-N vertical force and two 250-N horizontal
forces as shown. Replace this force and couple with a single force Fapplied at Point C, and determine the distance x from C to line BD.
(Point C is defined as the shear center of the section.)
250 N
SOLUTION
Replace the 250-N forces with a couple and move the 900-N force to Point C such that its moment about 11 is
equal to the moment of the couple
25"o M k^
O.tftwi ?JPh
iOoM•ZSoM
T^OO MD
Then
or
Mu = (0.18)(250 N)
= 45N-m
Mlf= x(900 N)
45N-m = x(900N)
x- 0.05 m
F = 900NJ x = 50.0 mm A
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246
X3.2 in'
2.05 I!)
2.8 in.
1)¥
PROBLEM 3,89
While tapping a hole, a machinist applies the horizontal forces
shown to the handle of the tap wrench. Show that these forces
are equivalent to a single force, and specify, if possible, the
point of application of the single force on the handle.
SOLUTION
Since the forces at A and B are parallel, the force at B can be replaced with the sum oftwo forces with one of
the forces equal in magnitude to the force atA except with an opposite sense, resulting in a force-couple.
Have FB - 2.9 lb- 2.65 lb = 0.25 lb, where the 2.65 lb force be part of the couple. Combining the two
parallel forces,
Couple = (2.65 lb)[(3.2in. + 2.8 in.)cos25°]
= 14.4103 lb -in.
and Mcouple
14.41 03 lb -in.
H 1A 6
*
A
f is^h
-Q
X
A single equivalent force will be located in the negative z-direction
Based on ZMn-1 4.4 1 03 lb • in. = [(.25 lb) cos 25°](a)
a = 63.600 in.
F'= (.25 lb)(cos 25°i + sin 25°k)
F'= (0.227 lb)i + (0. 1 057 lb)k and is applied on an extension of handle BD at a
distance of 63.6 in. to the right ofB
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247
48 lb
2o:20
40 in.
20 ib
PROBLEM 3.90
Three control rods attached to a lever ABC exert on it the
forces shown, (a) Replace the three forces with an equivalent
force-couple system at B. (b) Determine the single force that
is equivalent to the force-couple system obtained in Part a,
and specify its point of application on the lever.
SOLUTIONIK
(a) First note that the two 20-lb forces form A couple. Then a\6
where
F = 48 Ib ALd
= 180° -(60° + 55°) = 65°
*E>S°
and M = 2,MB
= (30in.)(481b)cos55°- (70 in.)(20 lb)cos20c
= -489.62 lb • in
The equivalent force-couple system at B is
F = 48.0 lb ^65° M == 490 lb- in.jH
0>) The single
between Aequivalent force F' is equa
and B. For equivalence.
I to F. Further, since the sense ofM is clockwise,F' must be applied
EM,: M = -aF' cos 55°
where a is the distance from B to the
-489.62 lb •
point of application of F'
in. = -tf(48.01b)cos55°
. Then
oi- o = 17.78 in. F' = 48.0 lb ^L 65.0° <
and is applied to the lever 17.78 in.
To the left of pin B <
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248
300 N
300 N
PROBLEM 3,91
A hexagonal plate is acted upon by the force P and the couple shown.Determine the magnitude and the direction of the smallest force P for
which this system can be replaced with a single force at is.
SOLUTION
From the statement ofthe problem, it follows that XME = for the given force-couple system. Further,
for Pmi„, must require that P be perpendicular to rB/E . Then
XME : (0.2 sin 30°+ 0.2)mx300N
+ (0.2m)sin30°x300N
~(0-4m)P- =0
3ooM
or R,, -300N
3<x>tv\
P., 300 N 30.0° <
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. AH rights reserved. No part of this Manual may he. displayed,
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249
PROBLEM 3,92
A rectangular plate is acted upon by the force and couple
shown. This system is to be replaced with a single equivalent
force, (a) For a - 40°, specify the magnitude and the line of
action of the equivalent force, (b) Specify the value of or if the line
of action of the equivalent force is to intersect line CD 300 mmto the right ofZX
SOLUTION
(a) The given force-couple system (fi\ M) at B is
F = 48Nj
and M =ZMB = (0.4 m)(l 5 N)cos 40° + (0.24 m)(1 5 N)sin 40
or M =6.91.03 N -m
The single equivalent force F' is equal to F. Further for equivalence
~ZMB : M = dF'
or 6.9.1.03 N -m = dx 48 N
or d~ 0.14396 m
and the line of action of/"7intersects line AB 144 mm to the right ofA.
(b) Following the solution to Part a but with d ~ 0. 1 m and aunknown, have
J.MB : (0.4 m)(l 5 N) cos a+ (0.24 m)(l 5 N) sin a
= (0.!m)(48N)
or 5 cosa + 3 sin a - 4
Rearranging and squaring 25 cos2 a ~ (4 - 3 sin. a)
2
Using cos2 a~ 1 -sin
2 a and expanding
25(1 -sin2a) -16 -24 sin «r + 9 sin
2or
or 34 sin2or ~ 24 sin or - 9 =
CJ O
F' = 48N A
Then sin a2(34)
sin a = 0.97686 or sin a = -0.27098
a = 77.7° or a = -15.72°
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250
100 mm
""60 mm
PROBLEM 3.93
An eccentric, compressive 1 220-N force P is applied to the end
of a cantilever beam. Replace P with an equivalent force-couple
system at G.
SOLUTION
0, trvn
We have
ZF: -(1220N)i=F
F = -(1220N)i <
Also, we have
1220
IMG : i>c xP =M
N-m = Mi J k
-.3 -.06
-1
M = (1 220 N m)[(~0.06)(-l)j- (~0. l)(-l)k]
or M=(73.2N-m)j-(122N-m)k <
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251
.67 mm
594 mm
PROBLEM 3.94
To keep a door closed, a wooden stick is wedged between the
floor and the doorknob. The stick exerts at B a 175-N force
directed along line AB. Replace that force with an equivalent
force-couple system at C.
1.00 mm
SOLUTION
We have Kt ^ (,!-*.«
where
IF: Plfi=Fc
* AB=
*"AB'AB
iat
I.3>(33 mm )i + (990mm)j-
1155.00 mm(594mm)k
()75N)
or Fc =(5.00N)i + (150N)j-(90.0N)k <
We have JM.C : rm:: xPAli=MC
» J k
Mc = 5 0.683 -0.860 N-m
1 30 -18
= (5){(-0.860)(-1 8)i - (0.683)(-l 8)j
+ [(0.683)(30)--(0.860)(l)]k}
or Mc = (77.4 N-m)i + (61.5 N •m)j + (106.8 N-m)k A
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252
16 ft
PROBLEM 3.95
An antenna is guyed by three cables as shown. Knowingthat the tension in cable AB is 288 lb, replace the force
exerted at A by cable AB with an equivalent force-couple
system at the center O of the base of the antenna.
SOLUTION
We have
Then
Now
dAB =yl(r64f +(-12Sf +{16)2 = 144 ft
T^=~^(-64i-128j + 16k)
= (32 1b)(-4i-8j + k)
M =M =r }/0 xT,„,
= 128jx32(-4i-8j + k)
- (4096 lb • ft)i + (1 6,384 lb • ft)k
The equivalent force-couple system at O is
F = -(1 28.0 lb)i - (256 lb) j + (32,0 lb)k -4
M = (4. 1 kip • fl)i + (1 6.38 kip • ft)k <
PROPRIETARY MATERIAL © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of (his Manual may be displayed,reproduced or distributed in anyform or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers andeducatorspermittedby McGraw-Hillfor their individual course preparation. Ifyou are a student using this Manual,you are using it without permission.
253
J.6ft
PROBLEM 3.96
An antenna is guyed by three cables as shown. Knowing
that the tension in cable AD is 270 lb, replace the force
exerted at A by cable AD with an equivalent force-couple
system at the center O ofthe base of the antenna.
SOLUTION
We have *ad = VC-*4)2+C-128)
2+(-128)
2
= 192 ft.
Then T^=~5Jb(~64i-128j + 128k)
= (90 1b)(-i-2j-2k)
Now M-M = r^xT^= 128jx90(-i~2j-2k)
= -(23,040 lb • ft)i + (1 1,520 lb • ft)k
The equivalent for ;e-couple system at O is
F = -(90.0 lb)i - (1 80.0 lb)j - (1 80.0 lb)k <
M = -(23 .0 kip • ft)i + (1 1 .52 kip • ft)k <
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254
PROBLEM 3.97
Replace the J50-N force with an equivalent force-couple
system at A.
120mm
Y\
20mm 60 mm
SOLUTION
Equivalence requires
where
LF: F = (150N)(~cos35 oj-sin35ok)
= -(122.873 N)j - (86.036 N)k
XMA : M=rm xF
rDIA = (0. 1 8 m)i - (0. 1 2 m)j + (0. 1 m)k
i J k
Then M=0.18 -0.12 0.1 N-m
-122.873 -86.036
- [(-0. 1 2)(-86.036) - (0. l)(-i 22,873)]i
+ [-(0.18)(-86.036)]j
+ [(0.18)(-122.873)jk
= (22.6 N • m)i + (1 5.49 N • m)j - (22. 1 N • m)k
The equivalent force-couple system atA is
F = -(1 22.9 N)j- (86.0 N)k <
M = (22.6 N • m)i + (1 5.49 N • m)j - (22. 1 N • m)k <4
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255
83.3 mm
PROBLEM 3.98
A 77-N force Ffand a 31-N • m couple Mi are
applied to corner E of the bent plate shown. If
Fj and Mi are to be replaced with an equivalent
force-couple system (F2, M2) at corner B and if
(M2)7_ - 0, determine (a) the distance d, (b) F2
and M2 .
SOLUTION
(a) We have XM& : M2z=0
M^xFJ +M^O (1)
where rHfB = (0.3 lm)i- (0,0233)j
_ (0.06 m)i + (0.06 m)j - (0.07 m)k (nn Kn
0. 1 1 m= (42 N)i + (42 N)j - (49 N)k
Ml2=k-M.,
M, = KuM\
~d\ + (0.03 m) j- (0.07 m)k „ . .VE .- _ ^ 1*—
i
J— (3 ] |s| . m)
V^2+0.0058 m
Then from Eq nation dX
1
0.31 -0.0233(-0.07m)(31N-m) n+ - J -0
Vc/2+0.0058
42 42 -49
Solving for d, Equat on (I ) reduces to
(13.0200 + 0.978 , }
2.17N,m _^
4d2+0.0058
From which d = 0.1350 in or d ~ 135.0mm A
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256
PROBLEM 3.98 (Continued)
(/>) F2 = F.= (42i + 42j-- 49k)N or F
2= (42 N)i
M2- VWB x.F, + M,
i
0.31
42
J*
-0.0233
42
k
-49
+(0.1350)1 + 0.03j -0.07k
0.155000(3IN-m)
M,
(1. 14
1
70i + 1 5. 1900J + 1 3.9986k) Nm+ (-27.000i + 6.0000j - 14.0000k) N • m-(25.858 N • m)i + (21.190 N • m)j
or M* =- (25.9N-m)i + (21.2N-m)j <
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced or distributed in anyform or by any means, without the prior mitten permission of the publisher, or used beyond the limiteddistribution to teachers and educatorspermittedby McGraw-Hillfor their individual coursepreparation. Ifyou are a student using this Manual,you are using it without permission.
257
SOLUTION
We have
Then
Also
Then.
Now
where
Then
PROBLEM 3.99
A 46-lb force F and a 2120-lb • in. couple M are
applied to corner A of the block shown. Replace
the given force-couple system with an equivalent
force-couple system at comer //.
3 in.
dAJ
F
18)z+(-14)-+(-3r =23 in.
46 1b
23-(18i-14j-3k)
d 4C
M
(36 tb)i - (28 lb)j - (6 1b)k
: 7(-45)2+(0)
2+(-28)
2 = 53 in.
2120 lb in.
53(-451 - 28k)
-(1800 1b-in.)i-(l120 1b-in.)k
rt +r^xF
r/J/y/
=(45in.)i + (14in..)]
M'^M +r^xF
M' = (-1800i~I120k) +
i J k
45 14
36 -28 -6
- (-1 8001 - 1 1 20k) + {[(1 4)(-6)]i + [-(45)(-6)]j + [(45)(-28) - (1 4)(36)]k}
= (- 1 800 - 84)i + (270)j + (-1120-1 764)k
= -(1 884 lb • in.)i + (270 lb • in.)j - (2884 lb • in.)k
= -(1.57 lb • ft)i + (22.5 lb • ft)j - (240 lb • ft)k
The equivalent force-couple system atH is F' = (36.0 lb)i - (28.0 lb)j - (6.00 lb)k <
M' = -(] 57 lb • ft)i + (22.5 lb • ft)j - (240 lb • ft)k 4
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258
PROBLEM 3.100
The handpiece for a miniature industrial grinder weighs 0.6 lb, andits center of gravity is located on the y axis. The head of the
handpiece is offset in the xz plane in such a way that line BC formsan angle of 25° with the x direction. Show that the weight of the
handpiece and the two couples M.j and M2 can be replaced with asingle equivalent force. Further, assuming that M
}
- 0.68 lb • in.
and M2 ~ 0.65 lb • in., determine (a) the magnitude and the direction
of the equivalent force, (/>) the point where its line of action
intersects the xz plane.
SOLUTION
First assume that the given forceW and couples M, and M2 act at the origin.
Now w = ™^
and M = M, +M2 M &
= ~(M2 cos 25°)i + (M, -M
2 sin 25°)k
Note that sinceW andM are perpendicular, it follows that they can be replaced
with a single equivalent force.
(a) We have F-W or V^-iV. =-(0.6 lb)j
(b) Assume that the line of action ofF passes through Point P(x, 0, z). Then for equivalence
M = r/vo xF
where rm = xi + zk
-(M2 cos 25°)i + (M, -M2 sin 25°)k
(Wz)\ -~ (Wx)k
Equating the I and k coefficients, z =~^ COs25°
and x =( M"'M* sin 25
'
ft
W M 2
or F = -(0.600 lb)j <
i J k
X z
-w
(/;) For
W { W
W = 0.61b Mx
= 0.68 lb • in. M2-0.65 lb -in.
0.68 -0.65 sin 25°
-0.6
•0.65 cos 25°
0.6
0.67550 in.
-0.98 1 83 in.
or x = 0.675 in, A
or 2 = -0.982 in. ^
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259
PROBLEM 3.101
A 4-m-Iong beam is subjected to a variety of loadings, (a) Replace each loading with an equivalent force-
couple system at end A of the beam, (b) Which of the loadings are equivalent?
-100 *
\
/
V
=SCK> N
•on \
4 ni
ifl)
(s)
Mi X mo n
i';00N-
2300 N-nt
.-'
100 N
800 N
•'i.iN 200 N- 1.
1
?-'\m Nmii
300 N
MJON-iu
900 N-ni
(W
200 N
.
(e)
100 N-tn
,100 N
300 N- us
(ft)
"A"
000 N
.100 N I
800 N300 N-m
: OON-iii
;i00N-ri)
2(Hi \
(/)
SOLUTION
(o) (a) We have s/y -400N-20()N = /?o
and £JW/. 1 800 N-m - (200 N)(4 in.) =Ma
(/?) We have £7y -600N = tf,,
and L/V/^ -900N-m = A/ft
(c) We have I/y 300N~900N^#,.
and Z/V/^ 4500 N-m - (900 N)(4 m) = Mc
&
M
A
A /w
or Rtt=600NH
m) = M„
or Mfl= lOOON-m^)^
or R6 =600NH
or M, -900 N -m)<
or Rc= 600 N J <
1 m) = Mc
or M, = 900 N • m ") <
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260
or R,,=400NH
or M,,=900N-m)^
PROBLEM 3.101 (Continued)
{d) We have £F„: -400 N + 800 N = Rd
and ZM,t: (800 N)(4m)- 2300 N-m = M
rf
(e) We have My - 400 N - 200 N = Re
or Re =600NJ^
and ZMA : 200 N • m + 400 N m - (200 N)(4 m) =Me
or Me =200N-m >)^
(,/') We have SFJt
: -800N + 200N = /?/
or R/ =600nJ^
and ZMA : - 300 N • m + 300 N • m + (200 N)(4 m) = M.f
or M/ =800N-m>
)^
(g) We have £F : -200 N- 800 N = tf,
or R^=1000n|^
and ZMA : 200 N •m + 4000 N • m - (800 N)(4 m) =M
or M£=1000N-m
>
)^(A) We have Z/y -300N-300N = ifA
or RA= 600 N
J <
and LMA : 2400 N • m - 300 N •m - (300 N)(4 m) = Mh
or M,,=900N-m^H(b) Therefore, loadings (c) and (h) are equivalent. -4
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261
200 N 100 N
4 mPROBLEM 3.102
\
100 N-m
SOLUTION
We have
and
or
: > \ a 4-m-long beam is loaded as shown. Determine the loading of
-' Problem 3.101 which is equivalent to this loading.:'S00N'in
XFV
: -200N-400N = fi
ZMA : -400 N • m + 2800 N-m - (400 N)(4 m) = M
M=800N-nO
Problem 3.101 Equivalent force-couples atA
M
or R =600NHR
HFMm.
Equivalent to case (/) ofProblem 3.101 -4
Case R M
(a) 600 N| 1000 N-m)
(*) 600 N 900 N • m )
(c) 600 N|
900 N • m ^)
id) 400 N|
900 N • m. ^)
(*) 600 NJ
200 N-m}
if) 600 N|
800 N • m ")
(g) 1000 N|
1.000 Nm")
(A) 600 N|
900 N • m ^)
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262
PROBLEM 3.103
Determine the single equivalent force and the distance from Point A to its line of action for the beam andloading of (a) Problem 3 . 1 1 Z>, (/?) Problem 3.10k/, (c) Problem 3. 1 1 e.
PROBLEM 3.101 A 4-m-long beam is subjected to a variety of loadings, (a) Replace each loading with anequivalent force-couple system at end A of the beam, (b) Which of the loadings are equivalent?
400 N
-J0U X
./lit) \
200 N-ii
4 in -Hit) X w.\ \
1 KOI) N i>>
(«)
i')(H)N-m101) N
(W
. cA Av.xt n 200 V:..
W)
800 NMJON-ii)
i,.\
11,'
""
1000 N-c ''
(g> (A)
>' \-lll
200 \r
'/:
"Oil \
300 N- mi
,'
NO!) N
i: '< \ in
»iN
« : N-ni
i,.
:/)
900 N
".'
! .-'MN-in
:«iu X-
.., vj
SOLUTION
I GVCH <\OOH«r\
^~~ c
a£h--4
w
W~t ^400n 2.2>oorvm
8<30rt
\*Y\
[v~d -^t'
200fc<W i%
til!
lm
(a) For equivalent single force at distance d fromA
We have ZF„: -600 N^^
ZOOiM
and
(b) We have
and
(c) We have
and
rf»c sje
or R ~ 600 N ^
XMC : (600N)(rf)-900N-m =
or d^ 1.500 m -«
£/y -400N + 800N = #
or R = 400Nf^
ZMC : (400N)(rf) + (800N)(4~</)
-2300N-m =
or c/ = 2.25 m «
lFy
: -400N-200N = /f
or R = 600 NJ^
ZMC : 200N-m + (400N)(t/)
-(200N)(4-c/) + 400Nm =^0
or tf = 0.333 m <
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263
PROBLEM 3.104
Five separate force-couple systems act at the corners of a piece of sheet metal, which has been bent into the
shape shown. Determine which of these systems is equivalent to a force F = (10 lb)i and a couple of moment
M - (15 lb • ft)] + (15 lb • ft)k located at the origin.
lb.lt
SOLUTION
First note that the force-couple system at F cannot be equivalent because of the direction of the force [The
force of the other four systems is (10 lb)i]. Next move each of the systems to the origin 0\ the forces remain
unchanged.
A: M^ =IM =(5 lb-ft)j + (15 lb-ft)k + (2 ft)kx(10 Ib)i
-(25 1b-fl)j + (15 1b-ft)k
D: MD = 2M = -(5 lb • ft)] + (25 lb • ft)k
+ [(4.5 ft)j + (l ft)] + (2 ft)k"jxl.O lb)i
= (151b-ft)i + (151b-ft)k
G : M6. =XM - (1 5 lb • ft)i + (1 5 lb • ft)
j
/: M;=SM
/=(151b-ft)]-(51b-ft)k
+ [(4.5ft)i + (lft)jjx(101b)j
= (15 1b-ft)]-(15 1b-ft)k
The equivalent force-couple system is the system at corner D, ^
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264
6 ft"
6 ft'
m B
fe& _ ,
,
PROBLEM 3,105
The weights oftwo children sitting at ends A and B of a seesaware 84 lb and 64 lb, respectively. Where should a third child sit
so that the resultant of the weights of the three children will
pass through C if she weighs (a) 60 lb, (/?) 52 lb.
SOLUTION
&\\b w„
K.^£± % w „
(a) For the resultant weight, to act at C, £Afc =0 Wc
. = 60 lb
Then (84 lb)(6 ft) - 60 \h{d) - 64 lb(6 ft) =
(b) For the resultant weight to act at C, ZMC = Wc - 52 lb
Then (84 lb)(6 ft) - 52 \h{d) - 64 lb(6 ft) =
feMlb
t
d = 2.00 ft to the right ofC
d = 2.31 ft to the right ofC
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265
kl ii
34 in.
P
PROBLEM 3.106
Three stage lights are mounted on a pipe as shown.
The lights at A and B each weigh 4.1 lb, while the one
at C weighs 3.5 lb. (a) If d = 25 in., determine the
distance from D to the line of action of the resultant
of the weights of the three lights, (b) Determine the
value of d so that the resultant of the weights passes
through the midpoint of the pipe.
SOLUTION
4.Wb LU \h 3.£
• A <b C-!
1
L K
3D E D
For equivalence
or
(b)
TFy
: -4.1-4.1-3.5 = -/? or R = 11.7 lb|
X/^: -(10 in.)(4.1 lb) -(44 in.)(4.1 lb)
-[(4.4 + rf)in.](3.5 lb)«-(£ in.)(11.7 lb)
375.4 + 3.5^ = 1 1.71 (rf.iinin.)
rf = 25 in.
We have 375.4+ 3.5(25) = ! 1.7 1 or I = 39.6 in.
The resultant passes through a Point 39.6 in. to the right ofD.
1 = 42 in.
We have 375.4 + 3.5</ = 11.7(42) or </ = 33.lin. A
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266
PROBLEM 3.107
A beam supports three loads of given magnitude and a fourth load whose magnitude is a function of position.
If b = 1 .5 m and the loads are to be replaced with a single equivalent force, determine (a) the value ofa so that
the distance from support A to the line of action of the equivalent force is maximum, (/?) the magnitude of the
equivalent force and its point of application on the beam.
U00N
9 in
400 N 1 I 600 N
400 ~ N
SOLUTION
For equivalence
liooN
4ooMj i toco N
or
4oo| N B.
r L ~iR
2.FV : -1300 + 400- - 400 - 600 = -R
b
/?= 2300-400- N
ZMA : ^400^A2 1 b
0)
*(400) ~(a + b)(6()Q) = -LR
or
1000a + 6006 -200
2300-400-
Then with 6 = 1.5 m L =10a + 9—
a
2
3
Where a, L are in m(a) Find value ofa to maximize L
dL=
da
s V3 ,
r*823- a3
23--al-M0a+9—-o ;
(2)
'<n8
23— a
\ 2
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267
PROBLEM 3.107 (Continued)
or„_ 184 80 64 ,80 n . 32 >
,,230 a a+—a~+—a + 24 a~ -03 3 9 3 9
or
Then
16a2 -276a + 1143 =
276± J(-276)2 - 4(1 6)(.l 143)
a~ —2(16)
or a = 1 0.3435 m and a ~ 6,9065 m
Since ^42? - 9 tn, a must be less than 9 m a = 6.9 1 m -4
(b) Using Eq. (1) .# = 2300 4006 "9065
or K-458N <«
1.5
1 0(6.9065) + 9 -- (6.9065)2
,t = 3 = 3.16 m23— (6.9065)
and using Eq. (2)
R is applied 3.16 m to the right ofA. M
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268
PROBLEM 3.108
Gear C is rigidly attached to arm AB. If the forces and couple
shown can be reduced to a single equivalent force at A,
determine the equivalent force and the magnitude of the
couple M.
SOLUTION
We have
For equivalence
or
or
Then
or
Also
YFX : -1 8 sin 30° + 25 cos40° = Rx
R =10.1511 lb
£F : -1 8 cos 30° - 40 - 25 sin 40° = R,.
ffv=-71.658 lb
R = yl(\QA5\\)2+(7l.65$y
= 72.416
71.658tan# =
10.1511
= 81.9° R = 72.4 lb "%" 81.9°^
IMA : M - (22 in.)(l 8 lb) sin 35° - (32 i.n.)(40 lb) cos 25°
~(48in.)(25 1b)sin65 = ()
M = 2474.8 lb -in. or A/ = 2061b-ft <
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269
•12 in.
U"(Y
!| j 8in.
i I j I
--* I'Mh
**>»» PROBLEM 3,109
A couple of magnitude M = 54 lb • in. and the three forces shown are
applied to an angle bracket, (a) Find the resultant of this system of
forces. (/?) Locate the points where the line of action of the resultant
intersects line AB and line BC.
SOLUTION
(a) We have LF : R = (-1 Oj) + (30 cos 60°)i
+ 30sin60°j + (-45i)
= -(30 lb)i + (15.9808 lb)
j
or R = 34.0 lb ^ 28.0° <«
(b) First reduce the given forces and couple to an equivalent force-couple system (R, M/f )
at B.
We have 2Affl
: Mfi=(54 lb -in) + (12 in.)(10 lb) -(8 in.)(45 lb)
Then with R at D
or
|\ iv_l— —
cLM
f}: -186 lb • in = a(.l 5.9808 lb)
t>
S>-
1*
a = U .64 in. ~ ^
.
and with R at E LMtf
: -1 86 lb in = C(30 lb) «Jc
or C - 6.2 in.
The line of action
below B.
ofR intersects line AB 1 1 .64 in. to the left of# and intersects line BC 6.20 in.
<
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270
10 lb
M
-12 in.
30 II
,60'
Sin.
'-^ ,!.3li.
PROBLEM 3.110
A couple M and the three forces shown are applied to an angle bracket.
Find the moment of the couple if the line of action ofthe resultant of the
force system is to pass through (a) Point A, (/?) Point B, (c) Point C.
SOLUTION
In each case, must have Mf -
{a) +)M*=2MA =M + (1 2 in.)[(30 lb)sin 60°] - (8 in.)(45 lb) =
M--= 4-48.231 lb -in. M = 48.21b -in.) <
(b) +)M*=!MB:= A/ + (12 in.)(10 lb) -(8 in.)(45 lb) =
M--= +240 lb in. M = 240 lb in.) <
(c) +)m£=z,mc =M + (12 in.)(l lb) - (8 in.)[{30 lb) cos 60°] =
M = M = <
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271
3 10 N500 N
X A u/I
375 mm
:>/.,
-500 mm
GOO N
200 mm
TfiON
PROBLEM 3.111
Four forces act on a 700 x 375-mm plate as shown, (a) Find
the resultant of these forces, (b) Locate the two points where
the line of action of the resultant intersects the edge of the
plate.
SOLUTION
(t>)
tan<9
= (-400 N + 160 N - 760 N)i
+(600N + 300N + 300N)j
= ~(1000N)i + (1200N)j
R = ^(1000 N)2+(1200 N)2
= 1562.09 N
1 200 N ^
1000N
-1.20000
-50.194°
M£=£rxF= (0.5m)ix(300N + 300N)j
= (300 N • m)k
(300N-m)k=xix(1200N)j
x = 0.25000 mx = 250 mm
(300 N m) = y\ x (-1 000 N)i
v = 0.30000 my - 300 mm
OOOM~V*Ot-\
R = 1562N ^50.2°^
-lOOO N
Intersection 250 mm to right ofC and 300 mm above C -4
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Ill
500 N;Vfo N
-i
/
/ s
/ j
/ I 375 mm;/':
C»
600 N
— 500 mm7<j0 N
200 mm
PROBLEM 3.112
Solve Problem 3.1 1 1, assuming that the 760-N force is directed
to the right.
PROBLEM 3.111 Four forces act on a 700 x 375-mm plate as
shown, (a) Find the resultant of these forces, (b) Locate the two
points where the line of action of the resultant intersects the edge
of the plate.
SOLUTION
(b)
= (-400N + J60N + 760N)i
+(600N + 300N + 300N)j
= (520N)i + (1200N)j
tan#
R = 7(520 N)2+(1200 N)2 = 1 307.82
/l200N
N
..520 N
= 66.5714'
2.3077
M «
or
c
v - - -
/ fc>
>U n/\H V J&5»h\ /sw\
Ti^OwC?0O M
R = 1308N ^66.6C
or
IrxF
= (0.5m)ix(300N + 300N)j
- (300 N • m)k
(300N-m)k = xix(1200N)j
x = 0.25000 m
x = 0.250 mm
(300 N • m)k - [/i + (0.375 m)j]x[(520 N)i + (1200 N)j]
= (1200x'~l95)k
/ = 0,41250 m
x' ~ 412.5 mm
*~ S20 N
Intersection. 41.2 mm to the right ofA and 250 mm to the right ofC
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273
4 If
240 lb\
70A&-~
8ft 8 ft-
j
1.00
\\
\\
8 It—*\*— 8 ft
180 ih
8 ft
300 II)
76ft
PROBLEM 3.113
A truss supports the loading shown. Determine the equivalent
force acting on the truss and the point of intersection of its
line of action with a line drawn through Points y4 and G.
SOLUTION
We have R = IF
R = (240 lb)(cos70°i - sin 70°j) - (1 60 lb)j
+ (300 lb)(- cos 40°i - sin 40°j)- (1 80 lb)
j
R = -(147.728 lb)i - (758.36 Ib)j
= 7(147.728)2+(758.36)
2
= 772.62 lb
v«*o lb
$ = tan
tan"
rA
^y-758.36
-147.728
= 78.977 cor R = 773 lb ^ 79.0° <
We have
where
YMA = dRv
ZM
d
-[240 lb cos 70°](6 ft) - [240 lb sin 70°](4 ft)
-(160 lb)(12 ft) + [300 lbcos40°](6 ft)
-[300 lb sin 40°](20 ft) - (1 80 lb)(8 ft)
-7232.5 lb -ft
-7232.5 lb -ft
-758.36 lb
9.5370 ft or rf = 9.54 ft to the right ofA <
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274
I] ill
- (> in
21011)
r 150 lb
PROBLEM 3.114
Pulleys A and B are mounted on bracket CDEF. The tension
on each side of the two belts is as shown. Replace the four
forces with a single equivalent force, and determine where its
line of action intersects the bottom edge of the bracket.
SOLUTION
Equivalent, force-couple atA due to belts on pulleyA
We have IF: -120 lb -160 lb = J"?
,
R/(=280 lb
We have IMA ; -40 lb(2 in.) =MA
M„=801b-.in.J)
Equivalent force-couple at B due to belts on pulley B
We have £F: (2 1.0 lb + 1. 50 lb)^L 25° = R„
R„=3601b^l!l25 <
We have 1M B : -601b(1.5in.) = MB
M g =90Ib-nO
Equivalent force-couple atF
R = 280 lb
AL«*HW<P^? ,b
A
We have LF: RF = (- 280 lb)] + (360 lb)(cos 25°i+ sin 25°j)
= (326.27 lb)i- (127.857 lb)j
R = RF
= 7(326.27)2+(127.857)
2
= 350.43 lb
r d \
= tan-i
tan"-127.857
326.27
f 1 in octN
-21.399'
Rr=f?
or R,, =R = 3501b^ 2.1.4° «
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275
PROBLEM 3.114 (Continued)
We have TMF : MF = ~(280 lb)(6 in.) - 80 lb in.
-[(360lb)cos25°](1.0in.)
+[(360 lb) sin 25°](l 2 in.) - 90 lb • in.
M F =-(350.56 lb -in.)k
To determine where a single resultant force will intersect line FE,
MF = clRy
M,
-127 -857 lb
2.74 18 in. or d = 2.1A in.
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276
420 N240 nun
•loo n h :
50 mm '- \ i • >
520 mm
fv
?u.\-m. :
50 nun
640 nun
SI! v
ISO mm
50mm
PROBLEM 3.115
A machine component is subjected to the forces and couples
shown. The component is to be held in place by a single rivet
that can resist a force but not a couple. For P — 0, determine the
location of the rivet hole if it is to be located (a) on line FG,(b) on line GIL
SOLUTION
We have
ten-™
tZO N
ftaKl
4oW«n^ .R,b
+ F
First replace the applied forces and couples with an equivalent force-couple system at G.
ZFX : 200cos 15° - 1 20cos 70° + P = Rx
tfv= (152.1.42 + .P)N
ZFr
: - 200sin 1 5°- 1 20sin 70° - 80 = R
Thus
or
or R. -244.53 N
or
Setting P-0 inEq.(l):
Now with R at /
or
and with R at J
or
ZMC : - (0.47 m)(200 N) cos 1 5° + (0.05 m)(200 N) sin .1.
5°
+ (0.47 m)(!20 N) cos 70° -(0.1 9 m)(120 N) sin 70°
- (0. 1 3 m)(P N) - (0.59 m)(80 N) + 42 N - m+40N-m = MG
MG =-(55.544 + 0.1 3P)N-m
ZMG : - 55.544 N • m = -a(244.53 N)
a- 0.227 m
SMC : - 55 .544 N m = -b(\ 52. 142 N)
b = 0.365 m
(I)
(a) The rivet hole is 0.365 m above G.
(b) The rivet hole is 0.227 m to the right of G. 4
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Ill
200
50.
520 mir
50 mm
240 mm'120 N
^
^.O-Nvm. ^kd
80 N
mm
640 mm
I
180 mm
- .50mm
PROBLEM 3.116
Solve Problem 3. 1 1.5, assuming that P = 60 N.
PROBLEM 3.115 A machine component is subjected to the
forces and couples shown. The component is to be held in
place by a single rivet that can resist a force but not a couple.
For P = 0, determine the location of the rivet hole if it is to
be located (a) on line FG, (b) on line GIL
SOLUTION
See the solution to Problem 3.115 leading to the development of Equation (1)
MG = -(55.544+ 0. 1 3P) N • mand /?, =(152.142 + /*) N
For P = 60 N
We have Rx=(152.142 + 60)
= 2.12.14N
A/c =-[55.544+ 0.13(60)]
= -63.344 N-m
Then with R at / ZMG : -63.344 N-m = -o(244.53N)
or a = 0.259 m
and with R at J XMG : -63.344 N-m = -£(212.14 N)
or b = 0.299 m
(a) The rivet hole is 0.299 m above G. A
(b) The rivet hole is 0.259 m to the right of G. A
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278
)
PROBLEM 3.117
A 32-lb motor is mounted on the floor. Find the resultant of the
weight and the forces exerted on the belt, and determine where
the line of action of the resultant intersects the floor.
f
2 in.
i-vi
r2 in.
eJpM'-- -Si
SOLUTION
l4-OSth30°
*> I4-Ocos30°
We have
£F: (60 lb)i - (32 lb)j+ (1 40 lb)(cos30°i + sin 30°j) = R
R = (181.244 lb)i + (38.0 lb)j
or R = ]85.2Ib^1J.84°^
We have ZM : XM =xRy
-[(140 lb)cos30°][(4 + 2cos30°)in.] -[(140 lb)sin30°][(2 in.)sin 30°]
~(60 lb)(2 in.) = x(38.0 lb)
1
and
38.0
x = -23.289 in.
(-694.97 -70.0 -120) in.
Or, resultant intersects the base (x axis) 23.3 in. to the left of
the vertical, centerline (y axis) of the motor.
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279
»-»('-$)PROBLEM 3.118
As followerAB rolls along the surface ofmember C, it exerts
a constant force F perpendicular to the surface, (a) Replace Fwith an equivalent force-couple system at the Point Dobtained by drawing the perpendicular from the point of
contact to the x axis, (b) For a = 1 m and b — 2 m, determine
the value ofx for which the moment of the equivalent force-
couple system at D is maximum.
SOLUTION
(a) The slope ofany tangent to the surface ofmember C is
~2bdy __ d
dx dxv * ;
Since the force F is perpendicular to the surface,
2
tana2b\x
For equivalence
where
IF: F = R
ZMD : (F cosa)(yA ) =MD
cos cr
:
2bx
^l(a2
)
2 +(2bxy
yA ~b'<4'
2Fb<
M,
.3\
Va4 +4*2jc2
Therefore, the equivalent force-couple system at £> is
Wb(.-s)
ZbX
R =F7 tan"
'«2 ^
2.F//
M
v2to
y
„3>
V 2 ..2^+4//x
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280
PROBLEM 3.118 (Continued)
(/?) To maximize M, the value ofx must satisfy —— =dx
where, for
M 8F(jk-*3
)
a = 1 m, /? = 2 m
dx
/i + 16jc2(1-3jc
2)-(x-;c
3)
= 8F-
1
(32x)(l + 16x2 )" 1/2
or
(l + .16;r)
(1 + 1 6,y2)(1 - 3x
2) -
1
6x(x - x3) =
32x4 +3x2 -l=0
-3±V9-4(32)(-l)
2(32)
Using the positive value of.r
0.13601 ] m? and -0.22976 nV
x = 0.36880 m or x = 369mm -^
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281
\i
, ::%, 200 mm
PROBLEM 3.119
Four forces are applied to the machine component
ABDE as shown. Replace these forces by an equivalent
force-couple system atA'
20 mm fijli 300 N
25! X \^f7-.^MM0'^ 1 60 i i i m'
j!
f :?;"| 100 mm
A
SOLUTION
R = -(50 N)j - (300 N)i - (1 20 N)i - (250 N)k
R = -(420 N)i - (50 N)j - (250 N)k
rw =(0.2m)i
rD =(0.2m)i + (0.16m)k
rB = (0.2 m)i - (0. 1 m)j+ (0. 1 6 m)k
M*=r;,x[-(300N)i-(50N)j]
+ rD x (-250 N)k + r x ( - 1 20 N)i
-(3oon)i
-(ztorji
i J k
0.2 m
-300 N -50 N
i J k
0.2 m 0.16m
-250 N
k» J
+ 0.2 m -0.1m 0.16 m
-120 N
= -(1.0 N • m)k + (50 N m)j - (1 9.2 N • m)j - (1 2 N m)k
Force-couple system atA is
R = -(420 N)i - (50 N)j - (250 N)k M* = (30.8 N • m)j -- (220 N • m)k <
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2S2
225 i»hi
'<.Qj
i(>" ^r i!>* s
'i[}(^r ii0N
:/|5 \
145 N
ISO mm
PROBLEM 3.120
Two ] 50-mm-diameter pulleys are mounted
on line shaft AD. The belts at B and C lie in
vertical, planes parallel to the yz plane. Replace
the belt forces shown with an equivalent force-
couple system at A.
SOLUTION
Equivalent force-couple at each pulley
Pulley B
Pulley C
Then
R/f= (i45 N)(-cos20°j + sin 20°k) - 2 1 5 Nj
= -(35 1 .26 N)j + (49.593 N)k
MB = -(21 5 N-.I45 NX0.075 m)i
= -(5.25N-m)i
Rc = (1 55 N + 240 N)(-sin 1 0°j - cos 1 0°k)
= -(68.591 N)j- (389.00 N)k
Mc = (240 N - 1 55 N)(0.075 m)i
= (6.3750 N • m)i
R = R/;+ Rc = - (4.1 9.85 N)j - (339.4 l)k
M.A =MB +Mc + xm x Rs + xcu xRc
- -(5.25 N • m)i + (6.3750 N m)i +
i
0.225
2lS>4
tS5e* (toi
2HOfi
&c
or R = (420N)j-(339N)k <
J k
-351.26 49.593
N-m
* J
0.45
-68.59:
k
-389.00
N-m
= (1.1 2500 N • m)i + (1 63.892 N • m)j - (1.09.899 N • m)k
or MA = (.1 . 125 N m)i + (1 63.9 N • m)j - (1 09.9 N • m)k <
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283
PROBLEM 3.121
..1! While using a pencil sharpener, a student applies the forces and
couple shown, (a) Determine the forces exerted at B and C knowing
that these forces and the couple are equivalent to a force-couple
system at A consisting of the force R = (2.61b)i +./?>,j~(0.71b)k
and the couple MRA =MX\ +(1.0 lb • ft)j - (0.72 lb • ft)k. (b) Find
the corresponding values of R and Mx .
SOLUTION
(a) From the statement ofthe problem, equivalence requires
B +C = R
# + C, = 2.6 lbor
and
or
or
Using Eq. (1)
or
and
or
(b) Eq.(2)=>
Using Eq. (3)
IF
ZFX
Y,Fy
XFZ
IMA :
LMV :
ZMy
:
ZM„
-C\,=Ry
-Cz=-0.7 1b or Cz
=0.7 1b
(rB/A xB + M.B ) + raA xC=M«
'1.75
(1)
(2)
(llb-ft) +
'3.75
12
12
ft (/?,) +
ft (CV )= M,
1.75
(3)
tt|(Cv H-|— (l
12 ' I 12(0.7 lb) = 1 lb • ft
1 +
3.75£T+ I.75C
V=9.55
3.755, + 1.75(2.65,) = 9.55
Bx = 2.5 lb
Cx =0.1. lb
'35 ^— ft (C,.) --0.721b- ft
12 )}
Cy= 2.4686 lb
B = (2.5 lb)i C = (0. 1000 lb)i - (2.47 lb)j- (0.700 lb)k <
Ry=*-2Al\b <
' .75^
12(2.4686) =Mx
or A/, =1.360 lb -ft ^
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284
PROBLEM 3,122
A mechanic uses a crowfoot wrench to loosen a bolt at C Themechanic holds the socket wrench handle at Points A and Band applies forces at these points. Knowing that these forces
are equivalent to a force-couple system at C consisting of the
force C - (8 lb)i + (4 lb)k and the couple Mc = (360 lb • in.)i,
determine the forces applied at A and at B when A — 2 lb.
SOLUTION
We have
or
0ibXCm.
or
From i-coefficient
j-coefficient
k-coefficient
£F: A +B==C
F- A +B =8 lb
Bx =-(Ax +8)b)
IF- A. + # =0
or Ay=-B
y
LFZ
: 2lb + ^=4lb
B = 2 lb
SMC : r8/c xB + r,/c xA = M c
0)
(2)
(3)
i J k
8 2 +
Bx *, 2
i j k
8 8
4 4- 2
lb -in. = (360 lb -in.)!
(25y- SA
y)i + (25, -16 + %AX - 1 6)
j
+(85J,+8^)k = (360lb-in.)i
2Z?j;-8^, =360 lb -in.
-2.fi,+8/fv=32UVin.
8.^+8^=0
(4)
(5)
(6)
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285
PROBLEM 3.122 (Continued)
From Equations (2) and (4): 2By-8(--5,) = 360
Z?v=361b A
r=36 lb
From Equations (1) and (5): 2("4~8) + 8,4t=32
4 = 1.6 lb
From Equation (1): Bx =-(1.6 + 8) = -9.6 lb
A = (1 .600 lb)i - (36.0 lb)] + (2.00 lb)k <
B = -(9.60 Ib)i + (36.0 lb)j + (2.00 Ib)k <4
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286
64 in.
lfl} ? 96 in.
PROBLEM 3.123
As an adjustable brace BC is used to bring a wall into plumb, the
force-couple system shown is exerted on the wall. Replace this
force-couple system with an equivalent force-couple system at Aif R = 2 1 .2 lb and M = 1 3.25 lb • ft.
SOLUTION5
We have 2F: R - R^ - RXBC
where , (42 in.)i - (96 in.)j - (1 6 in.)k
106 in.
A
r = (42i - 96j - 1 6k)aj06
J /
v<> \\ j
S* "v, \\ j
c
or R^, = (8.40 lb)i - (1 9.20 lb)j - (3.20 lb)k <
We have £MV,: rc//JxR +M=M
-4
where xCIA = (42 in.)i + (48 in..)k =— (421 + 48k)ft
= (3.5ft)i + (4.0ft)k
R = (8.40 lb)i - (19.50 lb)j - (3.20 lb)k
=-42i + 96j + 16k
106
= -(5.25 lb • ft)i + (1 2 lb • ft)j + (2 lb • ft)k
i j k
Then 3.5 4.0
8.40 -19.20 -3.20
lb-ft + (-5.25i + 12j-f 2k)lb-ft =Mi4
M.A = (71.55 lb • ft)i + (56.80 lb ft)j - (65.20 lb ft)k
or M ^ - (7 1 .6 lb • ft)i + (56.8 lb • ft)j - (65.2 lb - ft)k <
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287
PROBLEM 3,124
A mechanic replaces a car's exhaust system by firmly clamping the catalytic converter FG to its mounting
brackets H and ./ and then loosely assembling the mufflers and the exhaust pipes. To position the tailpipe AB,
he pushes in and up at A while pulling down at B. (a) Replace the given force system with an equivalent
force-couple system at D. (b) Determine whether pipe CD tends to rotate clockwise or counterclockwise
relative to muffler DE, as viewed by the mechanic.
« .<f
0.14 in0.33
:-:i('
SOLUTION
(a) Equivalence requires
SF: R^A + B
(100 N)(cos 30°j - sin 30° k)- (1 1 5 N)j
-(28.4N)j-(50N)k
and
where
Then
SM.D : MD
AID
'aidk ^a +ybid x ^b
l B/D
-(0.48 m)i - (0.225 m)j + (1 . 1.2 m)k
-(0.38 m)i + (0.82 m)k
M n =100
i j k i J k
0.48 -0.225 1.1.2 + 115 -0.38 0.82
cos30° -sin 30° „1
1 00[(0.225 sin 30° - 1 . 1 2 cos 30°)i + (-0.48 sin 30°)
j
+ (-0.48 cos 30°)k] + 1 1 5[(0.82)i + (0.38)k]
8.56i-24.0j + 2.13k
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288
PROBLEM 3.124 (Continued)
The equivalent force-couple system atD is
R = -(28.4 N)j - (50.0 N)k <
MD = (8.56 N • m)i - (24.0 N • m)j + (2. 1 3 N • m)k <
(b) Since (MD )Z is positive, pipe CD will tend to rotate counterclockwise relative to muffler DE. A
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2 SO
PROBLEM 3.125
For the exhaust system of Problem 3.124, (a) replace the given force system with an equivalent force-couple
system at F, where the exhaust pipe is connected to the catalytic converter, (b) determine whether pipe EFtends to rotate clockwise or counterclockwise, as viewed by the mechanic.
0..M m0.33 m
0.30 m
0.10 m
0.56 in
o \
SOLUTION
(a) Equivalence requires
and
where
Then
M
SF: R = A + B
= (100 N)(cos 30°j - sin 30° k) - (1 1 5 N)j
= -(28.4N)j-(50N)k
M^r^xA +r^xB
'A/F
y BIF
-(0.48 m)i - (0.345 m)j + (2.10 m)k
-(0.38 m)I - (0. 1 2 m)j + (1 .80 m)k
i J k i J k
0.48 -0.345 2.3.0 + 115 -0.38 0.12 1.80
cos 30° -sin 30° -1
M;, =100
MF = 1 00[(0.345 sin 30° - 2. 1 cos 30°)i + (-0.48 sin 30°)j
+ (-0.48 cos 30°)k] + 1 1 5[(1 .80)1 + (0.38)k]
= 42.4i-24.0j+ 2.13k
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290
PROBLEM 3.125 (Continued)
The equivalent force-couple system atF is
R = -(28.4N)j~(50N)k A
M ,, = (42.4 N • m)i - (24.0 N • m)j + (2. 1 3 N • m)k <
(b) Since (MF )Z is positive, pipe EF will tend to rotate counterclockwise relative to the mechanic. -4
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291
PROBLEM 3.126
The head-and-motor assembly of a radial drill press was originally
positioned with arm AB parallel to the z axis and the axis of the
chuck and bit parallel to they axis. The assembly was then rotated
25° about the y axis and 20° about the centerline of the horizontal
arm AB, bringing it into the position shown. The drilling process
was started by switching on the motor and rotating the handle to
bring the bit into contact with the workpiece. Replace the force
and couple exerted by the drill press with an equivalent force-
couple system at the center O of the base of the vertical column.
SOLUTION
We have
or
We have
where
R = F = (l 1 lb)[( sin 20° cos 25°)]i - (cos 20°)j - (sin 20° sin 25°)k]
= (3.4097 lb)i - (1 0.3366 lb)j - (1 .58998 lb)k
R = (3.4 1 lb)i - (1 0.34 lb)j - (1 .590 Ib)k
M =rB/0 x¥xM.c
rm) = [(l4in.)sin25°]i + (15 in.)j + [(14in.)cos25°]k
- (5.9167 in.)i + (15 in.)j + (12.6883 in.)k
Mc = (90 lb • in.)[(sin 20° cos 25°)i - (cos 20°)j - (sin 20° sin 25°)k]
= (27.898 lb • in.)i - (84.572 lb • in.)j - (13.0090 ib • in.)k
M.
i J k
5.9167 15 12.6883
3.4097 -10.3366 1.58998
lb -in.
p= ll tb
+(27.898 - 84.572 - 13.0090) lb - in.
= (1 35.202 lb • in.)* - (3 1 .90 1 lb • in.)j - (1 25.3 1 3 lb in.)k
or M = (1 35.2 lb in.)i - (3 1 .9 lb in.)j - (125.3 lb - in.)k
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292
PROBLEM 3.127
Three children are standing on a 5x5-tn raft. If the weights
of the children at Points A, B, and C are 375 N, 260 N, and400 N, respectively, determine the magnitude and the point
of application of the resultant of the three weights.
SOLUTION
We have
We have
We have
F f
XF: F,+Fa +FC =R
-(375 N)j - (260 N)j - (400 N)j =R-(1035N)j-R
or /? = 1035N <
^.v :FaM + F*(zb) + fc(zc) = R(zD )
(375 N)(3 m) + (260 N)(0.5 m) + (400 N)(4.75 m) = (1 035 N)(zD )
zD = 3.0483 m or Zf) = 3.05 m <
Z,MZ : PA (xA ) + FB (xB ) +Fc {xc ) = R(xD )
375 N(l m) + (260 N)( 1 .5 m) + (400 N)(4.75 m) = (1035 N)(.^ )
xD = 2.5749 m r xn = 2.57 m <
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293
0.5 in
0.25 ill" P ' 0.25 m
PROBLEM 3.128
Three children are standing on a 5x5-m raft. The weights
of the children at Points A, B, and C are 375 N, 260 N, and
400 N, respectively. If a fourth child of weight 425 Nclimbs onto the raft, determine where she should stand if
the other children remain in the positions shown and the
line of action of the resultant of the four weights is to pass
through, the center of the raft.
SOLUTION
We have
We have
We have
L.F: FA +FB +$C =R
-(375 N)j - (260 N)j - (400 N)j - (425 N)j - R
R=-(1460N)j
1MX : FA (zA ) +FB (zB ) + Fc (zc ) + FD {zD ) = R{zH )
(375 N)(3 m) + (260 N)(0.5 m) + (400 N)(4.75 m)
+(425 U)(zD ) = (1460 N)(2.5 m)
zD =1.16471 m
SMZ: FA (xA ) + FB (xa ) + Fc (xc ) + FD (xD ) = R{xH )
(375 N)(l m) + (260 M)(l .5 m) + (400 N)(4.75 m)
+(425 N)(x/} )
= (1460 N)(2.5 m)
xD = 2,3235 m
or Zt, =1.165 m M
or xn = 2.32 m A
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294
l'i><m f:-.i
feii )}•
8 If"
5 ft
<)ii
5.5 ft
"!Ufi
ir 2.5 ft
on.'ij
1:
PROBLEM 3.129
Four signs are mounted on a frame spanning a highway,
and the magnitudes of the horizontal wind forces acting on
the signs are as shown. Determine the magnitude and the
point of application of the resultant of the four wind forces
when a -\ ft and 6 = 1.2 ft.
SOLUTION
We have
.» 1
Soft,
Assume that the resultant R is applied at PointP whose coordinates are (x,y, 0).
Equivalence then requires
E/<; : - 1 05 - 90 - 1 60 - 50 = -R
XMX : (5 ft)(1 05 lb) - (1 ft)(90 lb) + (3 it)(l 60 lb)
+ (5.5 ft)(50 lb) = -j/(405 lb)
or y = -2.94 ft
ZMy : (5.5 ft)(105 lb) + (12 ft)(90 lb) + (14.5 ft)(160 lb)
+ (22.5 ft)(50 lb) = -j(405 lb)
or x~ 12.60 ft
R acts 1 2.60 ft to the right ofmemberAB and 2.94 ft below member BC.
or R - 405 lb 4
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295
5 ft
'>>><:.. p. i
5.5 it
9 ft
S
H).5ft
PROBLEM 3.130
Four signs are mounted on a frame spanning a highway,
and the magnitudes of the horizontal wind forces acting on
the signs are as shown. Determine a and b so that the point
of application ofthe resultant of the four forces is at G.
SOLUTION
Since R acts at G, equivalence then i equires that 1M of the applied system of forces also be zero . Then
at G :EMV : -(a + 3) ft x (90 lb) + (2 ft)(l 05 lb)
+ (2.5 ft)(50 lb) =
or a~ 0.722 ft <
XMy
: -(9ft)(105ft)-
+ (8ft)(501b) =
-(14.5-6) ftx (90 lb)
Ol b-= 20.6 ft A
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2%
224 X
.1,5 m
PROBLEM 3.131*
A group of students loads a 2x3.3-m flatbed trailer with two
0.66 x 0.66x 0.66-m boxes and one 0.66x 0.66x 1 .2-m box.
Each of the boxes at the rear of the trailer is positioned so that
it is aligned with both the back and a side of the trailer.
Determine the smallest load the students should place in a
second 0.66x0.66x1.2-m box and where on the trailer they
should secure it, without any part of the box overhanging the
sides of the trailer, if each box is uniformly loaded and the line
of action of the resultant of the weights of the four boxes is to
pass through the point of intersection of the centerlines of the
trailer and the axle. {Hint: Keep in mind that the box may be
placed either on its side or on its end.)
SOLUTION
For the smallest weight on the trailer so that the resultant force of the four weights acts over the axle at the
intersection with the center line of the trailer, the added 0.66x0.66x1.2-m box should be placed adjacent to
one of the edges of the trailer with the 0.66x 0.66-m side on the bottom. The edges to be considered are
based on the location of the resultant for the three given weights.
We have
We have
We have
EF : - (224 N)j - (392 N)j~(l 76 N)j = RR = -(792N)j
£MZ : -(224N)(0.33 m)-(392 N)(1.67 m) - (176 N)( 1.67 m) = (-792 N)(j)
xR =1.29101 m
XMX : (224 N)(0.33 m) + (392 N)(0.6 m) + (1 76 N)(2.0 m) = (792 N)(z)
z„= 0.83475 mFrom the statement of the problem, it is known that the resultant of R from the original loading and the
lightest load W passes through G> the point of intersection of the two center lines. Thus, IMa — 0.
Further, since the lightest load W is to be as small as possible, the fourth box should be placed as far from Gas possible without the box overhanging the trailer. These two requirements imply
(0.33 m < x< 1 m)(l .5 m < z < 2.97 m)
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297
PROBLEM 3.131* (Continued)
With XL = 0.33 m
at G: ZMZ : o- 0.33) mxWL -(1.29101-1) mx(792N) =
or wL~ 344.00 N
Now must check if this is physically possible,
at G: IMV : (Zz- 1 .5)mx344 N) - (1 .5 - 0.83475)mx (792 N) =
or ^ -3,032 m
which is not acceptable.
With ^ = 2.97 m:
at G: I,MX
(2.97 - 1 .5)m xWL- (1 .5 - 0.83475)m x (792 N) = )
or K = 358.42 N
Now check if thi s is physically possible
at G: XMZ
- (1- ^)mx(358.42 N)-(l .29101 -l)mx(792 N) =
or ^ -0.357 m ok!
The minimum weight, of the fourth box is WL = 358 N <
And it is placed on end (A 0.66x0.66-m
from side AD.
side down) along side j42? with the center of the box 0.357 m<
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298
224 N
392 N
1..S m
PROBLEM 3.132*
Solve Problem 3. 1 31 if the students want to place as muchweight as possible in the fourth box and at least one side of
the box must coincide with a side of the trailer.
PROBLEM 3.131* A group of students loads a 2x3.3-m
flatbed trailer with two 0.66 x().66x().66-m boxes and one
0.66 x 0.66 x 1 .2-m box. Each of the boxes at the rear of the
trailer is positioned so that it is aligned with both the back
and a side of the trailer. Determine the smallest load the
students should place in a second 0.66x0.66x1 .2-m box
and where on the trailer they should secure it, without any
part of the box overhanging the sides of the trailer, if each
box is uniformly loaded and the line of action of the
resultant of the weights of the four boxes is to pass through.
the point of intersection of the centerlines of the trailer and
the axle. (Hint: Keep in mind that the box may be placed
either on its side or on its end.)
SOLUTION
First replace the three known loads with a single equivalent force R applied at coordinate (XR , 0, Z/( )
Equivalence requires
224-392-176 = -/?£/y
or R = 792 N
Y
or
or
IMV
: (0.33 m)(224 N) + (0.6 m)(392 N)
+ (2m)(176N) = Zfl(792N)
zR -0.83475 m
ZMZ : -(0.33 m)(224 N)-(l .67 m)(392 N)
- (1 .67 m)(176 N) = xs (792 N)
xD =1.29101 m
A*»-tr
From the statement of the problem, it is known that the resultant of R and the heaviest loads W# passes
through G, the point of intersection of the two center lines. Thus,
£MC =
Further, since W/y is to be as large as possible, the fourth box should be placed as close to G as possible while
keeping one of the sides of the box coincident with a side of the trailer. Thus, the two limiting cases are
0.6 m or 2.7 m
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299
PROBLEM 3.132* (Continued)
Now consider these two possibilities
With x„ -0.6 m:
at G: X.M,: (\~0,6)mxWH - (1.29 101 -l)mx (792 N) =
or W„ = 576.20 N
Checking if this is physically possible
at G: SMV
: {zH - 1.5)mx (576.20 N)-(1.5-0.83475)mx(792 N) =
or zfI=2.414 m
which is acceptable.
With zn = 2.7 m
at G: TMX : (2.7 - 1 .5) Wn - (1 .5 - 0.83475)m x (792 N) =
or WH = 439 N
Since this is less than the first case, the maximum weight of the fourth box is
WH = 576 N <
and it is placed with a 0.66x1 .2-m side down, a 0.66-m edge along side AD, and the center 2.41 mfrom side DC. <
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300
PROBLEM 3.133
Three forces of the same magnitude P act on a cube of side a as
shown. Replace the three forces by an. equivalent wrench and
determine (a) the magnitude and direction of the resultant force
R, (/>) the pitch of the wrench, (c) the axis of the wrench.
SOLUTION
Force-couple system at O:
r =n + pj + Pk = P(i + j + k)
Mj = c/j x Pi + ak x Pj + c/i x Pk
i *^ H
!
<?;- -
m:
Pak-Pai-Paj
-Pa(i + j + k)
y
Since R and M^ have the same direction, they form a wrench with M, =M^. Thus, the axis of the wrench
is the diagonal OA. We note that
cos X- cos - cos Z
——t=t = ~y=r
/? = Pfi X= #,, = ez = 54.7
C
M}= Mo = -PoV3
Pitch = pM, -PaS
-—a
(6) -a
(c) Axis ofthe wrench is diagonal 0yf
tf PV3
r^pS ex ^eY~ez
^5A.T
4
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301
\Qj.\
PROBLEM 3.134*
A piece of sheet metal is bent into the shape shown and is acted
upon by three forces, if the forces have the same magnitude P,
replace them with an equivalent wrench and determine (a) the
magnitude and the direction of the resultant force R, (h) the pitch
of the wrench, (c) the axis of the wrench.
SOLUTION
First reduce the given forces to an equivalent force-couple system (R, Mj) at the origin.
We have
£F: -P\ + I>j + Pk = /?
or R = Pk /K|FA = J^K1M : -i*P)\ + -(*/>)! +
(-«/> k = "2 * 1 ^ r x
or M.g =qp(-\- j +V(a) Then for the wrench
R = P <
and /vaxis
~ ~ K
cos 6>v= cos 6
y= cos 9Z
= 1
or V^9O° 0,, =90° g =O° <
(b) Now
= k -opf-l-j+|kj
= *,/>
2
Then ^- Mi-!*P m. r>~ ,-, -4
rt /> 2
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302
PROBLEM 3.134* (Continued)
(c) The components of the wrench are (R, M(), where M, = M
xXaxi&>
and the axis of the wrench is
assumed to intersect ihexy plane at Point Q whose coordinates are (x,y, 0). Thus require
Where
Then
M z =rfixR*
M2 =M xM,
aP\ -i-j+ ~k -~aPk=;(xl + y$) + Fk
Equating coefficients
i : - aP — yP or y — -a
j: ~aP~-xP or x~a
The axis of the wrench is parallel to the z axis and intersects the xy plane at x = a, y — —a. -^
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303
PROBLEM 3.135*
The forces and couples shown are applied to two screws as a piece
of sheet metal is fastened to a block ofwood. Reduce the forces and
the couples to an equivalent wrench and determine (a) the resultant
force R, (b) the pitch of the wrench, (c) the point where the axis of
the wrench intersects the xz plane.
100 mm
SOLUTION
First, reduce the given force system to a force-couple system.
We have
We have
XF: -(20N)i-(15N)j = R # = 25N
IMa : S(r xF) +XMC = Mg
MJ = -20 N(0. 1 m)j - (4 N • m)i -(IN- m)j
= -(4N-m)i~(3N-m)j
(b) We have
R = -(20.0 N)i-(1 5.0 N)j <
M, =VM5 /U
(-0.81 - 0.6j) • [~(4 N • m)]i - (3 N m)j]
5N-m
Pitch
(c) From above note that
p =M, 5 N
m
R 25 N- 0.200 m
or p = 0.200 m <
M, =M /e
Therefore, the axis of the wrench goes through the origin. The line of action of the wrench lies in thcxy
plane with a slope of
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304
filh'in.
I .11 lit
PROBLEM 3,136*
The forces and couples shown are applied to two screws as a piece
of sheet metal is fastened to a block of wood. Reduce the forces and
the couples to an equivalent wrench and determine (a) the resultant
force R, (b) the pitch of the wrench, (c) the point where the axis of
the wrench intersects the xz plane.
SOLUTION
First, reduce the given force system to a force-couple at the origin.
We have
We have
(b) We have
IF: -(101b)J-(111b)j = R
R=-(211b)j
IM : Z(r x F) +ZMC =MJ
i J k
M£ = 20 lb- in. +
-10
= (351b-in.)i~(12lb-in.)j
R = -(21lb)j
R
and pitch
ij k
-.15
-11
lb in. - (12 lb- in)
j
or R = -(21.0 lb)j
M,=VM» kR =R
= (-j)-[(351b-in.)i-(121b-in.)j]
= 1 2 lb • in. and M , = -(1 2 lb in .)j
p=«L =H^i = 0.57I43i„.R 211b
or p~ 0.571 in.
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305
PROBLEM 3.136* (Continued)
(c) We have
Require
From i:
From k:
(35 lb
Mg*=M,+M2
M 2=Mj-M, =(35lb-in.)i
M 2 =re/0 xR
in.)i = (jri + zk)x[-(2Ilb)j]
35i = -(21x)k + (21z)i
35 = 21z
z = 1.66667 in.
= -21x
z =
The axis of the wrench is parallel to the y axis and intersects the xz plane at x — 0, z~= 1 .667 in. ^
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306
IJ
0.1 m
0.6 m
30 N'rtt
M \
jum
PROBLEM 3.137*
Two bolts at A and B are tightened by applying the forces
and couples shown. Replace the two wrenches with a
single equivalent wrench and determine (a) the resultant
R, (b) the pitch of the single equivalent wrench, (c) the
point where the axis of the wrench intersects the xz plane.
SOLUTION
3 M?
\RJ2 y
*
First, reduce the given force system to a force-couple at the origin.
We have EF: -(84N)j~(80N)k = R rt = ll6N
and £M : I(r x F) +SMC =MJ
i J k
0.4 0.3
80
i J k
0.6 .1 +
84
+ (-30j-32k)N-m =M;
Mi = -(1 5.6 N • m)i + (2 N m)j - (82.4 N m)k
(a)
(b) We have
R = -(84.0 N)j- (80.0 N)k ^
.M,=^-Mg *,,=
-84j-80k
116
55.379 N • m
[-(15.6 N • m)i + (2 N • m)j-(82.4 N • m)k]
and
Then pitch
MX=M
XAR --(40.102 N-m)j-(38.192N-m)k
M, 55.379 N-mP
R 116 N0.47741m or p = 0.477 m <
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307
PROBLEM 3,137* (Continued)
(c) We have mJ=Mj+m 2
M 2 = JVf£~M., = [(-J.5.6i + 2j-82.4k)-(40.102j-38.I92k)]N-m
= -(15.6 N • m)i + (42.102 N • m)j- (44.208 N -m)k
Require M 2 = rQIO X R
(- 1 5.6i + 42. 1 02j - 44.208k) = (xi. + 2k)x (84j - 80k)
= (84z)i + (80x)j-(84x)k
From i: -15.6 = 842
z = -0.185714 m
or 2 = -0.1 857 m
From k: -44.208 = -84x
x = 0.52629 m.
or jc = 0.526 m
The axis <if the wrench intersects the xz plane at
x = 0.526 m y = 2 = -0.1857 m <
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308
PROBLEM 3.138*
Two bolts at A and B are tightened by applying the forces and
couples shown. Replace the two wrenches with a single equivalent
wrench and determine (a) the resultant R, (b) the pitch of the
single equivalent wrench, (c) the point where the axis of the
wrench intersects the xz plane.
SOLUTION
E^b^&i
First, reduce the given force system to a force-couple at the origin at B.
(a) We have ZF: -(26.4 lb)k -(17 lb)| —i+— j |
= R
and 7? = 31.41b
R = -(8.00 lb)i - (1 5.00 lb) j - (26.4 lb)k <
We have XM B : rAIB x FA +M A +M B =M %
(/;) We have
i\r
i J k
-10
™26.4
220k -2381 — i + -^ji
j7 ]7j= 2641 - 220k - 1 4(8i + J 5j)
M J = (152 lb • in.)i - (21 lb • in.)j - (220 lb • in.)k
M^X.-M.^ XRRR
_-8.()Oi-15.00j- 26.4k
31.4
= 246.56 lb- in.
[(1 52 ib • in.)i -(210 lb • in.)j - (220 ib • in.)k]
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309
PROBLEM 3.138* (Continued)
and M, =MXXR = -(62.8 1 8 lb • in.)i - (1 1 7.783 lb • in.)j - (207.30 lb • in.)k
Then pitchM, 246.56 lb -in. „ ocoo . noc , ^p-—l- — = 7.8522 in. or » = 7.85 in. AR 31.41b '
(c) We have MJ=M,+M2
M2=Ml ~M, = (1 52i - 21 Oj -220k)™ (-62.8.181-1 17.783j- 207.30k)
= (214.82 lb in.)i - (92.21. 7 lb • in.)j - (12.7000 lb • in.)k
Require M2 =rp xR
i J *
214.821- 92.2 17J-12.7000k = x z
™8 -15 -26.4
= (15z)i-(8z)j + (26.4*)j-(15*)k
From i: 214.82 = 152 z = 14.3213 in.
From k: -12.7000 = -15* x = 0.84667 in.
The axis of the wrench intersects the xz plane at x~ 0.847 in. y - z~ 14.32 in, A.
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310
PROBLEM 3.139*
Two ropes attached at A and B are used to move the trunk of a
fallen tree. Replace the forces exerted by the ropes with an
equivalent wrench and determine (a) the resultant force R,
(/>) the pitch of the wrench, (c) the point where the axis of the
wrench intersects the yz plane.
SOLUTION
(a) First replace the given forces with an equivalent force-couple system f R, ftlf, ) at the origin.
We have
4,c=V(6)2+(2)
2+ (9)
2 ==llm
Then
TAC1650 NU
= (6i + 2j + 9k)
: (900 N)i + (300 N)j + (1 350 N)k
and
T,BO500 N15
(14i + 2j + 5k)
(1 400 N)i + (200 N)j + (500 N)k
Equivalence then requires
LF: R = T/JC+T/W
= (900i + 300j + 1350k)
+(I400i + 200j + 500k)
= (2300 N)i + (500 N)j + (1 850 N)k
£M : M.g-r,xT,c +rfixTfiD
- (12 m)kx[(900 N)i + (300 N)j + (1350 N)k]
+ (9 m)ix[(1400 N)i + (200 N)j + (500 N)k]
= -(3600)i + (1 0800 - 4500)j + (1 800)k
= -(3600 N • m)i + (6300 N • m)j + (1.800 N • m)k
The components of the wrench are (R, M,), where
R = (2300 N)i + (500 N)j + (1 850 N)k
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311
PROBLEM 3.139* (Continued)
(b) We have
Let
Then
Finally
(c) We have
or
Now
For equivalence
tf = 1007(23)2 + (5)
2 + (I8.5)2 = 2993.7 N
K R !
R 29.937(23i + 5j + 18,5k)
^i=^„i.-MS
29.937
1
(23i + 5j + 1 8.5k) • (-36001 + 6300j + 1 800k)
[(23X-36) + (5X63) + (18.5X1 8)]0.29937
-601.26 N-m
M, -601.26 N-mR 2993.7 N
or P = -0.201 m <«
M,=M}Xaxis
= (-601.26 N-m) x
—
: (23i + 5j + 18.5k)29.937
M. = -(461.93 N • m)i - (1 00.42 1 N • m)j - (37 1 .56 N • m)k
M 2 -Mj-M,= (-36001 + 6300J + 1800k)
-(-46
1
,93i - 1 00.42 1j - 37 1 .56k)
= -(3138.1 N m)i + (6400.4 N • m)j + (2 1 7 1 .6 N • m)k
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312
PROBLEM 3.139* (Continued)
Thus require M3=r,>xR r - (y\ + zk)
Substituting
i j k
-3 1 38. li + 6400.4j + 2 1 7 1.6k - y z
2300 500 1850
Equating coefficients
j: 6400.4 = 2300 z or z = 2.78 mk: 2171.6 = -2300^ or y = -0.944 m
The axis of the wrench intersects the yz plane at y ~ -0.944 m z - 2.78 in <
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313
PROBLEM 3.140*
A flagpole is guyed by three cables. If the tensions in the
cables have the same magnitude P, replace the forces exerted
on the pole with an equivalent wrench and determine (a) the
resultant force R, (/>) the pitch of the wrench, (c) the point
where the axis of the wrench intersects the xz plane.
SOLUTION
,iM,
(a) First reduce the given force system to a force-couple at the origin.
We have ZF: PXBA + PkDC + P
k
DE = R
R = P4. 3
+3, 4.^ (~9. 4. 12
i— il+ l— i— 1 +— k
5 5 J U5 5 25
R =—(2i-20j-k) A25
«=—V(2) +(2°) +(!)
We have
(-AP 3P x
EM: I(r xP) =M
3P 4P+ (20a)jx|—j-_j| + (20fl)jx
-9P. AP . YIP.1 1 +
25 5 25m;
m:24/J«
(-i-k)
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314
PROBLEM 3.140* (Continued)
(b) We have
where
Then
and pitch
(c)
MX=X8 >M*
X„
M,
P
(2I-20J-k)25
nfip 9V5R ~25
—M2i~20}~k) —(~i~k) = r9V5 5 15V5
(2i-20j-k)
My -SPa ( 25 ^
R \5yf5 .21Sp
M, = MX,-%Pa
-8a
81or p = ~ 0.0988a -*
15^5 ,9>/5
1 8 Pa
v.-'v-'/
(2i-20j-k) =—(-2l + 20j + k.)
675
Then M2=M*-M, =-?^H-k)-—(-21 + 20J + k)=—(-4301- 20j- 406k)
5 675 675
Require
( %Pa
I 675
M2~ r
e/(?xR
\ f
(-4031 - 20j - 406k) = (xi + 2k)x3P^l-(2l-20J-k)
^-Wi25j
[20zi + (x+ 2z)j-20xk]
From i:
Pa8(~403) = 20z
675
Pa
rw \
V 25y
8(_4()6)^l ^ _20jc675
From k:
The axis of the wrench intersects the jez plane at
3P
25
z = -1,990! 2a
x = 2.0049a
x = 2.00a, z~ -1.990a <
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315
60 mm
/" [M
.1 I4N-IH
40 mm
8M£ v" V .1
10 N-m ^^Pq160mm /^¥ ^
40 mm
PROBLEM 3.141*
Determine whether the force-and-couple system shown can be
reduced to a single equivalent force R. If it can, determine Rand the point where the line of action of R intersects the yz
plane. If it cannot be so reduced, replace the given system with
an equivalent wrench and determine its resultant, its pitch, and
the point where its axis intersects theyz plane.
SOLUTION
First, reduce the given force system to a force-couple at the origin.
We have XF: F^+FG =R"(40 mm)i + (60 mm)j - (120 mm)k
and
We have
R = (50N)k + 70N
= (20N)i + (30N)j-(10N)k
/? = 37.417 N
140 mm
1M : £(r xF)+2M(
Mi
M.g = [((). 12 m)j x (50 N)k] + {(0. 16 m)i x[(20 N)i + (30 N)j - (60 N)k]}
+ (I0N-m)
+ (14N-m)
(160mm)i-(120 mm)j
200mm
(40 mm)i - (1 20 mm) j + (60 mm)k
1 40 mmm; (1 8 N • m)i - (8.4N • m)j + (.1 0.8 N • m)k
To be able to reduce the original forces and couples to a single equivalent force, R and M must be
perpendicular. Thus, R •M = 0.
Substituting
(20i + 30j - 1 0k) - (1 8i - 8.4j + 1 0.8k) =
(20)(1 8) + (30)(-8.4) + (-10)(10.8) =or
or 0^0
R and M are perpendicular so that the given system can be reduced to the single equivalent force
R = (20.0 N)i + (30.0 N)j-(1 0.00 N)k <
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316
PROBLEM 3.141* (Continued)
Then for equivalence
Thus require
Substituting
MS=r_xR r=yl + zk
18i-8.4j + 10.8k
i j k
y z
20 30 -10
Equating coefficients
j: -8.4 = 20z or 2 = -0.42 mk: 10.8 = -20^ or y = -0.54 m
The line of action ofR intersects the yz plane at x = y = -0.540 m z = -0.420 m
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317
PROBLEM 3.142*
Determine whether the force-and-coupie system shown can be
reduced to a single equivalent force R. If it can, determine Rand the point where the line of action of R intersects the yz
plane. If it cannot be so reduced, replace the given system with
an equivalent wrench and determine its resultant, its pitch, and
the point where its axis intersects the yz plane.
SOLUTION
First determine the resultant of the force
dDA
is at D. We have
= VH2)2+(9)
2+(8)
2=17in.
dm :^y](-~6)2+(Q)
2+(-%)
2 =\0m.
Then
*DA==34lb
-( 12i + 9j + 8k)17
= -(241b)i + (18Ib)j + (16 1b)k
and
*w10
= -{18 1b)i-(241b)k
Then
If : R = FZM +FaD= (-241 + 1 8j+
1
6k + (-1 8i - 24k)
= -(42 1b)i + (18lb)j-(8!b)k
For the applied couple
dAK = V(-6)2+ ("6)
2 + o 8)2= 6^ in -
Then
M 1601b-in., .. ,. 10I ,_
—
(-61-61 + 18k)671
1
-1^ [ (lib- in,)" - lb • i«-)j + (3 lb in.)k]
To be able to rixiuce the original ibrces and couple to a single equivalent force R and Mmust be perpendicular. Thus
R-M=0
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318
Substituting
PROBLEM 3.142* (Continued)
(-42i + 18j-8k)-^E(-i-j + 3k) =
or
or
160[(-42)H) + (18)(~l)+ (-8)(3)] =
0^0
R and M are perpendicular so that the given system can be reduced to the single equivalent force
R = -(42.0 Ib)i + (18.00 lb)j -(8.00 lb)k
Then for equivalence
o
T5"
Thus require
where
Substituting
M - rp/D xR
vPID = -(12 in.)i + [{y - 3)in.Jj + (z in.)k
160i j k
H-J + 3k)= -12 (y-3) z
-42 1 8 -8
= [(>>~3X-8)-(z)(.l.8)]i
+ [(z)(-42)-(-12)(-8)]j
+ [(-12)(1.8)-(.y-3)(-42)]k
Equating coefficients
k:
160
480
-42z-96 or z = -1.137 in.
-216 + 42(^-3) or y = 11.59 in.
The line of action ofR intersects theyz plane at x — y - 1 1 .59 in. z ~ - 1 . 1 37 in.
PROPRIETARY MATERIAL © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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319
u PROBLEM 3.143*
A*
1.,
1-
Replace the wrench shown with an equivalent system consisting of two
forces perpendicular to they axis and applied respectively at A and B.
b
\ a
^J"S"s\
* X
SOLUTIONy
Express the forces at A and B asB^^
A =
B =
Ax i + Azk
Bxi + Bzk
Then, for equivalence to the given force system £^*
ZFX
Ax + Bx =0 0)
ZFZ
AZ+ B
Z=R (2)
ZMX Az(a) + Bz (a+ b) = (3)
ZMZ : -Ax(a)-Bx (a + b) = M (4)
From Equation (I ), BX ^~AX
Substitute into Equation (4)
-Ax(a) + Ax (a + b) = M
AM A D M
xb * b
From Equation (2), K~--R~AZ
and Equation (3), Aza+ (R~Az)(a + b) = Q
^K)and Bz
= -- R - RK b)
Bz=
b
Then A =[bj
i +Rn+-)k 4
~(2Mf>«
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320
PROBLEM 3.144*
Show that, in general, a wrench can be replaced with two forces chosen in such a way that one force passesthrough a given point while the other force lies in a given plane.
SOLUTION
First, choose a coordinate system so that the xy plane coincides with the given plane. Also, position the
coordinate system so that the line of action of the wrench passes through the origin as shown in Figure a.
Since the orientation of the plane and the components (R, M) of the wrench are known, it follows that the
scalar components ofR and M are known relative to the shown coordinate system.
A force system to be shown as equivalent is illustrated in Figure b. Let A be the force passing through the
given Point P and B be the force that lies in the given plane. Let. b be the x-axis intercept of B.
The known components of the wrench can be expressed as
R = Rxi + Ryj + Rzk and M = MJ + MJ+
M
zk
while the unknown forces A and B can be expressed as
A = Ax\ + Ay $ + Azk and B - Bx i + Bzk
Since the position vector of Point P is given, it follows that the scalar components (\%y, z) of theposition vector i> are also known.
Then, for equivalence ofthe two systems
(1)
(2)
(3)
(4)
bBz (5)
(6)
Based on the above six independent equations for the six unknowns (Ax ,Ayy Az , Bx , B , B2 , b), there
exists a unique solution for A and B.
From Equation (2) a. - $ -4
£/V ** == AX +BX
ZFy
: V= Ar
X/y /?,=-• A2 + Bz
£MT : Mx
= yAz -zAy
ZMy
; My~ zA
x- xAz
2M_: Mz= xA
v-yAx
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321
PROBLEM 3.144* (Continued)
r oEquation (6) 4 =
,yj(xR
y~Mz ) A
(OEquation (1) Bx
--= *,-,yj
(xRy-M2 ) 4
Equation (4) Az= flW+zR,) <
( i>
Equation (3) Bz~~-K-
,yj
(Mx + zRy ) <
Equation (5) b(xM\ +yM
y +zMJ(Mx -yRz +zRy )
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322
PROBLEM 3.145*
Show that a wrench can be replaced with two perpendicular forces, one ofwhich is applied at a given point.
SOLUTION
M
*- 6iv«*> point
u-Co) (b) y
First, observe that it is always possible to construct a line perpendicular to a given line so that the constructedline also passes through a given point. Thus, it is possible to align one of the coordinate axes of a rectangularcoordinate system with the axis of the wrench while one of the other axes passes through the given point.
See Figures a and b.
We have R = />j and M = JWj and are known.
The unknown forces A and B can be expressed as
A = AJ + AJ + AM and B = BJ + BJ+ Bsk
The distance a is known. It is assumed that force B intersects the xz plane at (x, 0, z). The for equivalence
tt?x : = Ax +Bx
XFy
: R = Ay+B
y
-LBV. 0^Az+B
z
-zB..
-aAz ~xBz +zB x
= aAy+ xB
XMV
:
2Af„: MXMZ :
Since A and B are made perpendicular,
A-B = or AXBX+ AVB V + AR
There are eight unknowns: Ax , Ay , A., Bx , B > Bz , x, z
But only seven independent equations. Therefore, there exists an infinite number ofsolutions.
Next consider Equation (4): 0~-zB
If Bv- 0, Equation (7) becomes A.B. + A,B ~0
(I)
(2)
(3)
(4)
(5)
(6)
(7)
Using Equations (I.) and (3) this equation becomes At + At=0
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323
PROBLEM 3.145* (Continued)
Since the components ofA must be real, a nontrivial solution is not possible. Thus, it is required that By * 0,
so that from Equation (4), z = 0.
To obtain one possible solution, arbitrarily let Ax - 0.
(Note: Setting Ay , Az , or Bz equal to zero results in unacceptable solutions.)
The defining equations then become
= 5.
Then Equation (2) can be written
Equation (3) can be written
Equation (6) can be written
Substituting into Equation (5)',
R = Ay+B
y
Q~AZ+ B
Z
M = ~aA~xB,
0-aAy+xB
y
AvB
¥+ AzBz
=0
Ay=R~B
y
or
aAy
M=~aAz
aR >
R-By
B(~AZ )
y J
Substituting into Equation (7)',
(R-By)B
y+
aR } [aR }
or B sa R
ya2R2 ±M 2
Then from Equations (2), (8), and (3)
A„=R-a R rm:
a2R2 +M 2
a2R2 +M :
A..
B.
aR
2»3 Aa R
a2 R 2 +M 7
aR2M
aRlMJ2. r>2cfR£ + M"
a2R2 +M2
or
(2)
(3)
(5/
(6)
(7)'
(8)
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324
PROBLEM 3.145* (Continued)
In summary
A
B
RM„2 n 2a'R
A + AT
aRl
a2R2 +M :
-(Mj-aRk)
-(aRj + Mk)
Which shows that it is possible to replace a wrench with two perpendicular forces, one of which is applied at
a given point.
Lastly, if R > and M > 0, it follows from the equations found for A and B that A > and B, > 0.
From Equation (6), x<0 (assuming a>0). Then, as a consequence of letting Ax -0, force A lies in a
plane parallel to theyz plane and to the right of the origin, while force B lies in a plane parallel to theyz planebut to the left to the origin, as shown in the figure below.
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced or distributed in anyform or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permittedby McGraw-Hillfor their individual coursepreparation. Ifyou are a student using this Manual,you are using it without permission.
325
PROBLEM 3.146*
Show that a wrench can be replaced with two forces, one ofwhich has a prescribed line of action.
SOLUTION
First, choose a rectangular coordinate system where one axis coincides with the axis of the wrench and
another axis intersects the prescribed line of action (AA'). Mote that it has been assumed that the line of action
of force B intersects the xz plane at Point P(x, 0, z). Denoting the known direction of line AA' by
it follows that force A can be expressed as
Force B can be expressed as
B = Bxi + Byj+ Bzk
Next, observe that since the axis of the wrench and the prescribed line of action AA' are known, it follows that
the distance a can be determined. In the following solution, it is assumed that a is known.
Then, for equivalence
IFX : 0=AAx +B (
ZIy. R = AAy+ B
y
IF' = AA, + B„
SM.,.: zB.
JMV : M = -aAX. + zBx- xB.
EMV: Q = -aAZ
y+ xB
y
Since there are six unknowns (A, Bx> By, Bz, x, z) and six independent equations, it will be possible to
obtain a solution.
(!)
(2)
(3)
(4)
(5)
(6)
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32(»
PROBLEM 3.146* (Continued)
Case /: Let z - to satisfy Equation (4)
Now Equation (2)
Equation (3)
Equation (6)
Substitution into Equation (5)
Substitution into Equation (2)
Then
In summary
and
AXy=R-B
y
Bz= ~AX
Z
aAX, ( \a
M
A
-aAX7-
J ( M
v^v(R-B
y )
a
KB>>
(R-BY-AX)
# = -
X2\aR
I ( M
B.
X.z V
)bi+bvaR) y y y
B.XaR<
1
XzaR~X
vM
MRA =—
XzaR~X
yM ^ aR
M X,
BX =-AXX
B=~AX,
XXMRXzaR-Xv
MXZMR
XzaR-XyM
BR
A = P
, aR ,"A
Av~ A.
y M
XA <
XaR-XM(XxM + X
zaR] + XzMk) <
x~ay J
\~R(
X
zaR~X
yM
XmR 2
KM .
or x =™— A.X,R
Note that for this case., the lines of action of both A and B intersect the a- axis.
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327
PROBLEM 3.146* (Continued)
Case 2: Let B ~ to satisfy Equation (4)
Now Equation (2)
Equation (1)
Equation (3)
Equation (6)
Substitution into Equation (5)
A
B=-RfX.^
B=-R( X^
\. y J
aAXy~ which requires a -
M=z R
r x?
K. y J
R(X^xxv
or X^x — Xr ihThis last expression is the equation for the line of action of force B.
In summary
A
B
r R^
v*yj
f R^
<. y J
K
M,i-4k)
Assuming that Xx , Xy , Xz > 0, the equivalent force system is as shown below.
3 A.
Note that the component ofA in the xz plane is parallel to B.
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328
0.6 m -*U_ 0.6 in
0.5 in
C\V' \\
O.V> m
PROBLEM 3.147
A crate of mass 80 kg is held in the position shown. Determine(a) the moment produced by the weightW of the crate about E,
(b) the smallest force applied at B that creates a moment ofequal magnitude and opposite sense about E.
SOLUTION
(a) By definition
We have
o.2Sm
W = mg = 80 kg(9.8 1 mis2) = 784.8 N
ZME : ME = (784.8 N)(0.25 m)
^r
W
M£ = 196.2N-m^(/?) For the force at B to be the smallest, resulting in a moment. (M£) about .£, the line of action of force Ffl
must be perpendicular to the line connecting E to B. The sense of F« must be such that the forceproduces a counterclockwise moment about E.
Note:
We have
and
or
d = ^(0.85 m)2 + (0.5 m)2 = 0.9861 5 m
ZME : 1 96.2 N • m = FB (0.986 1 5 m)
Fs =198.954 N
A
0,5 in
= tan_1 0.85 m
0.5 m59.534c
C £
1 99.0N^ 59,5° <
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329
PROBLEM 3.148
iXin (m.Ba™r
\%
21 mm
It is known that the connecting rod AB exerts on the crank BC a 1 .5-kN force
directed down and to the left along the centerline of AB. Determine the moment
of the force about C.
28 nm
SOLUTION
Using (a)
(&> MC =yl(FAB )x +X\(FAB)y
(0.028 m)
42N-m
— X1500N25
24+ (0.021 m)|—X1500N
1
25
or Mc =42.0N-m)*«
Using (b)
MC =y2(FAB)x
= (0.1 m)25
X1500N = 42N-m
or Mc = 42.0 N • m *) <
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330
PROBLEM 3.149
A 6-ft-long fishing rod AB is securely anchored in the sand of a beach. After a fish takes the bait, the resulting
force in the line is 6 lb. Determine the moment aboutA ofthe force exerted by the line at B.
F^^^S^^RSilSi^R®^/i ;->"- v --~-.r^t;~~-r~-^^i
^-~ -------
SOLUTION
We have
Then
Now
where
Then
or
TX2= (6 lb) cos 8° = 5.9416 lb
Tx = Txz sin 30° = 2.9708 lb
Tv= Tsc sin 8° = -0.83504 lb
T, = r cos 30<
™*A ~" rBfA X TffC
[/?//»
M
M
-5.14561b
(6sin45°)j-(6cos45°)k
__ 6 ft
6
6
(j-k)
i J k
1 -1
2.9708 -0.83504 -5.1456
._ (-5.1456~0.83504)i -~~ (2.9708)j-~s-(2.9708)k
~(25.4 lb-ft)i-(I2.60 lb-ft)j-(12.60 lb-ft)k
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331
3 m
x 0.16 mDetail of the stake at B
PROBLEM 3.150
Ropes AB and BC are two of the ropes used to
support a tent. The two ropes are attached to a
stake at B. If the tension in rope AB is 540 N,
determine (a) the angle between rope AB and
the stake, (b) the projection on the stake of the
force exerted by rope AB at Point B.
SOLUTION
First note BA = V("3)2 + Of + (-1 -5)
2 - 4.5 m
BD = ,J(rO.QS)2 +(0M)2
+(0.16)2 =0.42 m
Then TiM =^-(-3I + 3J-1.5k)
=Zk(_2i + 2j~k)
Xlin=:
BD=
]
( 0.081 + 0.38J + 0.1 6k)m BD 0.42
=—(-41 + 191 + 8k)21
J
(a) We have %li^BD= TBA COS0
or ^L(~2l + 2j - k) •—(-41 + 1 9j + 8k) = TBA cos 6
orcos - J-K-2X-4) + (2)(19) + (-1X8)]
63
-0,60317
or (9 = 52.9° ^
(b) We have
= TBA cos6
= (540N)(0.60317)
or (7,
JM )JD =326N ^
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332
PROBLEM 3.151
A former uses cables and winch pullers B and E to plumbone side of a small barn. If it is known that the sum of the
moments about the x axis of the forces exerted by the
cables on the barn at Points A and D is equal to 4728 lb • ft,
determine the magnitude ofTDE when TAti - 255 lb.
SOLUTION
The moment about the x axis due to the two cable forces can be found using the z components of eachforce acting at their intersection with the xy-plane (A and D). The x components of the forces are parallel to
the x axis, and the y components of the forces intersect the x axis. Therefore, neither the x oi-y componentsproduce a moment about the x axis.
We have
where
£M,: (TM )M (yA ) + (Tm )z (yD ) = Mx
\'ab)z ~*- '*AB
= k-{T4BAAB )
k 2551b-i-12j-H2k
17
801b
(TD/i )z ~k-%
= k
DE
1.5i-14j + 12k N
T,DE8.5
0.64865r,DE
yA
y»
12 ft
14 ft
4728 lb -ft
and
(1 80 Ib)(12 ft)+ (0.648657^ )(1 4 ft) = 4728 lb • ft
TDE = 282.79 lb or TnF -2831b <DE
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3M
/;,;:,..,,..,.. PROBLEM 3.152
Solve Problem 3.151 when the tension in cable AB is 306 lb.
H /' -{ Hft/hb - i"- ^
PROBLEM 3.151 A farmer uses cables and winch pullers Band E to plumb one side of a small barn. If it is known that the
sum of the moments about the x axis of the forces exerted by
the cables on the barn at Points A. and D is equal to 4728 lb • ft,
determine the magnitude ofT;^ when Tab ~ 255 lb.
SOLUTION
The moment about the x axis due to the two cable forces can be found using the z components of each force
acting at the intersection with the xy plane (A and D). The x components of the forces are parallel to the x axis,
and the y components of the forces intersect the x axis. Therefore, neither the x or y components produce a
moment about the x axis.
We have
Where
£M,: iTAB )z {yA ) + {Tm )z {yD ) =Mx
(TAB )x =k-rAB
M^sV)k 306 lb
-i-12j + 12k
17
= 21.6 lb
(TDE ) -k T
0.5i-14j + 12k^k T,DE
8.5
0.648657 DE
>>A
yu
12 ft
14 ft
and
MA.=47281b-ft
(2 1 6 lb)(12 ft) + (0.648657^ )(14 It) = 4728 lb - ft
TDE = 235.211b or 7V„, = 235 lb ^/j/;
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334
'3 Hi
HMt^fiffi^v
PROBLEM 3.153
A wiring harness is made by routing either two or
three wires around 2-in.-diameter pegs mounted on a
sheet of plywood. If the force in each wire is 3 lb,
determine the resultant couple acting on the plywood
when a ~ 1 8 in. and (a) only wires AB and CD are in
place, (//) all three wires are in place.
SOLUTION
In general, M - "ZdF, where d is the perpendicular distance between the lines of action of the two forces
acting on a given wire.
(a)
A»
2.4 »*•
We have M = dABFAB+da>FCD
(2 + 24)in.x3 1b + |2 + -x28 in.x3 1b or M = 151.2 lb- in. ") <
(b)
tlM
Af
We have M ^idab fab+ dCDFCD \ + dEFFEF
= 151.2 lb-in.-28in.x3 lb or M = 67.2 lb -in. )<
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335
I
PROBLEM 3.154
A worker tries to move a rock by applying a 360-N force
to a steel bar as shown, (a) Replace that force with an
equivalent force-couple system at D. (b) Two workers
attempt to move the same rock by applying a vertical force
at A and another force at D. Determine these two forces if
they are to be equivalent to the single force of Part a.
SOLUTION
(a) We have
We have
where
LF: 360 N(-sin40°i - cos40°j) = -(23 1.40 N)i - (275.78 N)j = F
or F = 360N ^50°^
JMD : r/;//)xR =M
re/D = -[(0.65 m) cos 30°]i + [(0.65 m)sin 30°].)
= -(0.56292 m)i + (0.32500 m)j
i J k
-0.56292 0.32500
-231.40 -275.78
= [1 55.240 + 75.206)N-m]k
- (230.45 N m)k
M N-m
or M = 230N-mX
) A
(b) We have
where
£MD : M = r,/D xF,
'hid=40 -05 m)cos30°]i + [(1 .05 m)sin30°]j
= -(0.90933 m)i + (0.52500 m)j
i J k
FA = -0.90933 0.52500 N-m
-I
= [230.45 N-m]k
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336
PROBLEM 3.154 (Continued)
or (0.90933F, )k = 230.45k
FA = 253.42 N or F^=253NH
We have XF: F^F^+F^
-(231.40 N)i -(275.78 N)j = -(253.42 N^ + FoC-cos^i-sin^j)
From i: 23 1 .40 N = FD cos (9 (D
j: 22.36 fi = FD sm0 (2)
Equation (2) divided by Equation (1)
tan = 0.096629
(9 = 5.5193° or #-5.52°
Substitution into Equation (1)
FD =mM
= 232.48 N° cos 5.5 193°
or FD = 232N ^5.52°^
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337
1.10 N
PROBLEM 3.155
A 1 10-N force acting in a vertical plane parallel to theyz plane
is applied to the 220-mm-long horizontal handle AB of a
socket wrench. Replace the force with an equivalent force-
couple system at the origin O of the coordinate system.
SOLUTION
We have
where
We have
where
SF: PS =F
yB = 11 N[~ (sin 1 5°)j + (cos 1 5°)k]
= -(28.470 N)j + (1 06.252 N)k
YM \ %,xPB
or F = -(28.5 N)j + (1 06.3 N)k <
hio = [(°-22 cos 35 °)» + (°- ] 5)J - (0.22sin 35°)k]m
= (0. 1.802 1 3 m)i + (0. 1 5 m)j - (0. 126 1 87 m)k
i j k
0.180213 0.15 0.126187
-28.5 106.3
N • m = ML
M o [(12.3487)i-(19.1566)j-(5.1361)k]Nm
or M = (1 2.35 N • m)i - (1 9. 1 6 N • m)j - (5 . 1 3 N • m)k <
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338
fiO Ht PROBLEM 3.156
Four ropes are attached to a crate and exert the forces shown.
If the forces are to be replaced with a single equivalent force
applied at a point on line AB, determine (a) the equivalent
force and the distance from A to the point of application of
the force when a - 30°, (b) the value of a so that the single
equivalent force is applied at Point B.
SOLUTION
We have
(a) For equivalence
tOOIfes
\U3 >t>
* 1 i_ .. Ka "1^&
f-
XFV
: -100 cos 30° + 400coi 65° + 90 cos K° = RX
or Rx = 120.480 lb
LFy
: 1 00 sin a + 1 60 + 400 sin 65°+ 90 sin 65° - Ry
or Ry= (604.09 + 100 sin a) lb 0)
With a -30°
Then R = 4
54.09 lb
„ 654.09tan =
120.480J20.480)
2+(654.09)
2
= 665 lb or = 79.6°
Also 1,MA : (46 in.)(l 60 lb) + (66 in.)(400 lb) sin 65°
+(26 in.)(400 lb) cos 65° + (66 in.)(90 lb)sin65°
+(36 in.)(90 lb) cos 65° = </(654.09 lb)
or LMA =42,435 lb -in. and c/ = 64.9 in. R~= 665 lb ^79.6°^
and R is applied 64.9 in. To the right ofA. <(b) We have d — 66 in.
Then LMA : 42, 435 lb • in = (66 m.)Ry
or Ry= 642.95 lb
Using Eq. (I) 642,95 = 604.09 + 100sina or a = 22.9° ^
PROPRIETARY MATERIAL. (!) 2010 The McGraw-Hill Companies, Inc. All rights reserved. /Vo /«/// o//Aw Mww/ may be displayed,
reproduced or distributed in anyform or by any means, without the prior written permission of the publisher, or used beyond the limited
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339
200 mm
H 200 mm
PROBLEM 3.157
A blade held in a brace is used to tighten a screw at A.
(a) Determine the forces exerted at B and C, knowing
that these forces are equivalent to. a force-couple
system at A consisting of R = -(30 N)i + Rv \
+ Rzk
and Mj =-(l2N m)i. (b) Find the corresponding
values of Ryand Rz . (c) What is the orientation of the
slot in the head of the screw for which the blade is least
likely to slip when the brace is in the position shown?
SOLUTION
(") Equivalence requires IF; R = B +C
or -(30 N)i + Ryj+ Rzk = -j?k + (-Cri + C
y j+ C2k)
Equating the i coefficients i: -30N = -CV
or CV=30N
Also JMA : M*=r^xB + rc7,xC
-(l 2 N • m)i = [(0.2 m)i + (0. 1 5 m)j] x (~B)k
+(0.4 m)ix[-(30 N)i + C,,j + Czk]
Equating coefficients i: -12 N-m = -(0.15 m)/? or 5 = SON
k: = (0.4m)Cy
or Cy=0
j: = (0.2m)(80N)-(0.4m)C, or C,=40N
B = -(80.0N)k C = -(30.0 N)i + (40.0 N)k <
(b) Now we have for the equivalence of forces
-(30 N)i + RJ + Rzk = -(80 N)k + [(-30 N)i + (40 N)k]
Equating coefficients j: Ry=0 R
y^0 <
k : Rz= -80 + 40 or R
z= -40.0 N <
(c) First note that R = -(30 N)i -- (40 N)k. Thus, the screw is best able to resist the lateral force R.z
when the slot in the head of the screw is vertical. 4
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in anyform or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educatorspermittedby McGraw-Hillfor their individual course preparation. Ifyou are a student using this Manual,
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340
PROBLEM 3.158
A concrete foundation mat in the shape of a regular hexagon of side
12 ft supports four column loads as shown. Determine the magnitudes
of the additional loads that must be applied at B and F if the resultant
of all six loads is to pass through the center ofthe mat.
SOLUTION
From the statement ofthe problem it can be concluded that the six applied loads
R at 0. It then follows that
are equivalent to the resultant
£Mo =0 or 1MX= ZM
Z^Q
For the applied loads.
* J
f—E
/ \ 1K
\7t> x
—17. «L—
1
\
Then SM , = 0: (6>/3 &)FS + (6>/3 ft)(10 kips) - (6>/3 ft)(20 kips)
-(673 ft)/>=0
or FB -FF =10 (1)
SM, = 0: (1.2 ft)(15 kips) + (6 ft)FB -(6 ft)(10 kips)
-(12 ft)(30 kips) - (6 ft)(20 kips) + (6 ft)FF =
or Ffl+ 7^=60 (2)
Then (l) + (2)=> FB = 35.0 kips j 4
and Ff = 25.0 kips J -^
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. AH rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers andeducatorspermitted by McGraw-Hillfor their individual course preparation. Ifyou are a student using this Manual,you are using il without permission.
341