Vector Mechanics for Engineers Statics 7th - Cap 05

8/13/2019 Vector Mechanics for Engineers Statics 7th - Cap 05

1/166

PROBLEM 5.1

Locate the centroid of the plane area shown.

SOLUTION

2, inA , in.x , in.y 3, inxA 3, inyA

1 8 6 48 = 4 9 192 432

2 16 12 192 = 8 6 1536 1152

240 1344 1584

Then3

2

1344 in

240 in

xAX

A

= =

or 5.60 in.X =

and3

2

1584 in

240 in

yAY

A

= =

or 6.60 in.Y =


2/166

PROBLEM 5.2


SOLUTION

2, mmA , mmx , mmy 3, mmxA 3, mmyA

11

60 75 22502

= 40 25 90 000 56 250

2 105 75 7875 = 112.5 37.5 885 900 295 300

10 125 975 900 351 600

Then3

2

975 900 mm

10 125 mm

xAX

A

= =

or 96.4 mmX =

and3

2

351 600 mm

10 125 mm

yAY

A

= =

or 34.7 mmY =


3/166

PROBLEM 5.3


SOLUTION

For the area as a whole, it can be concluded by observation that

( )2

24 in.3

Y = or 16.00 in.Y =

2, inA , in.x 3, inxA

11

24 10 1202

= ( )2

10 6.6673

= 800

2 1

24 16 1922

= ( )1

10 16 15.3333

+ = 2944

312 3744

Then3

2

3744 in

312 in

xAX

A

= =

or 12.00 in.X =


4/166


5/166

PROBLEM 5.5


SOLUTION


1 120 200 24 000 = 60 120 1 440 000 2 880 000

2 ( )260 5654.9

2

= 94.5 120 534 600 678 600

18 345 905 400 2 201 400

Then3

2

905 400 mm

18 345 mm

xAX

A

= =

or 49.4 mmX =

and3

2

2 201 400 mm

18 345 mm

yAY

A

= =

or 93.8 mmY =


6/166

PROBLEM 5.6


SOLUTION

2, inA , in.x , in.y 3, inx A 3, iny A

1( )

29

63.6174

=

( )( )

4 93.8917

3

= 3.8917 243 243

2 ( )( )1

15 9 67.5

2

= 5 3 337.5 202.5

131.1 94.5 445.5

Then3

2

94.5 in

131.1 in

xAX

A

= =

or 0.721 in.X =

and3

2

445.5 in

131.1 in

yAY

A

= =

or 3.40 in.Y =


7/166

PROBLEM 5.7


SOLUTION

First note that symmetry implies X Y=

2, mmA , mmx 3, mmxA

1 40 40 1600 = 20 32 000

2

2(40)1257

4

= 16.98 21 330

343 10 667

Then3

2

10 667 mm

343 mm

xAX

A

= =

or 31.1 mmX =

and 31.1 mmY X= =


8/166

PROBLEM 5.8


SOLUTION

First note that symmetry implies 0X =

2, inA , in.y 3, inyA

1( )

24

25.132

= 1.6977 42.67

2 ( )

26

56.552

= 2.546 144

31.42 101.33

Then3

2

101.33 in

31.42 in

yAY

A

= =

or 3.23 in.Y =


9/166

PROBLEM 5.9

For the area of Problem 5.8,determine the ratio 2 1/r r so that 13 /4.y r=

SOLUTION

A y y A

12

12

r

14

3

r

31

2

3r

22

22

r

24

3

r

32

2

3r

( )2 22 12 r r

( )3 32 12

3 r r

Then Y A y A =

or ( ) ( )2 2 3 31 2 1 2 13 2

4 2 3r r r r r

=

2 3

2 2

1 1

91 1

16

r r

r r

=

Let2

1

r

p r=

[ ] 29

( 1)( 1) ( 1)( 1)16

p p p p p

+ = + +

or 216 (16 9 ) (16 9 ) 0p p + + =


10/166

PROBLEM 5.9 CONTINUED

Then2(16 9 ) (16 9 ) 4(16)(16 9 )

2(16)

p

=

or 0.5726 1.3397p p= =

Taking the positive root 2

1

1.340r

r=


11/166

PROBLEM 5.10

Show that as 1r approaches 2,r the location of the centroid approaches that

of a circular arc of radius ( )1 2 / 2.r r+

SOLUTION

First, determine the location of the centroid.

From Fig. 5.8A:( )

( ) ( )2 22 2 2 22

2

sin2

3y r A r

= =

( )2 2

2 cos

3r

=

Similarly( )

( ) 21 1 1 122

2 cos

3y r A r

= =

( ) ( ) ( ) ( )

( )

2 22 2 1 12 2

2 2

3 32 1

2 cos 2 cosThen3 3

2cos

3

y A r r r r

r r

=

=

( )

2 22 1

2 22 1

and2 2

2

A r r

r r

=

=

( ) ( )2 2 3 32 1 2 13 32 12 2

2 1 2

Now

2 cos

2 3

2 cos

3

Y A yA

Y r r r r

r rY

r r

=

=

=


12/166


Using Figure 5.8B, Y of an arc of radius ( )1 21

is2

r r+

( ) ( )( )

21 2

2

sin1

2Y r r

= +

( )1 2 21 cos

( )2

r r

= +

(1)

( )( )( )( )

2 23 32 1 2 1 2 12 1

2 22 1 2 12 1

2 22 1 2 1

2 1

Nowr r r r r r r r

r r r r r r

r r r r

r r

+ +=

+

+ +=

+

2

1

Let r r

r r

= + =

Then ( )1 21

2

r r r= +

( ) ( ) ( ) ( )( ) ( )

2 23 32 12 2

2 12 2

and

3

2

r r r r r r

r rr r

r

r

+ + + + =

+ +

+ =

1 2In the limit as 0 (i.e., ), thenr r =

3 32 12 2

2 1

1 2

3

2

3 1( )

2 2

r rr

r r

r r

=

= +

so that ( )1 22

2 3 cos3 4

Y r r

= +

or ( )1 22

1 cos2

Y r r

= +

Which agrees with Eq. (1).


13/166

PROBLEM 5.11


SOLUTION

First note that symmetry implies 0X =

2 2 2 in., 45r = =

( ) ( )( )

4

2

4

2 2 2 sin2 sin1.6977 in.

3 3

ry

= = =

2

, inA , in.y 3

, iny A

1 ( ) ( )1

4 3 62

= 1 6

2 ( )2

2 2 6.2834

= 2 0.3024y = 1.8997

3 ( ) ( )1

4 2 42

= 0.6667 2.667

8.283 5.2330

Then Y A yA =

( )2 38.283 in 5.2330 inY = or 0.632 in.Y =


14/166

PROBLEM 5.12


SOLUTION


1 (40)(90) 3600= 15 20 54 000 72 000

2 ( ) ( )40 60

21214

= 10 15 6750 10 125

3 ( ) ( )1

30 45 6752

= 25.47 19.099 54 000 40 500

6396 101 250 21 375

Then XA xA=

( )2 36396 mm 101 250 mmX = or 15.83 mmX = and YA yA=

( )2 36396 mm 21 375 mmY = or 3.34 mmY =


15/166

PROBLEM 5.13


SOLUTION


1 ( ) ( )2

40 80 21333

= 48 15 102 400 32 000

2 ( ) ( )1

40 80 16002

= 53.33 13.333 85 330 21 330

533.3 17 067 10 667

Then X A XA =

( )2 3533.3 mm 17 067 mmX = or 32.0 mmX = and Y A yA =

( )2 3533.3 mm 10 667 mmY = or 20.0 mmY =


16/166

PROBLEM 5.14


SOLUTION


1 ( )( )2

150 240 24 0003

= 56.25 96 1 350 000 2 304 000

2 ( ) ( )1

150 120 90002

= 50 40 450 000 360 000

15 000 900 000 1 944 000

Then X A xA =

( )2 315 000 mm 900 000 mmX = or 60.0 mmX =

and Y A yA =

( )215 000 mm 1 944 000Y = or 129.6 mmY =


17/166


18/166

PROBLEM 5.16


SOLUTION

2, inA , in.x , in.y 3, inxA 3, inyA

1 ( ) ( )2

8 8 42.673 = 3 2.8 128 119.47

2 ( ) ( )2

4 2 5.3333

= 1.5 0.8 8 4.267

37.33 120 123.73

Then X A x A =

( )2 337.33 in 120 inX = or 3.21 in.X = and Y A y A =

( )2 337.33 in 123.73 inY = or 3.31 in.Y =


19/166

PROBLEM 5.17

The horizontal xaxis is drawn through the centroid Cof the area shownand divides the area into two component areas A1 and A2. Determinethe first moment of each component area with respect to the xaxis, andexplain the results obtained.

SOLUTION

Note that xQ yA=

Then ( ) 21

5 1m 6 5 m

3 2xQ

=

or ( ) 3 31

25.0 10 mmxQ =

and ( ) 2 22

2 1 1 12.5 m 9 2.5 m 2.5 m 6 2.5 m

3 2 3 2xQ

= +

or ( ) 3 32

25.0 10 mmxQ = Now ( ) ( )

1 20x x xQ Q Q= + =

This result is expected since x is a centroidal axis ( )thus 0y =

and ( )0 0x xQ y A Y A y Q= = = =


20/166

PROBLEM 5.18

The horizontal xaxis is drawn through the centroid Cof the area shown

and divides the area into two component areas A1 and A2. Determinethe first moment of each component area with respect to the xaxis, and

explain theresults obtained.

SOLUTION

First, locate the position y of the figure.

2, mmA , mmy 3, mmyA

1 160 300 48 000 = 150 7 200 000

2 150 80 16 000 = 160 2 560 000

32 000 4 640 000

Then Y A y A =

( )2 3

32 000 mm 4 640 000 mmY =

or 145.0 mmY =


21/166


( )

I I

6 3

:

155 115 (160 155) 80 115

2 2 1.393 10 mm

A Q yA=

= +

=

( ) ( )

II II

6 3

:

145 85 160 145 80 85

2 2

1.393 10 mm

A Q yA=

=

=

( )area I IIQ 0x Q Q = + = Which is expected since xQ yA yA= = and 0y = ,since x is a centroidal axis.


22/166

PROBLEM 5.19

The first moment of the shaded area with respect to the xaxis is denoted

by .xQ (a) Express xQ in terms of rand . (b) For what value of is

xQ maximum, and what is the maximum value?

SOLUTION

( ) With and using Fig. 5.8 A,xa Q yA=

( )( ) ( )

( )

23 2 2 2

2 32

3 2

sin 1sin 2 cos sin

2

2cos cos sin

3

x

rQ r r r r

r

=

=

or 3 32

cos3

xQ r =

( )b By observation, is maximum whenxQ 0 =

and then 32

3xQ r=


23/166

PROBLEM 5.20

A composite beam is constructed by bolting four plates to four

2 2 3/8-in. angles as shown. The bolts are equally spaced along thebeam, and the beam supports a vertical load. As proved in mechanics of

materials, the shearing forces exerted on the bolts at A and B areproportional to the first moments with respect to the centroidal xaxis of

the red shaded areas shown, respectively, in parts a and bof the figure.Knowing that the force exerted on the bolt at A is 70 lb, determine theforce exerted on the bolt at B.

SOLUTION

From the problem statement: xF Q so that

( ) ( )A B

x xA B

F F

Q Q=

and ( )( )x BB Ax A

QF F

Q=

Now xQ yA=

So ( ) ( ) 30.375

7.5 in. in. 10 in. 0.375 in. 28.82 in2

x AQ

= + =

( ) ( ) ( )( )

( ) ( )( )

0.375and 2 7.5 in. in. 1.625 in. 0.375 in.

2

2 7.5 in. 1 in. 2 in. 0.375 in.

x xB AQ Q

= +

+

3 3 328.82 in 8.921 in 9.75 in= + +

347.49 in=

Then ( )3

3

47.49 in70 lb 115.3 lb

28.82 inBF = =


24/166

PROBLEM 5.21

A thin, homogeneous wire is bent to form the perimeter of the figure

indicated. Locate the center of gravity of the wire figure thus formed.

SOLUTIONFirst note that because the wire is homogeneous, its center of gravity will

coincide with the centroid of the corresponding line.

, in.L , in.x , in.y 2, inxL 2, inyL

1 16 8 0 128 0

2 12 16 6 102 72

3 24 4 12 96 288

4 6 8 9 48 54

5 8 4 6 32 48

6 6 0 3 0 18

72 336 480

Then X L x L =

( ) 272 in. 336 inX = or 4.67 in.X =

and Y L y L =

2(72 in.) 480 inY = or 6.67 in.Y =


25/166

PROBLEM 5.22



SOLUTION

First note that because the wire is homogeneous, its center of gravity will coincide with the centroid of the

corresponding line.

, mmL , mmx , mmy 2, mmxL 2, mmyL

1 165 82.5 0 13 612 0

2 75 165 37.5 12 375 2812

3 105 112.5 75 11 812 7875

4

2 2

60 75 96.05+ = 30 37.5 2881 3602 441.05 40 680 14 289

Then X L x L =

2(441.05 mm) 40 680 mmX = or 92.2 mmX =

and Y L y L =

2(441.05 mm) 14 289 mmY = 32.4 mmY =


26/166


27/166

PROBLEM 5.24



SOLUTIONFirst note that because the wire is homogeneous, its center of gravity will

coincide with the centroid of the corresponding line.

By symmetry 0X =

, in.L , in.y 2, inyL

1 2 0 0

2 ( )6 ( )2 6

3.820

= 72

3 2 0 0

4 ( )4 ( )2 4

2.546

= 32

35.416 104

Then Y L y L =

2(35.416 in.) 104 inY = or 2.94 in.Y =


28/166

PROBLEM 5.25

A 750 g= uniform steel rod is bent into a circular arc of radius 500 mmas shown. The rod is supported by a pin at Aand the cord BC. Determine(a) the tension in the cord, (b) the reaction at A.

SOLUTION

( )0.5 m sin30First note, from Figure 5.8B:

/6X

=

1.5m

=

( )( )2Then mg

0.75 kg 9.81 m/s

7.358 N

W =

=

=

Also note that ABDis an equilateral triangle.Equilibrium then requires

( ) ( )

( ) 0:

1.5 0.5 m m cos30 7.358 N 0.5 m sin 60 0

A

BC

a M

T

=

=

or 1.4698 NBCT = or 1.470 NBCT =

( )( ) 0: 1.4698 N cos60 0x xb F A = + =

or 0.7349 NxA =

( )0: 7.358 N 1.4698 N sin 60 0y yF A = + =

or 6.085 NyA = thus 6.13 N=A 83.1


29/166

PROBLEM 5.26

The homogeneous wire ABCDis bent as shown and is supported by a pin

at B. Knowing that 8 in.,l = determine the angle for which portionBCof the wire is horizontal.

SOLUTIONFirst note that for equilibrium, the center of gravity of the wire must lie

on a vertical line through B. Further, because the wire is homogeneous,

its center of gravity will coincide with the centroid of the corresponding

line.

Thus 0, which implies that 0BM x = = or 0xL =

Hence

( ) ( )2(6 in.) 8 in.

6 in. 8 in.2

+

( )6 in.

8 in. cos 6 in. 02

+ =

Then4

cos9

= or 63.6=


30/166

PROBLEM 5.27

The homogeneous wire ABCDis bent as shown and is supported by a pin

at B. Knowing that 30 , = determine the length lfor which portion CDof the wire is horizontal.

SOLUTIONFirst note that for equilibrium, the center of gravity of the wire must lie on

a vertical line through B. Further, because the wire is homogeneous, its

center of gravity will coincide with the centroid of the corresponding line.

Thus 0, which implies that 0BM x = = or 0i ix L =

Hence( )

( ) ( )2 6 in.

cos30 6 in. sin 30 6 in.

+

( )( )

in.cos30 in.

2

ll

+

( ) ( )6 in.

in. cos30 6 in. 0

2

l

+ =

or 2 12.0 316.16 0l l+ =

1with roots 12.77 and 24.77.l =

Taking the positive root

12.77 in.l =


31/166


32/166

PROBLEM 5.29

Determine the distance hso that the centroid of the shaded area is as close

to line BB as possible when (a) 0.2,k = (b) 0.6.k =

SOLUTION

ThenyA

yA

=

( ) ( )

( )

( )

2 2or

a haab kb a h

yba kb a h

+

=

( )2 2112 (1 )

a k kh

a k kh

+=

+

Let 1 andh

c ka

= =

Then

2

2

a c k

y c k

+

= + (1)

Now find a value of (or h) for which y is minimum:

( ) ( )( )

2

2

20

2

k c k k c k dy a

d c k

+ += =

+ or ( ) ( )22 0c k c k + + = (2)


33/166


Expanding (2) 2 22 2 0c c k + = or 2 2 0k c c + =

Then( ) ( ) ( )

22 2 4

2

c c k c

k

=

Taking the positive root, since 0h > (hence 0> )

( ) ( ) ( )2

2 1 4 1 4 1

2

k k k k h a

k

+ + =

(a) 0.2:k = ( ) ( ) ( )( )

( )

22 1 0.2 4 1 0.2 4 0.2 1 0.2

2 0.2h a

+ + = or 0.472h a=

(b) 0.6:k = ( ) ( ) ( )( )

( )

22 1 0.6 4 1 0.6 4 0.6 1 0.6

2 0.6h a

+ + = or 0.387h a=


34/166

PROBLEM 5.30

Show when the distance his selected to minimize the distance y fromline BB to the centroid of the shaded area that .y h=

SOLUTION

From Problem 5.29, note that Eq. (2) yields the value of that minimizes h.

Then from Eq. (2)

We see

2

2 c k

c k

+

= + (3)

Then, replacing the right-hand side of (1) by 2, from Eq. (3)

We obtain ( )22

ay =

Buth

a=

So y h= Q.E.D.


35/166

PROBLEM 5.31

Determine by direct integration the centroid of the area shown. Express

your answer in terms of aand h.

SOLUTION

For the element of area (EL) shownh

y xa

= ( )and

1

dA h y dx

xh dx

a

=

=

( )

Then

1

2

12

EL

EL

x x

y h y

h x

a

=

= +

= +

2

0

0

1Then area 1

2 2

a

a x xA dA h dx h x ah

a a

= = = =

2 32

0 0

2 2

20 0

2 32

2

0

1and 1

2 3 6

1 1 12 2

1

2 33

a

a

EL

a a

EL

a

x x xx dA x h dx h a h

a a

h x x h xy dA h dx dx

a a a

h xx ah

a

= = =

= + =

= =

2

Hence

1 1

2 6

ELxA x dA

x ah a h

=

=

13

x a=

21 1

2 3

ELyA y dA

y ah ah

=

=

2

3y h=


36/166

PROBLEM 5.32



SOLUTIONFor the element (EL) shownAt 3, :x a y h h ka= = = or

3

hk

a=

Then 1/31/3ax yh=

1/3

1/3

Now dA xdy

ay dy

h

=

=

1/3

1/ 3

1 1,

2 2EL EL

ax x y y y

h= = =

Then ( )1/3 4/31/3 1/300

3 3

4 4

hh a a

A dA y dy y ahh h

= = = =

1/3 1/3 5/3 2

1/3 1/3 2/300

1 1 3 3and

2 2 5 10

hh

EL

a a ax dA y y dy y a h

h h h

= = =

1/3 7/3 2

1/3 1/300

3 3

7 7

hh

EL

a ay dA y y dy y ah

h h

= = =

Hence 23 3

:4 10

ELxA x dA x ah a h

= =

25

x a= 23 3:

4 7ELyA y dA y ah ah

= =

4

7y h=


37/166

PROBLEM 5.33



SOLUTIONFor the element (EL) shownAt 31, :x a y h h k a= = = or 1 3hk a=

32a k h= or 2 3ak h=

Hence, on line 13

3

hy x

a=

and on line 2 1/3

1/3

hy x

a=

Then 1/3 3 1/3 3

1/3 3 1/3 3

1 and

2EL

h h h hdA x x dx y x x

a a a a

= = +

1/3 3 4/3 4

1/3 3 1/3 300

3 1 1

24 4

aa h h

A dA x x dx h x x aha a a a

= = = =

1/3 3 7/3 5 2

1/3 3 1/3 300

3 1 8

357 5

aa

EL

h hx dA x x x dx h x x a h

a a a a

= = =

1/3 3 1/3 3

1/3 3 1/3 30

1

2

a

EL

h h h hy dA x x x x dx

a a a a

= +

2 2/3 6 2 5/3 62

2/3 6 5/3 60

0

3 1 8

2 2 5 7 35

a

ah x x h x xdx ah

a a a a

= = =

From 28

:2 35

EL

ahxA x dA x a h

= =

or 16

35x a=

and 28

:2 35

EL

ahyA y dA y ah

= =

or 16

35y h=


38/166

PROBLEM 5.34

Determine by direct integration the centroid of the area shown.

SOLUTIONFirst note that symmetry implies 0x =

For the element (EL) shown2

(Figure 5.8B)ELr

y

dA r d r

=

=

Then ( )2

2

1

1

22 2

2 12 2

r

r

r

r

rA dA r d r r r

= = = =

and ( ) ( )2

2

1

1

3 3 32 1

2 1 22

3 3

rr

EL rr

ry dA rd r r r r

= = =

So ( ) ( )2 2 3 32 1 2 12

:2 3

ELyA y dA y r r r r

= =

or

3 32 1

2 22 1

4

3

r r

y r r

=


39/166

PROBLEM 5.35


SOLUTIONFirst note that symmetry implies 0x =

For the element (EL) showncos , siny R x R = =

cosdx R d = 2 2cosdA ydx R d = =

Hence

( )2 2 2 20

0

sin 2 12 cos 2 2 sin 2

2 4 2A dA R d R R

= = = + =

( )

( )

2 2 3 2

00

32

1 22 cos cos cos sin sin

2 3 3

cos sin 2sin3

EL

Ry dA R d R

R

= = +

= +

But soELyA y dA= ( )( )

32

2

cos sin 2sin3

2 sin 22

R

yR

+=

+

or( )

( )

2cos 22sin

3 2 sin 2y R

+=

+

Alternatively,22 3 sin

sin3 2 sin 2

y R

=

+


40/166

PROBLEM 5.36


SOLUTIONFor the element (EL) shown

2 2by a xa

= ( )

( )2 2

and dA b y dx

ba a x dx

a

=

=

( ) ( )2 21;2 2

EL EL

bx x y y b a a x

a= = + = +

Then ( )2 20abA dA a a x dxa= = To integrate, let 2 2sin : a cos , cosx a x a dx a d = = =

( )( )/2

0

/2

2 2

0

Then cos cos

2sin sin 12 4 4

bA a a a d

a

b a a aba

=

= + =

( ) ( )/2

3/22 2 2 2 2

0

0

3

1and

2 3

1

6

a

EL

b b ax dA x a a x dx x a x

a a

a b

= = +

=

( ) ( )

( )

2 2 2 2

0

2 2 32 2

2 20

0

2

1

3 62 2

a

EL

a

a

b by dA a a x a a x dx

a a

b b xx dx ab

a a

= +

= = =

21: 14 6

ELxA x dA x ab a b

= =

or ( )2

3 4

ax

=

21: 14 6

ELyA y dA y ab ab

= =

or ( )2

3 4

by

=


41/166

PROBLEM 5.37


your answer in terms of aand b.

SOLUTIONFor the element (EL) shown on line 1 at

22,x a b k a= = or 2 2

bk

a=

2

2

by x

a =

On line 2 at3

1, 2x a b k a= = or 2 32b

k a

= 3

3

2

by x

a

=

2 3

2 3

2b bdA x x dx

a a

= +

3 3 4

2

2 20

0

2 2Then

3 4

1 1 5

3 2 6

a

a b x b x xA dA x dx

x aa a

ab ab

= = + = +

= + =

4 52 3 2

2 3 20

0

2

2 2 1 2and

4 5 4 5

13

20

a

a

EL

b b b x xx dA x x x dx a b

aa a a

a b

= + = + = +

=

2 3 2 3

2 3 2 30

2 2 2 52 3 7

2 3 4 20

0

2 5 2

1 2 2

2

1 2 2

2 52 7

1 2 13

10 7 70

a

EL

a

a

b b b by dA x x x x dx

a a a a

b b b xx x dx x

a a a a

b a ab

= +

= =

= =

Then 25 13 :6 20

ELxA x dA x ab a b

= =

or 3950

x a= 25 13:

6 70ELyA y dA y ab ab

=

or 39

175y b=


42/166

PROBLEM 5.38



SOLUTIONAt 0,x y b= =

( )2

0b k a= or2

bk

a=

Then ( )22

b

y x aa

= Now ( )

2

2,

2 2EL EL

y bx x y x a

a= = =

and ( )2

2

bdA ydx x a dx

a= =

( ) ( )2 3

2 200

1Then

33

aa b b

A dA x a dx x a aba a

= = = =

( ) ( )

( ) ( ) ( )

2 3 2 2

2 20 0

4 23 2 2

2

22 2 5

2 2 400

2

and 2

2 1

4 3 2 12

1

52 2

1

10

a a

EL

aa

EL

b bx dA x x a dx x ax a x dx

a a

b x a

ax x a ba

b b by dA x a x a dx x a

a a a

ab

= = +

= + =

= =

=

Hence 21 1

:3 12

ELxA x dA x ab a b

= =

14

x a= 21 1:


= =

3

10y b=


43/166

PROBLEM 5.39


SOLUTION2

2

2

2

Have

1 1

2 2

1

EL

EL

x x

a x xy y

L L

x xdA ydx a dx

L L

=

= = +

= = +

Then

22 2 32

2 20

0

81

2 33

LL x x x x

A dA a dx a x aLL LL L

= = + = + =

22 2 3 4

2

2 20

0

2

2 22

2 20

2 2 3 4

2 3 40

2 2 3 4

2 3

and 12 3 4

10

3

1 12

1 2 3 22

2 2

L

L

EL

L

EL

EL

x x x x xx dA x a dx a

L LL L

aL

a x x x xy dA a dx

L LL L

a x x x xdx

L L L L

a x x xx

L L L

= + = +

=

= + +

= + +

= + +

25

2

4

0

11

55

Lx

a LL

=

Hence 28 10

:3 3

ELxA x dA x aL aL

= =

54

x L= 21 11:

8 5ELyA y dA y a a

= =

33

40y a=


44/166

PROBLEM 5.40



SOLUTIONFor 1 at , 2y x a y b= = 22b ka= or 22bk a= Then 21 2

2by x

a=

By observation ( )2 2 2b x

y x b ba a

= + = Now

and for 0 :

ELx x

x a

=

2 21 12 2

1 2 and

2EL

b by y x dA y dx x dx

a a= = = =

For 2 :a x a 2 2

12 and 2

2 2EL

b x xy y dA y dx b dx

a a

= = = =

22

20

223

2

0 0

2Then 2

2 72

3 2 6

a a

a

aa

b xA dA x dx b dx

aa

b x a xb ab

aa

= = +

= + =

( ) ( ){ ( ) ( )

22

20

24 3

2

2

0 0

2 2 32 2

2

2and 2

2

4 3

1 12 2

2 3

7

6

a a

EL a

a a

b xx dA x x dx x b dx

aa

b x xb x

aa

a b b a a a aa

a b

= +

= +

= + +

=


45/166


22 2

2 20 0

232 5 2

4 0

2

22 2

2

22

5 2 3

17

30

a a

EL

aa

a

b b b x xy dA x x dx b dx

a aa a

b x b a x

aa

ab

= +

= +

=

Hence 27 7

:6 6

ELxA x dA x ab a b

= =

x a= 27 17:


= =

17

35y b=


46/166

PROBLEM 5.41



SOLUTION2For y at , :x a y b= =

2a kb= or2

ak

b=

Then 1/22b

y xa

=

1/2 1/22

2

Now

and for 0 : ,2 2 2

EL

EL

x x

a y b x xx y dA y dx b dx

a a

=

= = = =

For ( )1/2

1 2

1 1:

2 2 2 2EL

a b x xx a y y y

a a

= + = +

( )1/2

2 1

1

2

x xdA y y dx b dx

aa

= = +

( )

( ) ( )

1/2 1/2/2

0 /2

/2 3/2 23/2

0 /2

3/2 3/23/2

22

1

Then 2

2 2 1

3 3 2 2

2

3 2 2

1 1

2 2 2 2

a a

a

aa

a

x x x

A dA b dx b dxaa a

b x xx b x

aa a

b a aa

a

a ab a a

a

= = + +

= + +

= +

+ +

1324

ab

=


47/166


( )

( ) ( )

1/2 1/2/2

0 /2

/2 5/2 3 45/2

0 /2

5/2 5/25/2

33 2

1and

2

2 2

5 5 3 4

2

5 2 2

1 1

3 2 4

a a

EL a

aa

a

x x xx dA x b dx x b dx

aa a

b x x xx b

aa a

b a aa

a

ab a a

a

= + +

= + +

= +

+ +

2

2

1/2 1/2/2

0

1/2 1/2

/2

/2 32 2 22

0/2

2

71

240

2

1 1

2 2 2

1 1 1

2 2 2 2 3 2

4

a

EL

a

a

aa

a

a

a b

b x xy dA b dx

a a

b x x x xb dx

a aa a

b b x xx

a a a a

b

=

=

+ + +

= +

=

( )2 2 32

2

2

1

2 2 6 2 2

11

48

a a b aa

a a

ab

+

=

Hence 213 71

:24 240

ELxA x dA x ab a b

= =

17

0.546130

x a a= =

213 11:24 48

ELyA y dA y ab ab

= =

11

0.42326

y b b= =


48/166

PROBLEM 5.42

A homogeneous wire is bent into the shape shown. Determine by direct

integration the xcoordinate of its centroid. Express your answer in termsof a.

SOLUTIONFirst note that because the wire is homogeneous, its center of gravity

coincides with the centroid of the corresponding line

Have at 2, :x a y a a ka= = = or1

ka

=

Thus 21 2 andy x dy xdxa a

= =

( )

2 2

2 22

2 2 20

0

2

2

2Then 1 1

4 4 2 4 1 1 ln 1

2 4

5 ln 2 5 1.47892 4

4 1

a

a

EL

dydL dx x dx

dx a

x x a xL dL x dx x

aa a a

a aa

xx dL x dxa

= + = +

= = + = + + + +

= + + =

= +

( )

3/22 2

20

0

23/2 2

2 413 8

5 1 0.848412

a

a a xa

aa

= +

= =

Then ( ) 2: 1.4789 0.8484ELxL x dL x a a= = 0.574x a=


49/166

PROBLEM 5.43


integration the xcoordinate of its centroid.



Now cosELx r = and dL rd =

Then [ ]7 /47 /4

/4/4

3

2

L dL rd r r

= = = =

and ( )7 /4

/4cosELx dL r rd

=

[ ]7 /42 2 2

/4

1 1sin 2

2 2r r r

= = =

23Thus : 22

xL xdL x r r

= =

2 2

3x r

=


50/166

PROBLEM 5.44


integration the xcoordinate of its centroid.



Now 3cosELx a = and2 2dL dx dy= +

Where 3 2cos : 3 cos sinx a dx a d = =

3 2sin : 3 sin cosy a dy a d = =

( ) ( )

( )

1/22 2

2 2

1/22 2

/ 2

0

Then 3 cos sin 3 sin cos

3 cos sin cos sin

3 cos sin

1 3 cos sin 3 sin

2

dL a d a d

a d

a d

L dL a d a

= +

= +

=

= = =

/22

0

3

2a

=

( )/ 2 3

0

/22 5 2

0

and cos 3 cos sin

1 33 cos

5 5

ELx dL a a d

a a

=

= =

Hence 23 3

:2 5

ELxL x dL x a a

= =

2

5x a=


51/166

PROBLEM 5.44 CONTINUEDAlternative solution

2/33 2

2/3

3 2

cos cos

sin sin

xx a

a

yy a a

= =

= =

( )

( ) ( )

2/3 2/33/2

2/3 2/3

1/22/3 2/3 1/3

1 or

Then

x yy a x

a a

dya x x

dx

+ = =

=

( ) ( )1/22 2

1/22/3 2/3 1/3

Now

and 1 1

ELx x

dydL dx a x x dx

dx

=

= + = +

1/31/3 2/3

1/ 300

3 3Then

2 2

aa a

L dL dx a x ax

= = = =

and1/3

1/3 5/3 2

1/300

3 3

5 5

aa

EL

ax dL x dx a x a

x

= = =

Hence 23 3

:2 5

ELxL x dL x a a

= =

2

5x a=


52/166

PROBLEM 5.45


SOLUTION2 2

Have cos cos3 3

2 2sin sin

3 3

EL

EL

x r ae

y r ae

= =

= =

( )( ) 2 21 1

and2 2

dA r rd a e d = =

Then

( )2 2 2 2 2 2 200

1 1 1 11 133.623

2 2 2 4A dA a e d a e a e a

= = = = =

and 2 2 3 30 0

2 1 1cos cos

3 2 3ELx dA ae a e d a e d

= =

To proceed, use integration by parts, with

3 3 and 3

cos and sin

u e du e d

dv d v

= =

= =

Then ( )3 3 3cos sin sin 3e d e e d =

Now let 3 3 then 3u e du e d = =

sin , then cosdv d v = =

Then ( )( )3 3 3 3sin sin 3 cos cos 3e d e e e d =

So that ( )3

3 cos sin 3cos10

ee d

= +

( ) ( )3 3

3 3 3

0

1 sin 3cos 3 3 1239.26

3 10 30EL

e ax dA a e a

= + = =

Also 2 2 3 30 0

2 1 1sin sin

3 2 3ELy dA ae a e d a e d

= =


53/166

PROBLEM 5.45 CONTINUEDUsing integration by parts, as above, with

3 3 and 3u e du e d = =

sin and cosdv d v = =

Then ( ) ( )3 3 3sin cos cos 3e d e e d =

So that ( )3

3 sin cos 3sin10

ee d

= +

( ) ( )3 3

3 3 3

0

1 cos 3sin 1 413.09

3 10 30EL

e ay dA a e a

= + = + =

Hence ( )2 3: 133.623 1239.26ELxA x dA x a a= = or 9.27x a=

( )2 3: 133.623 413.09

ELyA y dA y a a= =

or 3.09y a=


54/166


55/166


56/166

PROBLEM 5.47

Determine the volume and the surface area of the solid obtained byrotating the area of Problem. 5.2 about (a) the x axis, (b) the line

165x = mm.

SOLUTION

From the solution to Problem 5.2:

( )2 area area10 125 mm , 96.4 mm, 34.7 mm AreaA X Y= = =


( )line line441.05 mm 92.2 mm, 32.4 mm LineL X Y= = =

Applying the theorems of Pappus-Guldinus, we have

(a) Rotation about thexaxis:( )( ) 3 2lineArea 2 2 32.4 mm 441.05 mm 89.786 10 mmY L = = =

3 289.8 10 mmA=

( )( ) 6 3areaVolume 2 2 34.7 mm 10125 mm 2.2075 10 mmY A = = =

6 32.21 10 mmV =

(b) Rotation about 165 mm:x =

( ) ( ) ( )5 2

lineArea 2 165 2 165 92.2 mm 441.05 mm 2.01774 10 mmX L = = =

6 20.202 10 mmA =

( ) ( ) ( ) 6 3areaVolume 2 165 2 165 96.4 mm 10 125 mm 4.3641 10 mmX A = = = 6 34.36 10 mmV =


57/166

PROBLEM 5.48

Determine the volume and the surface area of the solid obtained byrotating the area of Problem 5.4 about (a) the line 22y= mm, (b) theline 12x= mm.

SOLUTION


2area area399 mm , 1.421 mm, 12.42 mm (Area)A X Y= = =


line line77.233 mm, 1.441 mm, 12.72 mm (Line)L X Y= = =


(a) Rotation about the line 22 mm:y =

( ) ( ) ( ) 2lineArea 2 22 2 22 12.72 mm 77.233 mm 4503 mmY L = = = 3 24.50 10 mmA =

( ) ( ) ( )2 3areaVolume 2 22 2 22 12.42 mm 399 mm 24 016.97 mmY A = = = 3 324.0 10 mmV =

(b) Rotation about line 12 mm:x =

( ) ( ) ( ) 2lineArea 2 12 2 12 1.441 mm 77.233 mm 5124.45 mmX L = = = 3 25.12 10 mmA =

( ) ( ) ( )2 3Volume 2 12 1.421 2 12 1.421 mm 399 mm 26 521.46 mmA = = = 3 326.5 10 mmV =


58/166

PROBLEM 5.49

Determine the volume and the surface area of the solid obtained byrotating the area of Problem 5.1 about (a) the x axis, (b) the line

16 in.x =

SOLUTION


2area area240 in , 5.60 in., 6.60 in. (Area)A X Y= = =


line line72 in., 4.67 in., 6.67 in.L X Y= = =


( ) Rotation about the axis:a x

( )( ) 2line2 2 6.67 in. 72 in. 3017.4 inxA Y L = = =

23020 inA=

( )( )2 3area2 2 6.60 in. 240 in 9952.6 inxV Y A = = = 39950 inV =

( ) Rotation about 16 in.:b x =

( ) ( ) ( ) 216 line2 16 2 16 4.67 in. 72 in. 5125.6 inxA X L = = = = 2

16 5130 inxA = =

( ) ( ) ( )2 316 area2 16 2 16 5.60 in. 240 in 15 682.8 inxV X A = = = = 3 3

16 15.68 10 inxV = =


59/166

PROBLEM 5.50

Determine the volume of the solid generated by rotating the semielliptical

area shown about (a) the axis ,AA (b) the axis ,BB (c) theyaxis.

SOLUTION

Applying the second theorem of Pappus-Guldinus, we have

(a) Rotation about axis :AA

( ) 2 2Volume 2 22

abyA a a b

= = =

2 2V a b=

(b) Rotation about axis :BB

( ) 2 2Volume 2 2 2 22

abyA a a b

= = =

2 22V a b=

(c) Rotation abouty-axis:

24 2Volume 2 23 2 3

a abyA a b

= = =

22

3V a b=


60/166

PROBLEM 5.51

Determine the volume and the surface area of the chain link shown,

which is made from a 2-in.-diameter bar, if 3R= in. and 10L= in.

SOLUTION

First note that the areaAand the circumference Cof the cross section of the bar are

2 and4

A d C d

= =

Observe that the semicircular ends of the link can be obtained by rotating the cross section through a horizontal

semicircular arc of radiusR. Then, applying the theorems of Pappus-Guldinus, we have

( ) ( ) ( ) ( ) ( )side endVolume 2 2 2 2 2V V AL RA L R A = + = + = +

( ) ( )2

3

2 10 in. 3 in. 2 in.4

122.049 in

= +

=

3122.0 inV =

( ) ( ) ( ) ( ) ( )side endArea 2 2 2 2 2A A CL RC L R C = + = + = +

( ) ( )2 10 in. 3 in. 4 in. = +

2

488.198 in= 2488 inA =


61/166

PROBLEM 5.52

Verify that the expressions for the volumes of the first four shapes in

Figure 5.21 on page 261 are correct.

SOLUTION

Following the second theorem of Pappus-Guldinus, in each case aspecific generating areaAwill be rotated about thexaxis to produce thegiven shape. Values of y are from Fig. 5.8A.

(1) Hemisphere: the generating area is a quarter circle

Have 242 2

3 4

aV yA a

= =

32or3

V a=

(2) Semiellipsoid of revolution: the generating area is a quarter ellipse

Have 4

2 23 4

aV yA ha

= =

22or3

V a h=

(3) Paraboloid of revolution: the generating area is a quarter parabola

Have 3 2

2 28 3

V yA a ah

= =

21or2

V a h=

(4) Cone: the generating area is a triangle

Have 1

2 23 2

aV yA ha

= =

21or3

V a h=


62/166

PROBLEM 5.53

A 15-mm-diameter hole is drilled in a piece of 20-mm-thick steel; the

hole is then countersunk as shown. Determine the volume of steelremoved during the countersinking process.

SOLUTION

The required volume can be generated by rotating the area shown about theyaxis. Applying the secondtheorem of Pappus-Guldinus, we have

5 12 2 7.5 mm 5 mm 5 mm

3 2V xA

= = +

3or 720 mmV =


63/166

PROBLEM 5.54

Three different drive belt profiles are to be studied. If at any given time

each belt makes contact with one-half of the circumference of its pulley,determine the contact area between the belt and the pulley for each

design.

SOLUTION

Applying the first theorem of Pappus-Guldinus, the contact area CA of abelt is given by

CA yL yL = =

Where the individual lengths are the Lengths of the belt cross section

that are in contact with the pulley

( )

( ) ( )

1 1 2 2Have 2

2.5 2.5 mm2 60 mm

2 cos 20

60 2.5 mm 12.5 mm

CA y L y L

= +

=

+

3 2or 3.24 10 mmCA =

Have ( )1 12CA y L =

7.5 7.5 mm2 60 1.6 mm

2 cos20

=

3 2or 2.74 10 mmCA =

Have ( ) ( )1 12 5

60 mm 5 mmCA y L

= =

3 2or 2.80 10 mmCA =


64/166

PROBLEM 5.55

Determine the capacity, in gallons, of the punch bowl shown if 12 in.R=

SOLUTION

The volume can be generated by rotating the triangle and circular sector shown about theyaxis. Applying the

second theorem of Pappus-Guldinus and using Fig. 5.8A, we have

( )

( )

1 1 2 2

2

3 33

3 3

2 2 2

1 1 1 1 3 2 sin 302 cos30

3 2 2 2 2 636

3 32

816 3 2 3

3 312 in. 3526.03 in

8

V xA xA x A x A

RR R R R

R RR

= = = +

= +

= + =

= =

3Since 1 gal 231in=

3

3

3526.03 in15.26 gal

231in /galV = =

15.26 galV =


65/166

PROBLEM 5.56

The aluminum shade for a small high-intensity lamp has a uniform

thickness of 3/32 in. Knowing that the specific weight of aluminum is30.101lb/in , determine the weight of the shade.

SOLUTION

The weight of the lamp shade is given by

W V At = =

where Ais the surface area of the shade. This area can be generated by rotating the line shown about the

xaxis. Applying the first theorem of Pappus-Guldinus, we have

( )1 1 2 2 3 3 4 42 2 2A yL yL y L y L y L y L = = = + + +

( ) ( ) ( )2 20.6 mm 0.60 0.75

2 0.6 mm mm 0.15 mm 1.5 mm

2 2

+

= + +

( ) ( )2 20.75 1.25

mm 0.50 mm 0.40 mm2

+ + +

( ) ( )2 2

2

1.25 1.5mm 0.25 mm 1.25 mm

2

22.5607 in

+ + +

=

Then3 2 3lb/in0.101 22.5607 in in. 0.21362 lb

32W = =

0.214 lbW =


66/166

PROBLEM 5.57

The top of a round wooden table has the edge profile shown. Knowingthat the diameter of the top is 1100 mm before shaping and that the

density of the wood is 3690 kg/m , determine the weight of the wastewood resulting from the production of 5000 tops.

SOLUTION

All dimensions are in mm

( ) ( )

waste blank top

2 6 3blank

top 1 2 3 4

Have

550 mm 30 mm 9.075 10 mm

V V V

V

V V V V V

=

= =

= + + +

Applying the second theorem of Pappus-Guldinus to parts 3 and 4

( ) ( ) ( ) ( )

( )

( )

( )

2 2

top

2

2

6 3

6 3

529 mm 18 mm 535 mm 12 mm

4 122 535 mm 12 mm

3 4

4 182 529 mm 18 mm

3 4

5.0371 3.347 0.1222 0.2731 10 mm

8.8671 10 mm

V

= +

+ +

+ +

= + + +

=

( ) 6 3waste

3 3

9.0750 8.8671 10 mm

0.2079 10 m

V

=

=

( ) ( )waste wood waste tops

3 3 3 2

Finally

690 kg/m 0.2079 10 m 9.81 m/s 5000 tops

W V g N

=

=

wasteor 2.21 kNW =


67/166


68/166

PROBLEM 5.59

The escutcheon (a decorative plate placed on a pipe where the pipe exits

from a wall) shown is cast from yellow brass. Knowing that the specificweight of yellow brass is

30.306 lb/in . determine the weight of the

escutcheon.

SOLUTION

The weight of the escutcheon is given by (specific weight)W V=

where Vis the volume of the plate. Vcan be generated by rotating the area Aabout the xaxis.

Have 3.0755 in. 2.958 in. 0.1175 in.a = =

and

0.5sin 0.16745 R 9.5941

3 = = =

Then 2 26 9.5941 16.4059 or 8.20295 0.143169 rad = = = =

The area Acan be obtained by combining the following four areas, as indicated.

Applying the second theorem of Pappus-Guldinus and then using Figure 5.8A, we have

2 2V yA yA = =


69/166


2, inA , in.y

3, inyA

1 ( )( )1

3.0755 1.5 2.3066

2

= ( )1

1.5 0.5

3

= 1.1533

2 ( )2

3 1.28851 = ( )

( )2 3 sin

sin 0.609213

+ = 0.78497

3 ( )( )1

2.958 0.5 0.73952

= ( )1

0.5 0.166673

= 0.12325

4 ( )( )0.1755 0.5 0.05875 = ( )1

0.5 0.252

= 0.14688

30.44296 inyA =

Then ( )3 32 0.44296 in 1.4476 inV = =

so that ( )

3 31.4476 in 0.306 lb/in 0.44296 lbW = =

0.443 lbW =


70/166

PROBLEM 5.60

The reflector of a small flashlight has the parabolic shape shown.

Determine the surface area of the inside of the reflector.

SOLUTION

First note that the required surface area Acan be generated by rotatingthe parabolic cross section through2 radians about the xaxis. Applying

the first theorem of Pappus-Guldinus, we have

2A yL=

( )22Now, since , at : 7.5x ky x a a k= = =

or 56.25a k= (1)

At ( ) ( )2

15 mm: 15 12.5x a a k= + + =

or 15 156.25a k+ = (2)

Then Eq. (2) 15 156.25

: or 8.4375 mmEq. (1) 56.25

a ka

a k

+= =

1Eq. (1) 0.15

mmk =

20.15 and 0.3dxx y ydy

= =

2

2Now 1 1 0.09dx

dL dy y dydy

= + = +

So 2 andA yL yL ydL= =

( )

12.5 2

7.5

12.53/2

2

7.5

2 1 0.09

2 1 2 1 0.09

3 0.18

A y y dy

y

= +

= +

21013 mm= 2or 1013 mmA =


71/166

PROBLEM 5.61

For the beam and loading shown, determine (a) the magnitude andlocation of the resultant of the distributed load, (b) the reactions at the

beam supports.

SOLUTION

( ) ( )

( )( )

1 2

1

2

Resultant

( ) Have 40 lb/ft 18 ft 720 lb

1120 lb/ft 18 ft 1080 lb

2

R R R

a R

R

= +

= =

= =

or 1800 lbR =

The resultant is located at the centroid Cof the distributed load x

Have ( ) ( )( )( ) ( )( ) ( )1

: 1800 lb 40 lb/ft 18 ft 9 ft 120 lb/ft 18 ft 12 ft2

AM x = +

or 10.80 ftx =

1800 lb

10.80 ft

R

x

=

=

( )b 0: 0x xF A = =

0: 1800 lb 0, 1800 lby y yF A A = = = 1800 lb =A

( )( )0: 1800 lb 10.8 ft 0A AM M = =

19.444 lb ftAM = or 19.44 kip ftA = M


72/166

PROBLEM 5.62

For the beam and loading shown, determine (a) the magnitude andlocation of the resultant of the distributed load, (b) the reactions at the

beam supports.

SOLUTION

( )( )

( )( )

I

II

( ) Have 300 N/m 6 m 1800 N

16 m 900 N/m 1800 N

3

a R

R

= =

= =

Then I II:yF R R R =

or 1800 N 1800 N 3600 NR = + =

( ) ( )( ) ( )( ): 3600 N 3 m 1800 N 4.5 m 1800 NAM x =

or

3.75 mx =

3600 NR = 3.75 mx =

(b) Reactions

0: 0x xF A = =

( ) ( )( )0: 6 m 3600 N 3.75 m 0A yM B = =

or 2250 NyB = 2250 N=B

0: 2250 N 3600 Ny yF A = + =

or 1350 NyA = 1350 N=A


73/166


74/166

PROBLEM 5.64

Determine the reactions at the beam supports for the given loading.

SOLUTION

( )( )

( )( )

I

II

Have 9 ft 200 lb/ft 1800 lb

13 ft 200 lb/ft 300 lb

2

R

R

= =

= =

Then 0: 0x xF A = =

( )( ) ( )( ) ( )0: 4.5 ft 1800 lb 10 ft 300 lb 9 ft 0A yM B = + =

or 1233.3 lbyB = 1233 lb=B

0: 1800 lb 300 lb 1233.3 lb 0y yF A = + =

or 866.7 lbyA = 867 lb=A


75/166

PROBLEM 5.65


SOLUTION

( )( )I 1Have 200 N/m 0.12 m 12 N2

R = =

( )( )II 200 N/m 0.2 m 40 NR = =

Then 0: 0x xF A = =

0: 18 N 12 N 40 N 0y yF A = + =

or 34 NyA = 34.0 N=A

( )( ) ( )( ) ( )( )0: 0.8 m 12 N 0.22 m 40 N 0.38 m 18 NA AM M = +

or 2.92 N mAM = 2.92 N mA = M


76/166

PROBLEM 5.66


SOLUTION

First, replace the given loading with the loading shown below. The two loadings are equivalent because both

are defined by a linear relation between load and distance, and the values at the end points are the same.

( )( )IHave 1.8 m 2000 3600 NN/mR = =

( )( )II1

1.8 m 4500 N/m 4050 N2

R = =

Then 0: 0x xF A = =

( ) ( )( ) ( )( )0: 3 m 2.1 m 3600 N 2.4 m 4050 NB yM A = +

or 270 NyA = 270 N=A

0: 270 N 3600 N 4050 N 0y yF B = + =

or 720 NyB = 720 N=B


77/166

PROBLEM 5.67


SOLUTION

( )( )I1

Have 4 m 2000 kN/m 2667 N3

R = =

( )( )II1

2 m 1000 kN/m 666.7 N3

R = =

Then 0: 0x xF A = =

0: 2667 N 666.7 N 0y yF A = =

or 3334 NyA = 3.33 kN=A

( )( ) ( ) ( )0: 1 m 2667 N 5.5 m 666.7 NA AM M =

or 6334 N mAM = 6.33 kN mA = M


78/166

PROBLEM 5.68


SOLUTION

First, replace the given loading with the loading shown below. The two loadings are equivalent because bothare defined by a parabolic relation between load and distance, and the values at end points are the same.

( )( )IHave 8 ft 100 lb/ft 800 lbR = =

( )( )II2

8 ft 600 lb/ft 3200 lb3

R = =

Then

0: 0x x

F A = =

( )( ) ( )( )0: 11 5 ft 800 lb 4 ft 3200 lb 0AM B = + =

or 800 lb=B

0: 3200 lb 800 lb 800 lb 0y yF A = + + =

or 1600 lb=A


79/166

PROBLEM 5.69

Determine (a) the distance aso that the vertical reactions at supports Aand Bare equal, (b) the corresponding reactions at the supports.

SOLUTION

( )( ) ( )I 1( ) Have ft 120 lb/ft 60 lb2

a R a a= =

( )( ) ( )II1

12 40 lb/ft 240 20 lb2

R a a= =

Then ( )0: 60 240 2 0y y yF A a a B = + =

or 240 40y yA B a+ = +

Now 120 20y y y yA B A B a= = = + (1)

Also ( ) ( ) ( ) ( )1

0: 12 m 60 lb 12 ft 12 ft 240 20 lb 03 3B ya

M A a a a

= + + =

or2140 1080

3 9yA a a= (2)

Equating Eqs. (1) and (2)

2140 10120 20 803 9

a a a+ =

or240 320 480 0

3a a + =

Then

1.6077 ft, 22.392a a= =

Now 12 fta 1.608 fta =

( ) Haveb 0: 0x xF A = =

Eq. (1) ( )120 20 1.61 152.2 lby yA B= = + =

152.2 lb= =A B


80/166

PROBLEM 5.70

Determine (a) the distance aso that the vertical reaction at support Bisminimum, (b) the corresponding reactions at the supports.

SOLUTION

( )( )I1

( ) Have ft 120 lb/ft 60 lb2

a R a a= =

( ) ( ) ( )II1

12 ft 40 lb/ft 240 20 lb2

R a a = =

Then ( ) ( ) ( )0: ft 60 lb 240 20 lb 8 ft 12 ft 03 3A ya a

M a a B

= + + =

or210 20 160

9 3yB a a= + (1)

Then 20 20

09 3

ydBa

da= = or 3.00 fta =

( ) Eq. (1)b ( ) ( )210 20

3.00 3.00 1609 3

yB = +

150 lb= 150.0 lb=B

and 0: 0x xF A = =

( ) ( )0: 60 3.00 lb 240 20 3.00 lb 150 lb 0y yF A = + =

or 210 lbyA = 210 lb=A


81/166

PROBLEM 5.71

Determine the reactions at the beam supports for the given loading when

0 1.5 kN/m.w =

SOLUTION

( )( )I1

Have 9 m 2 kN/m 9 kN2

R = =

( )( )II 9 m 1.5 kN/m 13.5 kNR = =

Then 0: 0x xF C = =

( )( ) ( )( ) ( )0: 50 kN m 1 m 9 kN 2.5 m 13.5 kN 6 m 0B yM C = + =

or

15.4583 kNyC =

15.46 kN=C

0: 9 kN 13.5 kN 15.4583 0y yF B = + =

or 7.0417 kNyB = 7.04 kN=B


82/166

PROBLEM 5.72

Determine (a) the distributed load 0w at the end Dof the beam ABCDfor which the reaction at Bis zero, (b) the corresponding reactions at C.

SOLUTION

( ) ( ) ( )I 0 01

Have 9 m 3.5 kN/m 4.5 3.5 kN2

R w w = =

( )( )II 0 09 m kN/m 9 kNR w w= =

( ) Thena ( ) ( ) ( )( )0 00: 50 kN m 5 m 4.5 3.5 kN 3.5 m 9 kN 0CM w w = + + =

or 09 28.75 0w + =

so 0 3.1944 kN/mw = 0 3.19 kN/mw =

Note: the negative sign means that the distributed force 0w is upward.

( )b 0: 0x xF C = =

( ) ( )0: 4.5 3.5 3.19 kN 9 3.19 kN 0y yF C = + + + =

or 1.375 kNyC = 1.375 kN=C


83/166

PROBLEM 5.73

A grade beam ABsupports three concentrated loads and rests on soil and

the top of a large rock. The soil exerts an upward distributed load, andthe rock exerts a concentrated load RRas shown. Knowing that P=4 kN

and 12

,B Aw w= determine the values of wA and RR corresponding toequilibrium.

SOLUTION

( )( )IHave 1.2 m kN/m 1.2 kNA AR w w= =

( )II1 1

1.8 m kN/m 0.45 kN

2 2

A AR w w

= =

( )III1

1.8 m kN/m 0.9 kN2

A AR w w

= =

Then ( ) ( ) ( ) ( )0: 0.6 m 1.2 kN 0.6 m 0.45 kN/mC A AM w w = +

( ) ( ) ( )( )0.9 m 0.9 kN/m 1.2 m 4 kN/mAw +

( ) ( ) ( )( )0.8 m 18 kN/m 0.7 m 24 kN/m 0 + =

or 6.667 kN/mAw = 6.67 kN/mAw =

and ( ) ( ) ( )( )0: 1.2 m 6.67 kN/m 0.45 m 6.67 kN/my RF R = + + ( )( )0.9 m 6.67 kN/m 24 kN 18 kN 4 kN+

or 29.0 kNRR = 29.0 kNRR =


84/166

PROBLEM 5.74

A grade beam ABsupports three concentrated loads and rests on soil and

the top of a large rock. The soil exerts an upward distributed load, andthe rock exerts a concentrated load RR as shown. Knowing that wB=

0.4wA, determine (a) the largest value of P for which the beam is inequilibrium, (b) the corresponding value of wA.

In the following problems, use =62.4 lb/ft3for the specific weight offresh water and c=150 lb/ft

3for the specific weight of concrete if U.S.

customary units are used. With SI units, use = 103 kg/m

3 for the

density of fresh water and c= 2.40 103 kg/m

3 for the density of

concrete. (See the footnote on page 222 for how to determine the specificweight of a material given its density.)

j

SOLUTION

( )( )IHave 1.2 m kN/m 1.2 kNA AR w w= =

( )( )II1

1.8 m 0.6 kN/m 0.54 kN2

A AR w w= =

( ) ( )III 1.8 m 0.4 kN/m 0.72 kNA AR w w= =

( ) Thena ( ) ( ) ( ) ( ) ( )0: 0.6 m 1.2 kN 1.2 m 1.8 m 0.54 kNA A R AM w R w = + +

( ) ( ) ( )( )2.1 m 0.72 kN 0.5 m 24 kNAw +

( ) ( ) ( )2.0 m 18 kN 2.4 m 0P + =

or 3.204 1.2 2.4 48A Rw R P+ = (1)

and 0: 1.2 0.54 0.72 24 18 0y R A A AF R W W W P = + + + =

or 2.46 42R AR W P+ = (2)

Now combine Eqs. (1) and (2) toeliminate :Aw

( ) ( )3.204 Eq. 2 2.46 Eq. 1 0.252 16.488 2.7RR P =

Since RR must be 0, the maximum acceptable value of Pis that for which 0,R =

or 6.1067 kNP= 6.11 kNP=

( ) Then, from Eq. (2):b 2.46 6.1067 42AW = or 19.56 kN/mAW =


85/166

PROBLEM 5.75

The cross section of a concrete dam is as shown. For a dam section of

unit width, determine (a) the reaction forces exerted by the ground on thebase ABof the dam, (b) the point of application of the resultant of the

reaction forces of part a, (c) the resultant of the pressure forces exertedby the water on the face BCof the dam.

In the following problems, use =62.4 lb/ft3for the specific weight of

fresh water and c=150 lb/ft3for the specific weight of concrete if U.S.

customary units are used. With SI units, use = 103 kg/m

3 for the

density of fresh water and c= 2.40 103 kg/m

3 for the density of

concrete. (See the footnote on page 222 for how to determine the specific

weight of a material given its density.)

SOLUTION

The free body shown consists of a 1-m thick section of the dam and the

triangular section BCDof the water behind the dam.

Note: 1 6 mX =

( )2 9 3 m 12 mX = + =

( )3 15 2 m 17 mX = + =

( )4 15 4 m 19 mX = + =

( ) Now so thata W gV =

( )( ) ( )( )( )3 211

2400 kg/m 9.81 m/s 9 m 15 m 1 m 1589 kN

2

W

= =

( )( ) ( )( )( )3 22 2400 kg/m 9.81 m/s 6 m 18 m 1 m 2543 kNW = =

( )( ) ( )( ) ( )3 231

2400 kg/m 9.81 m/s 6 m 18 m 1 m 1271 kN2

W

= =

( )( ) ( )( ) ( )3 241

2400 kg/m 9.81 m/s 6 m 18 m 1 m 529.7 kN2

W

= =

( )( ) ( )( )( )3 3 21 1Also 18 m 1 m 10 kg/m 9.81 m/s 18 m2 2

P Ap = =

1589 kN=

Then 0: 1589 kN 0xF H = =

or 1589 kNH = 1589 kN=H

0: 1589 kN 2543 kN 1271 kN 529.7 kNyF V =

or 5933 kNV = 5.93 MN=V


86/166


( ) Haveb ( ) ( )( )0: 5933 kN 6 m 1589 kNAM X = + ( )( ) ( )( )

( )( ) ( )( )

6 m 1589 kN 12 m 2543 kN

17 m 1271 kN 19 m 529.7 0

=

or 10.48 mX = 10.48 mX =

to the right of A

(c) Consider water sectionBCDas the free body.

Have 0 =F

Then 1675 kN =R 18.43

or 1675 kN=R 18.43

Alternative solution to part (c)

Consider the face BCof the dam.

2 2Have 6 18 18.9737 mBC = + =

6tan 18.43

18

= =

( ) ( )( ) ( )3 2

2

and 1000 kg/m 9.81 m/s 18 m

176.6 kN/m

p g h= =

=

( )( ) ( )21 1Then 18.97 m 1 m 176.6 kN/m2 2

1675 kN

R Ap = =

=

1675 kN =R 18.43


87/166

PROBLEM 5.76

The cross section of a concrete dam is as shown. For a dam section of

unit width, determine (a) the reaction forces exerted by the ground on thebase AB of the dam, (b) the point of application of the resultant of thereaction forces of part a, (c) the resultant of the pressure forces exerted

by the water on the faceBCof the dam.

SOLUTION

The free body shown consists of a 1-ft thick section of the dam and the parabolic section of water above (andbehind) the dam.

Note ( )15

16 ft 10 ft8

x = =

( )21

16 6 ft 19 ft2

x = + =

( )3 122 12 ft 25 ft4x = + =

( )45

22 12 ft 29.5 ft8

x = + =


88/166


Now W V=

( ) ( )( ) ( )312

150 lb/ft 16 ft 24 ft 1 ft 38,400 lb3

W

= =

( ) ( ) ( ) ( )32 150 lb/ft 6 ft 24 ft 1 ft 21,600 lbW = =

( ) ( )( ) ( )331

150 lb/ft 12 ft 18 ft 1 ft 10,800 lb3

W

= =

( ) ( )( ) ( )342

62.4 lb/ft 12 ft 18 ft 1 ft 8985.6 lb3

W

= =

( ) ( )2 31 1Also 18 1 ft 62.4 lb/ft 18 ft 10,108.8 lb2 2

P Ap = = =

( ) Thena 0: 10,108.8 lb 0xF H = =

or 10.11 kips=H

0: 38,400 lb 21,600 lb 10,800 lb 8995.6 lb 0yF V = =

or 79,785.6V = 79.8 kips=V

( )b ( ) ( )( ) ( )( ) ( )( )0: 79,785.6 lb 6 ft 38,400 lb 19 ft 21,600 lb 25 ft 10,800 lbAM X = ( )( ) ( )( )29.5 ft 8985.6 lb 6 ft 10,108.8 lb 0 + =

or 15.90 ftX =

The point of application of the resultant is 15.90 ft to the right ofA

(c) Consider the water sectionBCDas the free body.

Have 0 =F

13.53 kips

41.6

R

=

=

On the faceBDof the dam

13.53 kips=R 41.6


89/166

PROBLEM 5.77

The 9 12-ft sideAB of a tank is hinged at its bottom Aand is held inplace by a thin rodBC. The maximum tensile force the rod can withstandwithout breaking is 40 kips, and the design specifications require the

force in the rod not exceed 20 percent of this value. If the tank is slowlyfilled with water, determine the maximum allowable depth of water din

the tank.

SOLUTION

Consider the free-body diagram of the side.

( )1 1

Have

2 2

P Ap A d= =

Now ( )0: 9 ft 03Ad

M T P = =

maxThen, for :d

( ) ( )( ) ( )( ) ( )3 3max max max1

9 ft 0.2 40 10 lb 12 ft 62.4 lb/ft 03 2

dd d

=

or3 3 3

max216 10 ft 374.4 d =

or3 3max 576.92 ftd = max 8.32 ftd =


90/166


91/166

PROBLEM 5.79

The friction force between a 2 2-m square sluice gate AB and itsguides is equal to 10 percent of the resultant of the pressure forces

exerted by the water on the face of the gate. Determine the initial forceneeded to lift the gate that its mass is 500 kg.

SOLUTION

Consider the free-body diagram of the gate.

( ) ( )( )( )2 3 3 2I I1 1

Now 2 2 m 10 kg/m 9.81 m/s 3 m2 2

P Ap = =

58.86 kN=

( ) ( )( )( )2 3 3 2II II1 1

2 2 m 10 kg/m 9.81 m/s 5 m2 2

P Ap = =

98.10 kN=

( )I IIThen 0.1 0.1F P P P= = +

( )0.1 58.86 98.10 kN= +

15.696 kN=

Finally ( )( )20: 15.696 kN 500 kg 9.81 m/s 0yF T = =

or 20.6 kN=T


92/166

PROBLEM 5.80

The dam for a lake is designed to withstand the additional force caused

by silt which has settled on the lake bottom. Assuming that silt isequivalent to a liquid of density 3 31.76 10 kg/ms = and consideringa 1-m-wide section of dam, determine the percentage increase in theforce acting on the dam face for a silt accumulation of depth 1.5 m.

SOLUTION

First, determine the force on the dam face without the silt.

Have ( )1 1

2 2w wP Ap A gh= =

( )( ) ( )( )( )3 3 21 6 m 1 m 10 kg/m 9.81 m/s 6 m2

=

176.58 kN=

Next, determine the force on the dam face with silt.

( )( ) ( )( )( )3 3 21Have 4.5 m 1m 10 kg/m 9.81 m/s 4.5 m2

wP =

99.326 kN=

( ) ( )( ) ( )( )( )3 3 2I 1.5 m 1 m 10 kg/m 9.81 m/s 4.5 msP = 66.218 kN=

( ) ( )( ) ( )( ) ( )3 3 2II1

1.5 m 1 m 1.76 10 kg/m 9.81 m/s 1.5 m2

sP =

19.424 kN=

Then ( ) ( )I II 184.97 kNw s sP P P P = + + =

The percentage increase, % inc., is then given by

( )184.97 176.58% inc. 100% 100% 4.7503%

176.58

w

w

P P

P

= = =

% inc. 4.75%=


93/166

PROBLEM 5.81

The base of a dam for a lake is designed to resist up to 150 percent of the

horizontal force of the water. After construction, it is found that silt(which is equivalent to a liquid of density 3 31.76 10 kg/ms = ) issettling on the lake bottom at a rate of 20 mm/y. Considering a 1-m-wide

section of dam, determine the number of years until the dam becomesunsafe.

SOLUTION

From Problem 5.80, the force on the dam face before the silt is deposited, is 176.58 kN.wP = The maximumallowable force allowP on the dam is then:

( )( )allow 1.5 1.5 176.58 kN 264.87 kNwP P= = =

Next determine the forceP on the dam face after adepth dof silt has settled.

Have ( ) ( ) ( )( )( )3 3 21

6 m 1 m 10 kg/m 9.81 m/s 6 m2

wP d d =

( )2

4.905 6 kNd=

( ) ( ) ( )( )( )3 3 2I 1 m 10 kg/m 9.81 m/s 6 msP d d =

( )29.81 6 kNd d=

( ) ( ) ( )( ) ( )3 3 2

II

11 m 1.76 10 kg/m 9.81 m/s m

2sP d d =

28.6328 kNd=

( ) ( ) ( ) ( )2 2 2I II2

4.905 36 12 9.81 6 8.6328 kN

3.7278 176.58 kN

w s sP P P P d d d d d

d

= + + = + + +

= +


94/166


Now required that allowP P = to determine the maximum value of d.

( )23.7278 176.58 kN 264.87 kNd + = or 4.8667 md =

Finally 3 m4.8667 m 20 10

yearN=

or 243 yearsN =


95/166

PROBLEM 5.82

The square gateABis held in the position shown by hinges along its top

edge A and by a shear pin at B. For a depth of water 3.5d= m,determine the force exerted on the gate by the shear pin.

SOLUTION

First consider the force of the water on the gate.

Have ( )1 12 2P Ap A gh= =

Then ( ) ( )( )( )2 3 3 2

I

118 m 10 kg/m 9.81 m/s 1.7 m

2P =

26.99 kN=

( ) ( )( )( )2 3 3 2II1

18 m 10 kg/m 9.81 m/s 1.7 1.8cos30 m2

P =

51.74 kN=

Now ( ) ( )I II

1 20: 0

3 3A AB AB AB BM L P L P L F = + =

( ) ( )1 2

or 26.99 kN 51.74 kN 03 3

BF+ =

or 43.49 kNBF =

4.35 kNB =F 30.0


96/166

PROBLEMS 5.83 AND 5.84

Problem 5.83: A temporary dam is constructed in a 5-ft-wide fresh

water channel by nailing two boards to pilings located at the sides of thechannel and propping a third boardABagainst the pilings and the floor ofthe channel. Neglecting friction, determine the reactions at AandBwhen

ropeBCis slack.

Problem 5.84: A temporary dam is constructed in a 5-ft-wide fresh

water channel by nailing two boards to pilings located at the sides of thechannel and propping a third boardABagainst the pilings and the floor of

the channel. Neglecting friction, determine the magnitude and directionof the minimum tension required in ropeBCto move boardAB.

SOLUTION

First, consider the force of the water on the gate.

Have ( )

1 1

2 2P Ap A h= =

( )( ) ( )( )3I1

So that 1.5 ft 5 ft 62.4 lb/ft 1.8 ft2

P =

421.2 lb=

( ) ( ) ( )( )3II1

1.5 ft 5 ft 62.4 lb/ft 3 ft2

P =

702 lb=

5.83 Find the reactions atAandBwhen rope is slack.

( ) ( )( ) ( )( )0: 0.9 ft 0.5 ft 421.2 lb 1.0 ft 702 lb 0AM B = + + =

or 1014 lbB =

1014 lb=B

( ) ( )4 4

0: 2 421.2 lb 702 lb 05 5

x xF A = + + =

or 449.28 lbxA =

Note that the factor 2 (2 )xA is included since xA is the horizontal forceexerted by the board on each piling.

( ) ( )3 30: 1014 lb 421.2 lb 702 lb 05 5

y yF A = + =

or 340.08 lbyA =

563 lb =A 37.1


97/166


98/166


Problem 5.85:A 2 3-m gate is hinged at Aand is held in position byrod CD. End D rests against a spring whose constant is 12 kN/m. Thespring is undeformed when the gate is vertical. Assuming that the forceexerted by rod CD on the gate remains horizontal, determine the

minimum depth of water dfor which the bottom Bof the gate will moveto the end of the cylindrical portion of the floor.

Problem 5.86:Solve Problem 5.85 if the mass of the gate is 500 kg.

SOLUTION

First, determine the forces exerted on the gate by the spring and the

water when Bis at the end of the cylindrical portion of the floor.

Have 1

sin 30

2

= =

Then ( )1.5 m tan30spx =

and sp spF kx=

( )( )12 kN/m 1.5 m tan 30=

10.39 kN=

Assume 2 md

Have ( )1 1

2 2

P Ap A g h= =

Then ( )( ) ( )( )( )3 3 2

I

12 m 3 m 10 kg/m 9.81 m/s 2 m

2P d =

( )29.43 2 kNd=

( )( ) ( )( )( )3 3 2II1

2 m 3 m 10 kg/m 9.81 m/s 2 2cos30 m2

P d = +

( )29.43 0.2679 kNd=


99/166

PROBLEMS 5.85 AND 5.86 CONTINUED

5.85 Find mind so that gate opens, 0.W =

Using the above free-body diagrams of the gate, we have

( )2

0: m 29.43 2 kN3

AM d

=

( )4

m 29.43 0.2679 kN3

d

+

( )( )1.5 m 10.39 kN 0 =

or ( ) ( )19.62 2 39.24 0.2679 15.585d d + =

58.86 65.3374d =

or 1.1105 md = 1.110 md =

5.86 Find mind so that the gate opens.

( )( )29.81 m/s 500 kg 4.905 kNW = =

Using the above free-body diagrams of the gate, we have

( )2

0: m 29.43 2 kN3

AM d

=

( )4

m 29.43 0.2679 kN3

d

+

( )( )1.5 m 10.39 kN +

( )( )0.5 m 4.905 kN 0 =

or ( ) ( )19.62 2 39.24 0.2679 18.0375d d + =

or 1.15171 md = 1.152 md =


100/166


Problem 5.87:The gate at the end of a 3-ft-wide fresh water channel is

fabricated from three 240-lb, rectangular steel plates. The gate is hingedat Aand rests against a frictionless support atD. Knowing that d= 2.5 ft,determine the reactions at Aand D.

Problem 5.88:The gate at the end of a 3-ft-wide fresh water channel isfabricated from three 240-lb, rectangular steel plates. The gate is hinged

at Aand rests against a frictionless support at D. Determine the depth ofwater dfor which the gate will open.

SOLUTION

5.87 Note that in addition to the weights of the gate segments, the water exerts pressure on all submerged

surfaces ( ).p h=

( )( )3 20.5Thus, at 0.5 ft, 62.4 lb/ft 0.5 ft 31.2 lb/fth p= = =

( )( )3 2

2.52.5 ft, 6.24 lb/ft 2.5 ft 156.0 lb/fth p= = =

( )( ) ( )211

Then 0.5 ft 3 ft 31.2 lb/ft 23.4 lb2

P = =

( )( ) ( )22 2 ft 3 ft 31.2 lb/ft 187.2 lbP = =

( ) ( ) ( )231

2 ft 3 ft 31.2 lb/ft 93.6 lb2

P = =

( )( ) ( )241

2 ft 3 ft 156 lb/ft 468 lb2

P = =

and ( ) ( ) ( )( ) ( ) ( )( )1

0: 4 2 ft 240 lb 1 ft 240 lb 2 0.5 ft 23.4 lb 1 ft 187.2 lb3

AM D = + + +

( )( ) ( )( )2 1

2 ft 93.6 lb 2 ft 468 lb 03 3

=

or 11.325 lbD = 11.33 lb =D


101/166

PROBLEMS 5.87 AND 5.88 CONTINUED

0: 11.32 23.4 93.6 468 0x xF A = + + + + =

or 596.32 lbxA =

0: 240 240 240 187.2 0y yF A = + =

or 532.8 lbyA = 800 lb =A 41.8

5.88 At ( ) ( )2 3

22 ft, 2 lb/ft where 62.4 lb/ftdh d p d = = =

( ) 2ft, lb/ftdh d p d = =

( ) ( ) ( ) ( )23

1 1 2

1 1 3Then 2 ft 3 ft lb/ft 2 ft 2 lb

2 2 2dP A p d d d = = =

(Note: For simplicity, the numerical value of the density will be substituted into the equilibrium equations

below, rather than at this level of the calculations.)

( ) ( ) ( ) ( )2 2 2 2 ft 3 ft 2 ft 6 2 lbdP A p d d = = =

( )( ) ( ) ( )3 3 21 1

2 ft 3 ft 2 ft 3 2 lb2 2

dP A p d d = = =

( )( ) ( ) ( ) ( )4 41 1

2 ft 3 ft ft 3 lb 3 2 6 lb2 2

dP A p d d d = = = = +

As the gate begins to open, 0D

( )( ) ( )( ) ( ) ( )21 3

0: 2 ft 240 lb 1 ft 240 lb 2 ft 2 ft 2 lb3 2

AM d d

= + + +

( ) ( ) ( ) ( )2

1 ft 6 2 lb 2 ft 3 2 lb3

d d

( ) ( )1

2 ft 3 2 lb 6 lb 03

d

+ =

( ) ( ) ( )3 21 720

or 2 3 2 12 2 42

d d d

+ + =

7204

62.4=

7.53846=

Solving numerically yields 2.55 ftd =


102/166

PROBLEM 5.89

A rain gutter is supported from the roof of a house by hangers that are

spaced 0.6 m apart. After leaves clog the gutters drain, the gutter slowlyfills with rainwater. When the gutter is completely filled with water,determine (a) the resultant of the pressure force exerted by the water on

the 0.6-m section of the curved surface of the gutter, (b) the force-couplesystem exerted on a hanger where it is attached to the gutter.

SOLUTION

(a)Consider a 0.6 m long parabolic section of water.

Then ( )1 1

2 2P Ap A gh= =

( )( ) ( )( )( )3 3 21 0.08 m 0.6 m 10 kg/m 9.81 m/s 0.08 m2

=

18.84 N=

wW gV=

( )( ) ( )( )( )3 3 2 210 kg/m 9.81 m/s 0.12 m 0.08 m 0.6 m3

=

37.67 N=

Now ( )0: 0w = + + =F R P W

2 2

So that , tan

w

w

W

R P W P= + =

42.12 N, 63.4= = 42.1 N=R 63.4

(b) Consider the free-body diagram of a 0.6 m long section of water and

gutter.

Then 0: 0x xF B = =

0: 37.67 N 0y yF B = =

or 37.67 NyB =

( ) ( )0: 0.06 0.048 m 37.67 N 0B BM M = + =

or 0.4520 N mBM =

The force-couple system exerted on the hanger is then

37.7 N , 0.452 N m


103/166

PROBLEM 5.90

The composite body shown is formed by removing a hemisphere of

radius r from a cylinder of radius R and height 2R. Determine (a) theycoordinate of the centroid when r = 3R/4, (b) the ratio r/R for which

1.2 .y R=

SOLUTIONNote, for the axes shown

V y yV

1 ( )( )2 32 2R R R = R 42 R

2 32

3r

3

8r 4

1

4r

3323

rR

4428

rR

Then

4 4

3 3

1

81

3

R ry V

YV R r

= =

4

3

11

8

11

3

r

R

r

R

=

( )a

4

3

1 31

3 43:

4 1 31

3 4

r R y R

= =

or 1.118y R=

( )b

4

3

11

81.2 : 1.2

11

3

r

Ry R R R

r

R

= =

or

4 3

3.2 1.6 0r r

R R

+ =

Solving numerically 0.884r

R=


104/166

PROBLEM 5.91

Determine the ycoordinate of the centroid of the body shown.

SOLUTION

First note that the values of Y will be the same for the given body and the body shown below. Then

V y yV

Cone21

3a h

1

4h

2 21

12a h

Cylinder

221

2 4

ab a b

=

1

2b

2 21

8a b

( )2 4 3

12a h b

( )2 2 22 324 a h b

Have Y V yV =

( ) ( )2 2 2 2Then 4 3 2 312 24

Y a h b a h b

=

( )

2 2

2 3or2 4 3

h bYh b

=


105/166

PROBLEM 5.92

Determine the z coordinate of the centroid of the body shown. (Hint:

Use the result of Sample Problem 5.13.)

SOLUTION

First note that the body can be formed by removing a half-cylinder from a half-cone, as shown.

V V

Half-Cone 216

a h *a

316

a h

Half-Cylinder

22

2 2 8

ab a b

=

4 2

3 2 3

a a

=

31

12a b

( )2 4 3

24a h b

( )

31 212

a h b

From Sample Problem 5.13

Have Z V zV =

( ) ( )2 31

Then 4 3 224 12

Z a h b a h b

=

2 2or

4 3

a h bZ

h b

=


106/166


107/166


Problem 5.94: For the machine element shown, determine the

x coordinate of the center of gravity.

Problem 5.95: For the machine element shown, determine theycoordinate of the center of gravity.

SOLUTIONS

First, assume that the machine element is homogeneous so that its center of gravity coincides with thecentroid of the corresponding volume.

3, inV , in.x , in.y

4, inxV 4, inyV

I (4)(3.6)(0.75) 10.8= 2.0 0.375 21.6 4.05

II (2.4)(2.0)(0.6) 2.88=

3.7 1.95 10.656 5.616III 2(0.45) (0.4) 0.2545 = 4.2 2.15 1.0688 0.54711

IV 2(0.5) (0.75) 0.5890 = 1.2 0.375 0.7068 0.22089

13.3454 32.618 9.9922

5.94

Have X V xV =

( )3 413.3454 in 32.618 inX = or 2.44 in.X = 5.95

Have Y V yV =

( )3 413.3454 in 9.9922 inY = or 0.749 in.Y =

Vector Mechanics for Engineers Statics 7th - Cap 05

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Transcript of Vector Mechanics for Engineers Statics 7th - Cap 05