dynamics-ch1

DynaDEn

X(t) ∫∑∑=

=

∞=

=→∆

=

=

=∆=∆≈2

0101)()()( lim

tt

t

i

it

ni

idttFttFttF

t∆

F(t)

t

F(t)

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© 2004 The McGraw-Hill Companies, Inc. All rights reserved.

Vector Mechanics for Engineers: Dynamics

SeventhEdition

7 - 49

Rectilinear Motion: Position, Velocity & Acceleration

• Consider particle with motion given by326 ttx −=

2312 ttdtdxv −==

tdt

xddtdva 6122

2−===

• at t = 0, x = 0, v = 0, a = 12 m/s2

• at t = 2 s, x = 16 m, v = vmax = 12 m/s, a = 0

• at t = 4 s, x = xmax = 32 m, v = 0, a = -12 m/s2

• at t = 6 s, x = 0, v = -36 m/s, a = 24 m/s2

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A=const

V=at+v0

X=1/2at2+v0t+x0

V2- v02=2as, s=x - x0

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Phisical Meaning



SeventhEdition

11 - 49

Sample Problem 11.2

Determine:• velocity and elevation above ground at

time t, • highest elevation reached by ball and

corresponding time, and • time when ball will hit the ground and

corresponding velocity.

Ball tossed with 10 m/s vertical velocity from window 20 m above ground.

SOLUTION:

• Integrate twice to find v(t) and y(t).

• Solve for t at which velocity equals zero (time for maximum elevation) and evaluate corresponding altitude.

• Solve for t at which altitude equals zero (time for ground impact) and evaluate corresponding velocity.

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SeventhEdition

12 - 49

Sample Problem 11.2

( )( ) tvtvdtdv

adtdv

ttv

v81.981.9

sm81.9

00

2

0

−=−−=

−==

∫∫

( ) ttv

−= 2s

m81.9sm10

( )( ) ( ) 2

21

00

81.91081.910

81.910

0

ttytydttdy

tvdtdy

tty

y−=−−=

−==

∫∫

( ) 22s

m905.4sm10m20 ttty

−

+=

SOLUTION:• Integrate twice to find v(t) and y(t).

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SeventhEdition

13 - 49

Sample Problem 11.2• Solve for t at which velocity equals zero and evaluate

corresponding altitude.

( ) 0sm81.9

sm10 2 =

−= ttv

s019.1=t

• Solve for t at which altitude equals zero and evaluate corresponding velocity.

( )

( ) ( )22

22

s019.1sm905.4s019.1

sm10m20

sm905.4

sm10m20

−

+=

−

+=

y

ttty

m1.25=y

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SeventhEdition

14 - 49

Sample Problem 11.2• Solve for t at which altitude equals zero and

evaluate corresponding velocity.

( ) 0sm905.4

sm10m20 2

2 =

−

+= ttty

( )s28.3

smeaningles s243.1=−=

tt

( )

( ) ( )s28.3sm81.9

sm10s28.3

sm81.9

sm10

2

2

−=

−=

v

ttv

sm2.22−=v

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SeventhEdition

15 - 49

Sample Problem 11.3

Brake mechanism used to reduce gun recoil consists of piston attached to barrel moving in fixed cylinder filled with oil. As barrel recoils with initial velocity v0, piston moves and oil is forced through orifices in piston, causing piston and cylinder to decelerate at rate proportional to their velocity.

Determine v(t), x(t), and v(x).

kva −=

SOLUTION:

• Integrate a = dv/dt = -kv to find v(t).

• Integrate v(t) = dx/dt to find x(t).

• Integrate a = v dv/dx = -kv to find v(x).

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SeventhEdition

16 - 49

Sample Problem 11.3SOLUTION:

• Integrate a = dv/dt = -kv to find v(t).( ) ( ) kt

vtvdtk

vdvkv

dtdva

ttv

v−=−=−== ∫∫

00ln

0

( ) ktevtv −= 0

• Integrate v(t) = dx/dt to find x(t).

( )

( )( )

tkt

tkt

tx

kt

ek

vtxdtevdx

evdtdxtv

00

00

0

0

1

−==

==

−−

−

∫∫

( ) ( )ktekvtx −−= 10

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SeventhEdition

17 - 49

Sample Problem 11.3• Integrate a = v dv/dx = -kv to find v(x).

kxvv

dxkdvdxkdvkvdxdvva

xv

v

−=−

−=−=−== ∫∫

0

00

kxvv −= 0

• Alternatively,

( ) ( )

−=

0

0 1v

tvkvtx

kxvv −= 0

( ) ( )0

0 or v

tveevtv ktkt == −−

( ) ( )ktekvtx −−= 10with

and

then

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SeventhEdition

18 - 49

Uniform Rectilinear Motion

For particle in uniform rectilinear motion, the acceleration is zero and the velocity is constant.

vtxxvtxx

dtvdx

vdtdx

tx

x

+==−

=

==

∫∫

0

0

00

constant

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SeventhEdition

19 - 49

Uniformly Accelerated Rectilinear MotionFor particle in uniformly accelerated rectilinear motion, the acceleration of

the particle is constant.

atvv

atvvdtadvadtdv tv

v

+=

=−=== ∫∫

0

000

constant

( )

221

00

221

000

000

attvxx

attvxxdtatvdxatvdtdx tx

x

++=

+=−+=+= ∫∫

( ) ( )

( )020

2

020

221

2

constant00

xxavv

xxavvdxadvvadxdvv

x

x

v

v

−+=

−=−=== ∫∫

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SeventhEdition

20 - 49

Motion of Several Particles: Relative Motion• For particles moving along the same line, time

should be recorded from the same starting instant and displacements should be measured from the same origin in the same direction.

=−= ABAB xxx relative position of Bwith respect to A

ABAB xxx +=

=−= ABAB vvv relative velocity of Bwith respect to A

ABAB vvv +=

=−= ABAB aaa relative acceleration of Bwith respect to A

ABAB aaa +=

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SeventhEdition

21 - 49

Sample Problem 11.4

Ball thrown vertically from 12 m level in elevator shaft with initial velocity of 18 m/s. At same instant, open-platform elevator passes 5 m level moving upward at 2 m/s.

Determine (a) when and where ball hits elevator and (b) relative velocity of ball and elevator at contact.

SOLUTION:

• Substitute initial position and velocity and constant acceleration of ball into general equations for uniformly accelerated rectilinear motion.

• Substitute initial position and constant velocity of elevator into equation for uniform rectilinear motion.

• Write equation for relative position of ball with respect to elevator and solve for zero relative position, i.e., impact.

• Substitute impact time into equation for position of elevator and relative velocity of ball with respect to elevator.

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SeventhEdition

22 - 49

Sample Problem 11.4SOLUTION:• Substitute initial position and velocity and constant

acceleration of ball into general equations for uniformly accelerated rectilinear motion.

22

221

00

20

sm905.4

sm18m12

sm81.9

sm18

ttattvyy

tatvv

B

B

−

+=++=

−=+=


ttvyy

v

EE

E

+=+=

=

sm2m5

sm2

0

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SeventhEdition

23 - 49

Sample Problem 11.4• Write equation for relative position of ball with respect to

elevator and solve for zero relative position, i.e., impact.

( ) ( ) 025905.41812 2 =+−−+= ttty EB

( )s65.3


tt

• Substitute impact time into equations for position of elevator and relative velocity of ball with respect to elevator.

( )65.325+=Eym3.12=Ey

( )( )65.381.916

281.918

−=

−−= tv EB

sm81.19−=EBv

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PROBLEM 11.14 The acceleration of a particle is defined by the relation 20.15 m/s .a = Knowing that 10x = − m when 0t = and 0.15= −υ m/s when 2t = s, determine the velocity, the position, and the total distance traveled when

5t = s.

SOLUTION

Determine velocity. 0.15 2 2 0.15v t tdv a dt dt− = =∫ ∫ ∫

( ) ( )( )15 0.15 0.15 2v t− − = −

0.15 0.45 m/sv t= −

At 5 s,t = ( )( )5 0.15 5 0.45v = − 5 0.300 m/s v =

When 0,v = 0.15 0.45 0 3.00 st t− = = For 0 3.00 s,t≤ ≤ 0, is decreasing.v x≤ For 3.00 5 s,t≤ ≤ 0, is increasing.v x≥

Determine position. ( )10 0 0 0.15 0.45x t tdx v dt t dt− = = −∫ ∫ ∫

( ) ( )2 2

010 0.075 0.45 0.075 0.45

tx t t t t− − = − = −

20.075 0.45 10 mx t t= − −

At 5 s,t = ( )( ) ( )( )25 0.075 5 0.45 5 10 10.375 mx = − − = −

5 10.38 m x = −

At 0,t = 0 10 m (given)x = −

At 3.00 s,t = ( )( ) ( )( )23 min 0.075 3.00 0.45 3.00 10 10.675 mmx x= = − − = −

Distances traveled: Over 0 3.00 s,t≤ ≤ 1 0 min 0.675 md x x= − =

Over 3.00 s 5 s,t< < 2 5 min 0.300 md x x= − =

Total distance traveled: 1 2 0.975 m d d d= + =

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SeventhEdition

20 - 49

Motion of Several Particles: Relative Motion• For particles moving along the same line, time

should be recorded from the same starting instant and displacements should be measured from the same origin in the same direction.

=−= ABAB xxx relative position of Bwith respect to A

ABAB xxx +=

=−= ABAB vvv relative velocity of Bwith respect to A

ABAB vvv +=

=−= ABAB aaa relative acceleration of Bwith respect to A

ABAB aaa +=

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SeventhEdition

21 - 49

Sample Problem 11.4

Ball thrown vertically from 12 m level in elevator shaft with initial velocity of 18 m/s. At same instant, open-platform elevator passes 5 m level moving upward at 2 m/s.

Determine (a) when and where ball hits elevator and (b) relative velocity of ball and elevator at contact.

SOLUTION:

• Substitute initial position and velocity and constant acceleration of ball into general equations for uniformly accelerated rectilinear motion.


• Write equation for relative position of ball with respect to elevator and solve for zero relative position, i.e., impact.

• Substitute impact time into equation for position of elevator and relative velocity of ball with respect to elevator.

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SeventhEdition

22 - 49

Sample Problem 11.4SOLUTION:• Substitute initial position and velocity and constant

acceleration of ball into general equations for uniformly accelerated rectilinear motion.

22

221

00

20

sm905.4

sm18m12

sm81.9

sm18

ttattvyy

tatvv

B

B

−

+=++=

−=+=


ttvyy

v

EE

E

+=+=

=

sm2m5

sm2

0

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SeventhEdition

23 - 49

Sample Problem 11.4• Write equation for relative position of ball with respect to

elevator and solve for zero relative position, i.e., impact.

( ) ( ) 025905.41812 2 =+−−+= ttty EB

( )s65.3


tt

• Substitute impact time into equations for position of elevator and relative velocity of ball with respect to elevator.

( )65.325+=Eym3.12=Ey

( )( )65.381.916

281.918

−=

−−= tv EB

sm81.19−=EBv

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SeventhEdition

11 - 25

Sample Problem 11.5

Pulley D is attached to a collar which is pulled down at 75 mm/s. At t = 0, collar A starts moving down from Kwith constant acceleration and zero initial velocity. Knowing that velocity of collar A is 300 mm/s as it passes L, determine the change in elevation, velocity, and acceleration of block B when block A is at L.

SOLUTION:

• Define origin at upper horizontal surface with positive displacement downward.

• Collar A has uniformly accelerated rectilinear motion. Solve for acceleration and time t to reach L.

• Pulley D has uniform rectilinear motion. Calculate change of position at time t.

• Block B motion is dependent on motions of collar A and pulley D. Write motion relationship and solve for change of block B position at time t.

• Differentiate motion relation twice to develop equations for velocity and acceleration of block B.

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SeventhEdition

11 - 26

Sample Problem 11.5SOLUTION:• Define origin at upper horizontal surface with

positive displacement downward.

• Collar A has uniformly accelerated rectilinear motion. Solve for acceleration and time t to reach L.

( ) ( )[ ]( ) ( ) 22

020

2

s/mm225mm2002mm300

2

==

−+=

AA

AAAAA

aa

xxavv

( )s 333.1s/mm 225mm300 2

0

==

+=

t

tavv AAA

DynaDEn



SeventhEdition

11 - 27

Sample Problem 11.5• Pulley D has uniform rectilinear motion. Calculate

change of position at time t.

( )( ) ( )( ) mm 100s333.1s/mm 750

0

==−

+=

DD

DDD

xxtvxx

• Block B motion is dependent on motions of collar A and pulley D. Write motion relationship and solve for change of block B position at time t.

Total length of cable remains constant,

( ) ( ) ( )( )[ ] ( )[ ] ( )[ ]

( ) ( ) ( )[ ] 0mm1002mm20002

22

0

000

000

=−++

=−+−+−

++=++

BB

BBDDAA

BDABDA

xxxxxxxx

xxxxxx

( ) mm4000 −=− BB xx

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SeventhEdition

11 - 28

Sample Problem 11.5• Differentiate motion relation twice to develop

equations for velocity and acceleration of block B.

( ) ( ) 0mm752mm30002constant2

=++=++=++

B

BDA

BDA

vvvvxxx

s/mm 450=Bv

( ) 0s/mm 22502=+

=++

B

BDA

vaaa

2s/mm 225−=Ba

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Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

Fig. P11.47 and P11.48

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Fig. P11.49 and P11.50

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Fig. P11.51 and P11.52

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Fig. P11.53 and P11.54

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Fig. P11.55 and P11.56

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Fig. P11.57 and P11.58

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Fig. P11.59 and P11.60

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Fig. P11.127

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Fig. P11.128

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PROBLEM 11.49

Block C starts from rest and moves downward with a constant acceleration. Knowing that after 12 s the velocity of block A is 456 mm/s, determine (a) the accelerations of A, B, and C, (b) the velocity and the change in position of block B after 8 s.

SOLUTION

Let x be position relative to the fixed supports, taken positive if downward.

Constraint of cable on left: 2 3 constantA Bx x+ =

2 22 3 0, or , and 3 3A B B A B Av v v v a a+ = = − = −

Constraint of cable on right: 2 constantB Cx x+ =

1 1 12 0, or , and 2 3 3B C C B A C Av v v v v a a+ = = − = =

Block C moves downward; hence, block A also moves downward.

(a) Accelerations.

( ) ( ) 200

456 0 or 38.0 mm/s12

A AA A A A

v vv v a T a

t− −

= + = = =

238.0 mm/sA =a

( ) 22 2 38.0 25.3 mm/s3 3B Aa a = − = − = −

225.3 mm/sB =a

( ) 21 1 38.0 12.67 mm/s3 3C Aa a = = =

212.67 mm/sC =a

(b) Velocity and change in position of B after 8 s.

( ) ( )( )0 0 25.3 8 203 mm/sB B Bv v a t= + = + − = −

203 mm/sB =v

( ) ( ) ( )( )220 0

1 10 25.3 8 811 mm2 2B B B Bx x v t a t− = + = + − = −

811 mmBx∆ =

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텍스트 상자

unit vector: 단위벡터

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SeventhEdition

37 - 49

Rectangular Components of Velocity & Acceleration• Rectangular components particularly effective

when component accelerations can be integrated independently, e.g., motion of a projectile,

00 ==−==== zagyaxa zyx &&&&&&

with initial conditions,( ) ( ) ( ) 0,,0 000000 ==== zyx vvvzyx

Integrating twice yields

( ) ( )( ) ( ) 0

02

21

00

00

=−==

=−==

zgtyvytvx

vgtvvvv

yx

zyyxx

• Motion in horizontal direction is uniform.

• Motion in vertical direction is uniformly accelerated.

• Motion of projectile could be replaced by two independent rectilinear motions.

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Sample Problem 11.7a

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Sample Problem 11.7b

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Sample Problem 11.7c

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Sample Problem 11.8a

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Sample Problem 11.8b

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Sample Problem 11.8c

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Sample Problem 11.8d

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Fig. P11.97

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Fig. P11.98

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Fig. P11.99

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landing: physical meaning


Fig. P11.101

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Fig. P11.104

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Fig. P11.105

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Fig. P11.106

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Fig. P11.109

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Fig. P11.111

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Fig. P11.112

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Fig. P11.114

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Fig. P11.115

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PROBLEM 11.97

A baseball pitching machine “throws” baseballs with a horizontal velocity 0v . Knowing that height h varies between 788 mm and 1068 mm, determine (a) the range of values of 0,v (b) the values of α corresponding to 788h = mm and 1068h = mm.

SOLUTION

(a) Vertical motion: 0 1.5 m,y = ( )0 0yv =

( ) ( )020 0

21 or 2y

y yy y v t gt t

g−

= + − =

At point B, ( )02 or B

y hy h t

g−

= =

When 788 mm 0.788 m,h = = ( )( )2 1.5 0.7880.3810 s

9.81Bt−

= =

When 1068 mm 1.068 m,h = = ( )( )2 1.5 1.0680.2968 s

9.81Bt−

= =

Horizontal motion: ( )0 000, ,xx v v= =

0 0 or B

B

x xx v t vt t

= = =

With 12.2 m,Bx = 012.2we get 32.02 m/s

0.3810v = =

012.2and 41.11 m/s

0.2968v = =

032.02 m/s 41.11 m/sv≤ ≤ or 0115.3 km/h 148.0 km/h v ≤≤

(b) Vertical motion: ( )0y yv v gt gt= − = −

Horizontal motion: 0xv v=

( )( ) 0

tany BB

x B

vdy gtdx v v

α = − = − =

For 0.788 m,h = ( )( )9.81 0.3810tan 0.11673,

32.02α = = 6.66 α = °

For 1.068 m,h = ( )( )9.81 0.2968tan 0.07082,

41.11α = = 4.05 α = °

12.2 m

1.5 mA v0

h

a

B

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PROBLEM 11.98

While delivering newspapers, a girl throws a paper with a horizontal velocity 0.v Determine the range of values of 0v if the newspaper is to land between points B and C.

SOLUTION

Sketch the limiting trajectories.

0

0

Point : 00

A x xy y= =

= =

Point : 2.2 m1.0 m

B

B

B xy

=

= −

Point : 4.0 m0.6 m

c

c

C xy

=

= −

Vertical motion with ( ) ( ) 200 0

1 20: or 2y y

yv y y v t gt tg−

= = + − = (1)

Horizontal motion with ( ) 0 0 00 : or xxv v x v t vt

= = = (2)

Substituting equation (1) into equation (2) gives 0 2gv x

y=

− (3)

Applying equation (3):

at point B, ( ) ( )0

9.812.2 4.87 m/s2 1.0

v = =− −

at point C, ( ) ( )0

9.814.0 11.44 m/s2 0.6

v = =−

Range of 0 :v 04.87 m/s 11.44 m/s v≤ ≤

0.4 m

1.2 m

1 m

0.2 m0.2 m0.2 m

2.2 m

B

C

A v0

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PROBLEM 11.99

A ski jumper starts with a horizontal take-off velocity of 25 m/s and lands on a straight landing hill inclined at o30 . Determine (a) the time between take-off and landing, (b) the length d of the jump, (c) the maximum vertical distance between the jumper and the landing hill.

SOLUTION

(a) At the landing point, tan 30y x= − °

Horizontal motion: ( )0 00xx x v t v t= + =

Vertical motion: ( ) 2 20 0

1 12 2yy y v t gt gt= + − = −

from which 2 02 2 tan 30 2 tan 30y x v ttg g g

° °= − = =

Rejecting the 0t = solution gives ( )( )0 2 25 tan 302 tan 309.81

vtg

°°= = 2.94 s t =

(b) Landing distance: ( )( )0 25 2.94cos30 cos30 cos30

x v td = = =° ° °

84.9 m d =

(c) Vertical distance: tan 30h x y= ° +

or 20

1tan 302

h v t gt= ° −

Differentiating and setting equal to zero,

0tan 30tan 30 0 or odh vv gt t

dt g°

= ° − = =

Then, ( ) ( ) 20 0 0

maxtan 30 tan 30 1 tan 30

2v v vh g

g g° ° °

= −

( ) ( )( )( )

2 22 20 25 tan 30tan 30

2 2 9.81v

g°°

= = max 10.6 m h =

25 m/s

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PROBLEM 11.100

A golfer aims his shot to clear the top of a tree by a distance h at the peak of the trajectory and to miss the pond on the opposite side. Knowing that the magnitude of 0v is 25 m/s, determine the range of values of h which must be avoided.

SOLUTION

Horizontal motion: ( )0 000

, or xxx x v t v t tv

= + = =

Vertical motion: ( )2

2 20 0 0 20

0

1 1 or 2 2 2y

gxy y v t gt y gt y yv

= + − = − = −

At ground level, 0,y = so that 2

0 202

gxyv

=

At 45 m,x = ( ) ( )( )( )

2

0 29.81 45

15.89 m2 25

y = =

0 12 3.89 mh y= − =

At 48 m,x = ( )( )( )( )

2

0 29.81 48

18.08 m2 25

y = =

0 12 6.08 mh y= − =

Range to avoid: 3.89 m 6.08 m h< <

h

12 m

45 m3 m

Pond

v0

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PROBLEM 11.101

A golfer hits a golf ball with an initial velocity of 48 m/s at an angle of 25°with the horizontal. Knowing that the fairway slopes downward at an average angle of 5 ,° determine the distance d between the golfer and point B where the ball first lands.

SOLUTION

Horizontal motion: 0 0cos 25 cos 25Ax x v t v t= + ° = °

At point B, 0cos5 cos 25B Bx d v t= ° = °

Vertical motion: 2 20 0 0

1 1sin 25 sin 252 2

y y v t gt v t gt= + ° − = ° −

01sin 5 sin 252B B By d v gt t = − ° = ° −

Ratio: ( )1

0 20 0

0

sin 25sin 5 1 or cos 25 tan 5 sin 25cos 5 cos 25 2

B BBB

B B

v gt ty v v gtx v t

° −°= − = − ° ° = ° −

° °

or ( ) ( )( ) ( )0 2 482 cos 25 tan 5 sin 25 cos 25 tan 5 sin 25 4.912 s9.81B

vtg

= ° ° + ° = ° ° + ° =

Then, ( )( )( )0 48 cos 25 4.912cos 25cos5 cos5 cos5

B Bx v td°°

= = =° ° °

215 m d =

25°

d

AB 5°

v0

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PROBLEM 11.102

Water flows from a drain spout with an initial velocity of 0.76 m/s at an angle of 15° with the horizontal. Determine the range of values of the distance d for which the water will enter the trough BC.

SOLUTION

Vertical motion: ( ) 20 0

12yy y v t gt= + −

( ) ( )20 0

1 02 ygt v t y y− − − =

With 0 3 m and 0.36 m at level ,y y BC= = 0 2.64 my y− =

( )0 0.76sin15 0.19670 m/syv = − ° = −

Then, 24.905 0.19670 2.64 0t t+ − =

Solving the quadratic equation,

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

20.19670 0.19670 4 4.905 2.640.7139 s

2 4.905t

− + − −= =

Horizontal motion: ( ) ( )( )0 0.76cos15 0.7139 0.524 mxx v t= = ° =

For water impact at C, 0.524 ft with 0.62 0.096 mc cx d x= = − = −

For water impact at B, 0.524 mBx d= =

So, 0.096 m 0.524 md− < <

Owing to the foot of the wall, d cannot be negative; hence, the allowable range of d is 0 0.524 m d< <

15°

3 m

v0

A

B C0.36 m

0.62 md

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Fig. 11.14

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Fig. 11.15a

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Fig. 11.15b

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Fig. 11.15c

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Fig. 11.15d

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Photo 11.4

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071208

선

071208

선

071208

선


Fig. 11.23

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Fig. P11.177

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SeventhEdition

44 - 49

Sample Problem 11.10

A motorist is traveling on curved section of highway at 60 mph. The motorist applies brakes causing a constant deceleration rate.

Knowing that after 8 s the speed has been reduced to 45 mph, determine the acceleration of the automobile immediately after the brakes are applied.

SOLUTION:

• Calculate tangential and normal components of acceleration.

• Determine acceleration magnitude and direction with respect to tangent to curve.DynaDEn



SeventhEdition

45 - 49


ft/s66mph45ft/s88mph60

==

SOLUTION:• Calculate tangential and normal components of

acceleration.( )

( )2

22

2

sft10.3

ft2500sft88

sft75.2

s 8sft8866

===

−=−

=∆∆

=

ρva

tva

n

t

• Determine acceleration magnitude and direction with respect to tangent to curve.

( ) 2222 10.375.2 +−=+= nt aaa 2sft14.4=a

75.210.3tantan 11 −− ==

t

naaα °= 4.48α

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SeventhEdition

46 - 49


Rotation of the arm about O is defined by θ = 0.15t2 where θ is in radians and tin seconds. Collar B slides along the arm such that r = 0.9 - 0.12t2 where r is in meters.

After the arm has rotated through 30o, determine (a) the total velocity of the collar, (b) the total acceleration of the collar, and (c) the relative acceleration of the collar with respect to the arm.

SOLUTION:

• Evaluate time t for θ = 30o.

• Evaluate radial and angular positions, and first and second derivatives at time t.

• Calculate velocity and acceleration in cylindrical coordinates.

• Evaluate acceleration with respect to arm.

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SeventhEdition

47 - 49

Sample Problem 11.12SOLUTION:• Evaluate time t for θ = 30o.

s 869.1rad524.0300.15 2

==°==

ttθ

• Evaluate radial and angular positions, and first and second derivatives at time t.

2

2

sm24.0

sm449.024.0m 481.012.09.0

−=

−=−==−=

r

trtr

&&

&

2

2

srad30.0

srad561.030.0rad524.015.0

=

==

==

θ

θθ

&&

& tt

DynaDEn



SeventhEdition

48 - 49

Sample Problem 11.12• Calculate velocity and acceleration.

( )( )

rr

r

vvvvv

rvsrv

θθ

θ

β

θ

122 tan

sm270.0srad561.0m481.0m449.0

−=+=

===

−==&

&

°== 0.31sm524.0 βv

( )( )

( )( ) ( )( )

rr

r

aaaaa

rra

rra

θθ

θ

γ

θθ

θ

122

2

2

2

22

2

tan

sm359.0

srad561.0sm449.02srad3.0m481.0

2sm391.0

srad561.0m481.0sm240.0

−=+=

−=

−+=

+=

−=

−−=

−=

&&&&

&&&

°== 6.42sm531.0 γa

DynaDEn



SeventhEdition

49 - 49

Sample Problem 11.12• Evaluate acceleration with respect to arm.

Motion of collar with respect to arm is rectilinear and defined by coordinate r.

2sm240.0−== ra OAB &&

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PROBLEM 11.145 From a photograph of a homeowner using a snowblower, it is determined that the radius of curvature of the trajectory of the snow was 9 m as the snow left the discharge chute at A. Determine (a) the discharge velocity

Av of the snow, (b) the radius of curvature of the trajectory at its maximum height.

SOLUTION

(a) The acceleration vector is 9.81 m/s2 .

At point A. Tangential and normal components of a are as shown in the sketch.

2cos 40 9.81cos 40 7.51 m/sna a= ° = ° =

( ) ( )( )2 2 29 7.51 67.59 m /sA A A nv aρ= = =

8.22 m/sA =v 40°

8.22 cos 40 6.30 m/sxv = ° =

(b) At maximum height, 6.30 m/sxv v= =

29.81 m/s ,na g= =

( )22 6.309.81n

va

ρ = = 4.05 mρ =

vA

A 40°

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PROBLEM 11.184 Assuming a uniform acceleration of 23.5 m/s and knowing that the speed of a car as it passes A is 50 km/h, determine (a) the time required for the car to reach B, (b) the speed of the car as it passes B.

SOLUTION

Let the origin lie at point A. 0, 50 mA Bx x= =

250 km/h 13.889 m/s, 3.5 m/sAv a= = =

(a) 212B A Ax x v t at= + +

2.69 st =

(b)

83.9 km/hBv =

352. t 2 + 13.889t – 50 = 0 t = –10.63s, 2.69s

vB = vA + at = 13.889 + (3.5) (2.69) = 23.3 m/s

vA = 50 km/h

50 m

A B

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PROBLEM 11.185 Block B starts from rest and moves downward with a constant acceleration. Knowing that after slider block A has moved 400 mm its velocity is 4 m/s, determine (a) the accelerations of A and B, (b) the velocity and the change in position of B after 2 s.

SOLUTION

Constraint of cable with Ax increasing to the left and By increasing downward.

3 constant, 3 0A B A Bx y x y+ = ∆ + ∆ =

3 0, 3 0A B A Bv v a a+ = + =

Block B moves downward so that Bv and Ba are positive.

Since, 3 , A B Av v v= − is negative, i.e. to the right.

For block A, 2 2A A Av a x= ∆

(a) ( ) ( )2 2

2( 4) 20 m/s2 2 0.4

AA

A

vax

−= = = −

∆ − or 20 m/s2

A =a

21 6.67 m/s3B Aa a= − = 6.67 m/s2

B =a

(b) At 2 s,t =

( )( )6.67 2B Bv a t= = 213.33 m/sBv =

( )( )221 1 6.67 22 2B By a t∆ = = 13.33 m2

By∆ =

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