Solution manual for engineering mechanics dynamics 13th edition by russell c hibbeler

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13–1.

5t2i - 4tj - 1k6 lb, and F3 = 5 - 2ti6 lb, where t is in

seconds Determine the distance the ball is from the origin 2 s

after being released from rest

z

F

3

SOLUTION

©F = ma; (2i + 6j - 2tk) + (t2i - 4tj - 1k) - 2ti - 6k = ¢ 32 6 2 ≤(axi + ay j + azk)

Equating components:

¢ 6 ≤ax=t2 -2t+2 ¢

6 ≤ay= -4t+6 ¢

6 ≤az= -2t-7

32.2 32.2 32.2

Since dv = a dt, integrating from n = 0, t = 0, yields

6 t3 6 6

¢ 32.2≤vx= 3 - t2 + 2t ¢ 32.2≤vy= - 2t2 + 6t ¢ 32.2≤vz = - t2 - 7t laws or

¢ 32.2 ≤ sx = 12 - 3 + t ¢ 32.2 ≤ sy= - 3 + 3t ¢ 32.2≤sz = - 3 teaching Web)

2

Dissemination Since ds = v dt, integrating from s = 0, t = 0 yields copyright Wide

instructors permitted 6 t 4 t 3 6 2t 3 6

2 2

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learning not When t = 2 s then, sx = 14.31 ft, sy = 35.78 ft sz = - 89.44of ft on the is United

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s = (14.31)2 + (35.78)2 + ( - 89.44)2 = 97.4 ft Ans. Thus, protected by the of the

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Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,

or transmission in any form or by any means, electronic, mechanical,

photocopying, recording, or likewise For information regarding permission(s), write to:

Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458

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The 10-lb block has an initial velocity of 10 ft>s on the smooth

plane If a force F = 12.5t2 lb, where t is in seconds, acts on the

block for 3 s, determine the final velocity of the block and the

distance the block travels during this time

a =8.05t

dv =a dtn

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If the coefficient of kinetic friction between the 50-kg crate and

the ground is mk = 0.3, determine the distance the crate travels

and its velocity when t = 3 s The crate starts from rest, and P =

200 N

P

308

SOLUTION

motion of the crate which is to the right, Fig a.

Equations of Motion: Here, ay = 0 Thus,

+ c ©Fy = 0; N - 50(9.81) + 200 sin 30° = 0

N = 390.5 N

+ ©Fx= max; 200 cos 30° - 0.3(390.5) = 50a

:

a = 1.121 m>s2

Kinematics: Since the acceleration a of the crate is constant,

laws or +

A : B v = v0 + act teaching Web)

v = 0 + 1.121(3) = 3.36 m>s copyrightAns. Wide

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of instructors not 2 the A + B s = s 0 + v 0 t + a c t Uniteduse learning is : 2 on and 1

(1.121) A32B = 5.04 m by the student work s = 0 + 0 + Ans. 2

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If the 50-kg crate starts from rest and achieves a velocity of v =

4 m>s when it travels a distance of 5 m to the right, determine

the magnitude of force P acting on the crate The coefficient of

kinetic friction between the crate and the ground is mk = 0.3

to the left to oppose the motion of the crate which is to the right, Fig a.

Using the results of N and a,

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13–5.

The water-park ride consists of an 800-lb sled which slides

from rest down the incline and then into the pool If the

frictional resistance on the incline is Fr = 30 lb, and in

the pool for a short distance Fr = 80 lb, determine how

fast the sled is traveling when s = 5 ft

100 ft SOLUTION

+ b aF x = ma x; 800 sin 45° - 30 = 800a

32.2

100 ft

a = 21.561 ft>s2

s

v12 = v02 + 2a c (s - s0)

v12 = 0 + 2(21.561)(100 2

2 - 0))

v1 = = 78.093 ft>s

+ 800

; aF x = max; -80 = 32.2a

laws

a = -3.22 ft>s2

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v2 = (78.093) + 2( -3.22)(5 - 0) permitted

v22= v12 + 2a c (s2 - s1)

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v 2 = 77.9 ft>s Aisns. use and

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photocopying, recording, or likewise For information regarding permission(s), write to:Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

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If P = 400 N and the coefficient of kinetic friction between the

50-kg crate and the inclined plane is mk = 0.25, determine the

velocity of the crate after it travels 6 m up the plane The crate

starts from rest

directed down the plane to oppose the motion of the crate which is assumed to be

directed up the plane The acceleration a of the crate is also assumed to be directed

up the plane, Fig a.

©Fy¿ = may¿; N + 400 sin 30° - 50(9.81) cos 30° = 50(0)

a = 0.8993 m>s

©Fx¿ = may¿;

.

400 cos 30° - 50(9.81) sin 30° - 0.25(224.79) = 50a

is solely work Ans.

P

30°

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If the 50-kg crate starts from rest and travels a distance of 6 m

up the plane in 4 s, determine the magnitude of force P acting

on the crate The coefficient of kinetic friction between the

crate and the ground is mk = 0.25

SOLUTION 30°

Kinematics: Here, the acceleration a of the crate will be determined first since its motion is known.

s = s0 + v0t + 1act2

2

6 = 0 + 0 + 1a(42)

2

a = 0.75 m>s2

Free-Body Diagram: Here, the kinetic friction Ff = mkN = 0.25N is required to be laws Web) teaching directed down the plane to oppose the motion of the crate which is directed up the or Equations of Motion: Here, ay¿ = 0 Thus, Dissemination copyright Wide plane, Fig a.

©Fy¿ = may¿; N + P sin 30° - 50(9.81) cos 30° = 50(0) instructors World permitted States N = 424.79 - 0.5P learning on is not

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Using the results of N and a, for student

by the ©Fx¿ = max¿; P cos 30° - 0.25(424.79 - 0.5protectedP) - 50(9.81) sinof30° = 50(0.75) assessing the is solely work

P = 392 N work provided and of thisintegrity Ans. This is

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P

30°

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The speed of the 3500-lb sports car is plotted over the 30-s time

period Plot the variation of the traction force F needed to cause

assessing

109 lb is solely work Ans

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The crate has a mass of 80 kg and is being towed by a chain

which is always directed at 20° from the horizontal as shown If

the magnitude of P is increased until the crate begins to slide,

determine the crate’s initial acceleration if the coefficient of

static friction is ms = 0.5 and the coefficient of kinetic friction

: ©Fx = max ; 353.29 cos 20° - 0.3(663.97) = 80a instructors not permitted

Aisns

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The crate has a mass of 80 kg and is being towed by a chain

which is always directed at 20° from the horizontal as shown

Determine the crate’s acceleration in t = 2 s if the coefficient of

static friction is ms = 0.4, the coefficient of kinetic friction is

mk = 0.3, and the towing force is P = (90t2) N, where t is in

Since Ff 7 (Ff)max = ms N = 0.4(661.67) = 264.67 N, the crate accelerates

Equations of Motion: The friction force developed between the crate and its contacting

surface is Ff = mkN = 0.3N since the crate is moving From FBD(b),

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The safe S has a weight of 200 lb and is supported by the rope

and pulley arrangement shown If the end of the rope is given to

a boy B of weight 90 lb, determine his acceleration if in the

confusion he doesn’t let go of the rope Neglect the mass of the

pulleys and rope

S

SOLUTION

B

Equation of Motion: The tension T developed in the cord is the same throughout

the entire cord since the cord passes over the smooth pulleys

From FBD(a),

+ c ©F y 90

(1)

= ma y; T-90 = - a b a B

32.2

From FBD(b),

+ c ©F y = ma y ; 2T - 200 = - a 200b a S

(2)

32.2

laws or

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copyright Wide Taking time derivative twice yields = l permitted 2sS +s B instructors

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1 + T2 2a S + a B = 0 is(3)

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a B = -2.30 ft>s by the of the Ans. = 2.30 ft>sprotectedsolely Solving Eqs.(1),(2), and (3) yields a S = 1.15 ft>s2 T T = 96.43 lb assessing

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The boy having a weight of 80 lb hangs uniformly from the bar

Determine the force in each of his arms in t = 2 s if the bar is

moving upward with (a) a constant velocity of 3 ft>s, and (b) a

speed of v = 14t22 ft>s, where t is in seconds.

SOLUTION

(b) v = 4t2

a = 8t

+ c aF y = ma y ; 2T - 80 = 8018t2

32.2

At t = 2 s.

T = 59.9 lb Ans. laws or teaching Web) Dissemination copyright Wide instructors World permitted States

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The bullet of mass m is given a velocity due to gas pressure

caused by the burning of powder within the chamber of the gun

Assuming this pressure creates a force of F = F0 sin 1pt>t02

on the bullet, determine the velocity of the bullet at any instant

it is in the barrel What is the bullet’s maximum velocity? Also,

determine the position of the bullet in the barrel as a function of

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The 2-Mg truck is traveling at 15 m>s when the brakes on all its

wheels are applied, causing it to skid for a distance of 10 m

C

before coming to rest Determine the constant horizontal force

developed in the coupling C, and the frictional force developed

between the tires of the truck and the road during this time The

total mass of the boat and trailer is 1 Mg

SOLUTION

Kinematics: Since the motion of the truck and trailer is known, their common

acceleration a will be determined first.

a :+

b

0 = 152 + 2a(10 - 0)

a = -11.25 m>s2 = 11.25 m>s2 ;

Free-Body Diagram:The free-body diagram of the truck and trailer are shown in Figs (a)

and (b), respectively Here, F representes the frictional force developed when the truck

skids, while the force developed in coupling C is represented by T.

Using the results of a and T and referring to Fig (b),

United of learning the is

not on

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A freight elevator, including its load, has a mass of 500 kg It is

prevented from rotating by the track and wheels mounted along

its sides When t = 2 s, the motor M draws in the cable with a

speed of 6 m>s, measured relative to the elevator If it starts

from rest, determine the constant acceleration of the elevator

and the tension in the cable Neglect the mass of the pulleys,

motor, and cables

M

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The man pushes on the 60-lb crate with a force F The force is

always directed down at 30° from the horizontal as shown, and

its magnitude is increased until the crate begins to slide

Determine the crate’s initial acceleration if the

coefficient of static friction is ms = 0.6 and the coefficient of

N = 91.80 lb F = 63.60 lbSince N = 91.80 lb,

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The double inclined plane supports two blocks A and B, each

having a weight of 10 lb If the coefficient of kinetic friction

between the blocks and the plane is mk = 0.1, determine the

acceleration of each block

SOLUTION

60

30

Equation of Motion: Since blocks A and B are sliding along the plane, the friction forces

developed between the blocks and the plane are (Ff)A = mk NA = 0.1 NA and (Ff)B = mk

NB = 0.1NB Here, aA = aB = a Applying Eq 13–7 to FBD(a), we have

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ramp Determine the point where it strikes the ground at C.

How long does it take to go from A to C?

SOLUTION

40a32.2

2

1

of sale

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the ramp of vA = 10 ft>s and the coefficient of kinetic

friction along AB is mk = 0.2

SOLUTION

40a32.2

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The 400-kg mine car is hoisted up the incline using the cable

and motor M For a short time, the force in the cable is F =

13200t22 N, where t is in seconds If the car has an initial

velocity v1 = 2 m>s when t = 0, determine its velocity when t =

2 s

SOLUTION

17 8 15

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The 400-kg mine car is hoisted up the incline using the cable

and motor M For a short time, the force in the cable is F =

13200t22 N, where t is in seconds If the car has an initial

velocity v1 = 2 m>s at s = 0 and t = 0, determine the distance it

moves up the plane when t = 2 s

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