link full download solution manual for engineering mechanics dynamics 13th edition by russell c hibbeler

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.. P 308 SOLUTION Free-B

Trang 1

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

v = 10 ft/s

F = (2.5t) lb

SOLUTION

+

¢

10

≤a : ©Fx = max; 2.5t = 32.2

a =8.05t

dv =a dt n

L0

t

L10

8.05t dt

v =4.025t2 + 10 When t = 3 s,

v = 46.2 ft>s

ds = v dt

L0 ds =L0(4.025t2 + 10) dt

s =1.3417t3 + 10t

When t = 3 s,

s = 66.2 ft

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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

Free-Body Diagram: The kinetic friction Ff = mkN is directed to the left to oppose the

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

Dissemination and States . World permitted 1

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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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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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

P

308

SOLUTION

Kinematics: The acceleration a of the crate will be determined first since its motion is

known

-( : ) v v0 2ac(s s0)

42 = 02 + 2a(5 - 0) a =

1.60 m>s2 :

Free-Body Diagram: Here, the kinetic friction Ff = mkN = 0.3N is required to be directed

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

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

Web)

N = 490.5 - 0.5P

copyright

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Using the results of N and a,

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: ©F x

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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

Trang 4

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

Web)

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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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Trang 5

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

Free-Body Diagram: Here, the kinetic friction Ff = mkN = 0.25N is required to be

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.

Equations of Motion: Here, ay¿ = 0 Thus,

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

N = 224.79 N

teaching

or

a = 0.8993 m>s

©Fx¿ = may¿;

.

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

2 2 United of learning the is

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

and

v = v0 + 2ac(s - s0)

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v = 3.29 m>s

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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

the motion

SOLUTION

Kinematics: For 0 … t 6 10 s v =

60

t = {6t} ft>s Applying equation a =

dv

,

a =

dv

= 6 ft>s2

For 10 6 t … 30 s, v - 60 = 80- 60 , v = {t + 50} ft>s Applying equation

30- 10

F

80

60

10

t (s)

30

v

Equation of Motion:

For 0 … t 6 10 s

;+

aFx = max ; F =

For 10 6 t … 30 s

;+

aFx = max ; F =

dt =

¢ 350032.2 ≤(6) =

¢ 350032.2 ≤(1) =

1 ft>s2

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Trang 8

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

is mk = 0.3

p

20

SOLUTION

Equations of Equilibrium: If the crate is on the verge of slipping, Ff = ms N = 0.5N

From FBD(a),

+

= 0; P cos 20° - 0.5N = 0

(2)

Solving Eqs.(1) and (2) yields

P = 353.29 N N = 663.97 N

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

contacting surface is F f = m k N = 0.3N since the crate is moving From FBD(b),

lear ning

a = 1.66 m>s2

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Trang 9

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

seconds

p

20

SOLUTION

Equations of Equilibrium: At t = 2 s, P = 90 A 22B = 360 N From FBD(a)

+

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),

©Fx = max ;

N - 80(9.81) + 360 sin 20° = 80(0)

laws

N = 661.67 N Web)

teaching

360 cos 20° - 0.3(661.67) = 80a copyright Wide

World

a = 1.75 m>s2

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