Schaum's Outline of College Physics, Twelfth Edition (Schaum's Outlines)
Schaum's Outline of College Physics, Twelfth Edition (Schaum's Outlines)
12th Edition
ISBN: 9781259587399
Author: Eugene Hecht
Publisher: McGraw-Hill Education
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Chapter 3, Problem 79SP

In Fig. 3-28, how large a force F is needed to give the blocks an acceleration of 3.0

m/s2 if the coefficient of kinetic friction between blocks and table is 0.20? How large

a force does the 1.50-kg block then exert on the 2.0-kg block?

Chapter 3, Problem 79SP, 3.79 [III] In Fig. 3-28, how large a force F is needed to give the blocks an acceleration of 3.0

Expert Solution & Answer
Check Mark
To determine

The maximum force F that is needed to accelerate the blocks by 3 m/s2, and the then reaction force between the blocks 2 kg and 1.5 kg while the coefficient of friction between blocks and table is 0.20 using the Fig. 3.28.

Answer to Problem 79SP

Solution:

22 N and 15 N

Explanation of Solution

Given data:

Arrangement of the blocks of mass 1.5 kg, 2 kg, and 1 kg isprovided in the Fig. 3.28.

Coefficient of friction between blocks and table is 0.20.

The needed acceleration of the blocksis 3 m/s2.

Formula used:

Write the expression for the first condition of the force’s equilibrium:

Fx=0Fy=0

Here, Fx is the net force in x- direction or horizontal direction and Fy is the net force in y-direction or vertical direction.

From the Newton’ s second law of the motion, the expression of the force is

Fx=max

Here, Fx is the netforce in the x direction, a is the acceleration of the blocks in x-direction, and m is the mass of the block.

Write the expression for the friction force:

Ff=μFN

Here, Ff is the friction force and FN is the normal force.

Sign convention: Consider the direction of the upward forces is positive and the direction of the downward forces is negative. Also, consider the leftward force as negative and rightward forces are positive.

Explanation:

Draw the free body diagram of a 1 kg block:

Schaum's Outline of College Physics, Twelfth Edition (Schaum's Outlines), Chapter 3, Problem 79SP , additional homework tip  1

In 1 kg block, W3 is the weight of the block 1 kg, FN3 is the normal force that act on the block 3, Ff3 is the friction force that oppose the motion of the block 3, and a is acceleration.

Recall the expression for the first condition of the force’s equilibrium:

Fy=0

From the free body diagram:

FN3m3g=0

Substitute 1 kg for m3 and 9.81 m/s2 for g

FN3(1 kg)(9.81 m/s2)=0FN3=9.81 N

Recall the expression for the friction force:

Ff3=μFN3

Substitute 9.81 N for FN3 and 0.20 for μ

Ff3=(0.20)(9.81 N)=1.962 N

Recall the expression for Newton’s second law of the motion:

Fx=ma

From free body diagram, the force equation is

N2Ff3=m3a

Substitute 1 kg for m3, 1.962 N for Ff3, and 3 m/s2 for g

N21.962 N=(1 kg)(3 m/s2)N2=4.9 N

Draw the free body diagram of the all the blocks:

Schaum's Outline of College Physics, Twelfth Edition (Schaum's Outlines), Chapter 3, Problem 79SP , additional homework tip  2

In 2 kg block, W2 is the weight of the 2 kg block, FN2 is the normal force that act on the block, N2 is the contact force between the 2 kg block and 1 kg block, Ff2 is the friction force that oppose the motion of the block, and a is acceleration.

Now, consider the 2 kg block and recall the expression for the first condition of the force’s equilibrium:

Fy=0

From free body diagram:

FN2m2g=0

Substitute 2 kg for m2 and 9.81 m/s2 for g

FN2(2 kg)(9.81 kg/m2)=0FN3=19.62 N

Recall the expression of the friction force:

Ff2=μFN2

Substitute 19.62 N for FN2 and 0.20 for μ

Ff2=(0.20)(19.62 N)=3.924 N

Recall the expression for Newton’s second law of the motion:

Fx=ma

From free body diagram:

N1N2Ff2=m2a

Substitute 2 kg for m3, 3.924 N for Ff3, 4.9 N for N2, and 3 m/s2 for a

N14.9 N3.924 N=(2 kg)(3 m/s2)N1=14.8 N15 N

Draw the free body diagram of the all the blocks:

Schaum's Outline of College Physics, Twelfth Edition (Schaum's Outlines), Chapter 3, Problem 79SP , additional homework tip  3

In the above diagram, 1.5 kg mass of the block, W1 is the weight of the block, FN1 is the normal force that act on the block, N1 is the contact force between the 1.5 kg block and 2 kg block, Ff1 is the friction force that oppose the motion of the block, and a is acceleration.

Now, consider the 1.5 kg block 1 andrecall the expression for the first condition of the force’s equilibrium:

Fy=0

From free body diagram:

FN1m1g=0

Substitute 1 kg for m1 and 9.81 m/s2 for g

FN1(1.5 kg)(9.81 m/s2)=0FN1=14.71 N

Recall the expression for the friction force:

Ff1=μFN1

Substitute 14.71 N for FN1 and 0.20 for μ

Ff1=(0.20)(14.71 N)=2.9 N

Recall the expression for Newton’s second law of the motion:

Fx=ma

From free body diagram:

FN1Ff1=m1a

Substitute 1.5 kg for m1, 2.9 N for Ff1, 14.8 N for N1, and 3 m/s2 for a

F14.8 N2.9 N=(1.5 kg)(3 m/s2)F=22.2 N22 N

Conclusion:

Therefore, the required force F will be 22 N to accelerate the blocks and the reaction force between the 1.5 kg block and 2 kg block is 15 N.

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Chapter 3 Solutions

Schaum's Outline of College Physics, Twelfth Edition (Schaum's Outlines)

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