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 4, Problem 16SP

The load in Fig. 4-7 is hanging at rest. Take the ropes to all be vertical and the pulleys to be weightless and frictionless. (a) How many segments of rope support the combination of the lower pulley and load? (b) What is the downward force on the lowest pulley (the “floating” one)? (c) What must be the total upward force exerted on the floating pulley by the two lengths of rope? (d) What is the upward force exerted on the floating pulley by each length of rope supporting it? (e) What is the tension in the rope wound around the two pulleys? (f) How much force is the man exerting? (g) What is the net downward force acting on the uppermost pulley? (h) How much force acts downward on the hook in the ceiling?

(a)

Expert Solution
Check Mark
To determine

The segments of the rope that support the combination of the lower pulley given inFigure 4.7.

Answer to Problem 16SP

Solution:

2

Explanation of Solution

Given data:

The weight of the hanging block is 200 N.

The load is hanging on rest, so there will be no acceleration.

Formula used:

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

Fx=0Fy=0

Here, Fx is the sum of the forces in the x- direction or horizontal direction and Fy is the sum of the forces in a y-direction or vertical direction.

Explanation:

Draw the free body diagram of the system:

Schaum's Outline of College Physics, Twelfth Edition (Schaum's Outlines), Chapter 4, Problem 16SP , additional homework tip  1

In the above diagram, point 1 is denoting the lowermost pulley, point 2 is denoting the uppermost pulley, 200 N is the load or weight of the hanging block, which is at rest, T2, T3, and T4 are the tension in the rope, which is passing over the upper mostpulley, T1 is the tension in the rope on the lowest pulley, T5 is the tension in the rope, which is a hook in the ceiling, and F is the force applied by a man on the rope.

Since the same rope is passing over theupper pulleyand lower pulley, the magnitude of tensionthroughout the rope should be same. Therefore, T2, T3, and T4 will be same. Hence,

T2=T3=T4

Refer to the above diagram and consider the pulley 1. In this case, T2 and T3 act along the upward direction and T1 acts along the downward direction. Therefore, apply the first equilibrium condition along the vertical direction:

Fy=0

Consider the upward forces as positive and the downward forces as negative. Therefore,

T2+T3T1=0T1=T2+T3

Substitute T3 for T2

T1=T3+T3=2T3

Conclusion:

Therefore, two segments of the rope support the combination of the lower pulley and load.

(b)

Expert Solution
Check Mark
To determine

The magnitude of the downward force applied to the lowest pulleyin the Figure 4.7.

Answer to Problem 16SP

Solution:

200 N

Explanation of Solution

Given data:

The load is 200 N.

The load is hanging on rest, so there will be no acceleration.

Formula used:

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

Fx=0Fy=0

Here, Fx is the sum of the forces in the x- direction or horizontal direction and Fy is the sum of the forces in the y-direction or vertical direction.

Explanation:

Draw the free body diagram when 200 N block or hanging load:

Schaum's Outline of College Physics, Twelfth Edition (Schaum's Outlines), Chapter 4, Problem 16SP , additional homework tip  2

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

Fy=0

Consider the direction of the upward forces is positive and the direction of the downward forces is negative. Therefore,

T1W=0

Substitute 200 N  for W

T1200 N=0T1=200 N

The tension forces of 200 N, acting in the rope, pule the load in upward direction and weight of the 200 N block moves along the downward direction. Sum of all the forces is zero because theload is motionless. So, the downward force applied to the lowest pulley is equal to 200 N.

Conclusion:

Therefore, the magnitude of the downward force on the lowest pulley is 200 N.

(c)

Expert Solution
Check Mark
To determine

The magnitude of the total upward force exertedon the floating pulley by the two lengths of ropein the Figure 4.7.

Answer to Problem 16SP

Solution:

200 N

Explanation of Solution

Given data:

The load is 200 N.

The load is hanging on rest, so there will be no acceleration.

Formula used:

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

Fx=0Fy=0

Here, Fx is the sum of the forces in the x- direction or horizontal direction and Fy is the sum of the forces in the y-direction or vertical direction.

Explanation:

Draw the free body diagram of the lowermost pulley:

Schaum's Outline of College Physics, Twelfth Edition (Schaum's Outlines), Chapter 4, Problem 16SP , additional homework tip  3

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

Fy=0

Consider the direction of the upward forces is positive and the direction of the downward forces is negative. Hence,

T3+T2T1=0

Substitute 200 N for T1

T3+T2200 N=0T3+T2=200 N

Conclusion:

The magnitude of the total upward force applied on the pulley by the two lengths of the rope is 200 N.

(d)

Expert Solution
Check Mark
To determine

The magnitude of the upward force applied on the floating pulley by each length of rope supporting the pulleyin the figure 4.7.

Answer to Problem 16SP

Solution:

100 N

Explanation of Solution

Given data:

The weight of the person is 200 N.

The load is hanging on rest, so there will be no acceleration.

Formula used:

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

Fx=0Fy=0

Here, Fx is the sum of the forces in the x- direction or horizontal direction and Fy is the sum of the forces in the y-direction or vertical direction.

Explanation:

Draw the free body diagram of the lowermost pulley:

Schaum's Outline of College Physics, Twelfth Edition (Schaum's Outlines), Chapter 4, Problem 16SP , additional homework tip  4

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

Fy=0

Consider the direction of the upward forces is positive and the direction of the downward forces is negative. Hence,

T3+T2T1=0

Substitute 200 N for T1 and T2 for T3

2T2200 N=0T2=200 N2=100 N

Conclusion:

The magnitude of the upward force applied on the floating pulley by each length of the rope is 100 N.

(e)

Expert Solution
Check Mark
To determine

The magnitude of the tensionin the rope wound around the two pulleysin the Figure 4.7.

Answer to Problem 16SP

Solution:

100 N

Explanation of Solution

Given data:

The load is 200 N.

The load is hanging on rest, so there will be no acceleration.

Formula used:

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

Fx=0Fy=0

Here, Fx is the sum of the forces in the x- direction or horizontal direction and Fy is the sum of the forces in the y-direction or vertical direction.

Explanation:

Refer to the diagram drawn in the subpart (a). A single rope is wounded on the both the polis. So, the magnitude of tension should be same throughout the rope. Therefore,

T2=T3=T4

Substitute 100 N for T2

T4=100 N

Conclusion:

The magnitude of the tension in the rope wound around the pulley is 100 N.

(f)

Expert Solution
Check Mark
To determine

The maximum force applied by the man to pull the ropein the Figure 4.7.

Answer to Problem 16SP

Solution:

100 N

Explanation of Solution

Given data:

The load is 200 N.

The load is hanging on rest, so there will be no acceleration.

Formula used:

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

Fx=0Fy=0

Here, Fx is the sum of the forces in the x- direction or horizontal direction and Fy is the sum of the forces in the y-direction or vertical direction.

Explanation:

Draw the free body diagram of the unwound rope, which is in the right side of the upper pulley, when a man exerts a force F to pull the rope.

Schaum's Outline of College Physics, Twelfth Edition (Schaum's Outlines), Chapter 4, Problem 16SP , additional homework tip  5

To observed the schematic diagram of the problem, F is the force applied by a man on the rope.

The force F is the downward force of the hand on the rope. It has the same magnitude as the tension of the rope on the hand, which is T4.

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

Fy=0

Consider the direction of the upward forces is positive and the direction of the downward forces is negative. Hence,

T4F=0

Substitute 100 N for T4

100 NF=0F=100 N

Conclusion:

The magnitude of the maximum force applied by the man is 100 N.

(g)

Expert Solution
Check Mark
To determine

The magnitude of the net downward force acting on the uppermost pulleyin the Figure 4.7.

Answer to Problem 16SP

Solution:

300 N

Explanation of Solution

Given data:

The load is 200 N.

The load is hanging on rest, so there will be no acceleration.

Formula used:

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

Fx=0Fy=0

Here, Fx is the sum of the forces in the x- direction or horizontal direction and Fy is the sum of the forces in the y-direction or vertical direction.

Explanation:

Draw the free body diagram of the upper pulley:

Schaum's Outline of College Physics, Twelfth Edition (Schaum's Outlines), Chapter 4, Problem 16SP , additional homework tip  6

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

Fy=0

Consider the direction of the upward forces is positive and the direction of the downward forces is negative. Hence,

T5T4T3T2=0

Substitute 100 N for T4, 100 N for T2, and 100 N for T3

T5100 N100 N100 N=0T5=300 N

The net downward force acting on the uppermost pulleyhas the same magnitude as the tension of the rope, which is T5.

Conclusion:

Therefore, the total downward force acting on the upper pulley is 300 N.

(h)

Expert Solution
Check Mark
To determine

The maximum force that is acting downward on the hook in the ceilingin the Figure 4.7.

Answer to Problem 16SP

Solution:

300 N

Explanation of Solution

Given data:

The load is 200 N.

The load is hanging on rest, so there will be no acceleration.

Formula used:

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

Fx=0Fy=0

Here, Fx is the sum of the forces in the x- direction or horizontal direction and Fy is the sum of the forces in the y-direction or vertical direction.

Explanation:

Refer to the schematic diagram of the problem from the subpart (a).

The maximum force applied downward on the hook in the ceiling is equal to the same magnitude of the tension T5, which is equal to 300 N.

Conclusion:

Therefore, themaximum force applied downward on the hook in the ceilingis 300 N.

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Chapter 4 Solutions

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

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