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?

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 $200\text{ N}$.

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:

$\begin{array}{l}{\displaystyle \sum F_x}=0\\ {\displaystyle \sum F_y}=0\end{array}$

Here, ${\displaystyle \sum F_x}$ is the sum of the forces in the x- direction or horizontal direction and ${\displaystyle \sum F_y}$ is the sum of the forces in a y-direction or vertical direction.

Explanation:

Draw the free body diagram of the system:

In the above diagram, point $1$ is denoting the lowermost pulley, point $2$ is denoting the uppermost pulley, $200\text{ N}$ is the load or weight of the hanging block, which is at rest, $T_2$, $T_3$, and $T_4$ are the tension in the rope, which is passing over the upper mostpulley, $T_1$ is the tension in the rope on the lowest pulley, $T_5$ 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, $T_2$, $T_3$, and $T_4$ will be same. Hence,

$\begin{array}{c}{T}_2=T_3\\ =T_4\end{array}$

Refer to the above diagram and consider the pulley 1. In this case, $T_2$ and $T_3$ act along the upward direction and $T_1$ acts along the downward direction. Therefore, apply the first equilibrium condition along the vertical direction:

${\displaystyle \sum F_y}=0$

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

$\begin{array}{c}{T}_2+T_3-T_1=0\\ T_1=T_2+T_3\end{array}$

Substitute $T_3$ for $T_2$

$\begin{array}{c}{T}_1=T_3+T_3\\ =2T_3\end{array}$

Conclusion:

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

To determine

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

Answer to Problem 16SP

Solution:

$200\text{ N}$

Explanation of Solution

Given data:

The load is $200\text{ N}$.

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:

$\begin{array}{l}{\displaystyle \sum F_x}=0\\ {\displaystyle \sum F_y}=0\end{array}$

Here, ${\displaystyle \sum F_x}$ is the sum of the forces in the x- direction or horizontal direction and ${\displaystyle \sum F_y}$ is the sum of the forces in the y-direction or vertical direction.

Explanation:

Draw the free body diagram when $200\text{ N}$ block or hanging load:

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

${\displaystyle \sum F_y}=0$

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

$T_1-W=0$

Substitute $200\text{ N }$ for $W$

$\begin{array}{c}{T}_1-200\text{ N}=0\\ T_1=200\text{ N}\end{array}$

The tension forces of 200 N, acting in the rope, pule the load in upward direction and weight of the $200\text{ N}$ 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 $200\text{ N}$.

Conclusion:

Therefore, the magnitude of the downward force on the lowest pulley is $200\text{ N}$.

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:

$200\text{ N}$

Explanation of Solution

Given data:

The load is $200\text{ N}$.

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:

$\begin{array}{l}{\displaystyle \sum F_x}=0\\ {\displaystyle \sum F_y}=0\end{array}$

Here, ${\displaystyle \sum F_x}$ is the sum of the forces in the x- direction or horizontal direction and ${\displaystyle \sum F_y}$ is the sum of the forces in the y-direction or vertical direction.

Explanation:

Draw the free body diagram of the lowermost pulley:

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

${\displaystyle \sum F_y}=0$

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

$T_3+T_2-T_1=0$

Substitute $200\text{ N}$ for $T_1$

$\begin{array}{c}{T}_3+T_2-200\text{ N=0}\\ T_3+T_2=200\text{ N}\end{array}$

Conclusion:

The magnitude of the total upward force applied on the pulley by the two lengths of the rope is $200\text{ N}$.

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:

$100\text{ N}$

Explanation of Solution

Given data:

The weight of the person is $200\text{ N}$.

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:

$\begin{array}{l}{\displaystyle \sum F_x}=0\\ {\displaystyle \sum F_y}=0\end{array}$

Here, ${\displaystyle \sum F_x}$ is the sum of the forces in the x- direction or horizontal direction and ${\displaystyle \sum F_y}$ is the sum of the forces in the y-direction or vertical direction.

Explanation:

Draw the free body diagram of the lowermost pulley:

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

${\displaystyle \sum F_y}=0$

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

$T_3+T_2-T_1=0$

Substitute $200\text{ N}$ for $T_1$ and $T_2$ for $T_3$

$\begin{array}{c}2T_2-200\text{ N}=0\\ T_2=\frac{200\text{ N}}{2}\\ =100\text{ N}\end{array}$

Conclusion:

The magnitude of the upward force applied on the floating pulley by each length of the rope is $100\text{ N}$.

To determine

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

Answer to Problem 16SP

Solution:

$100\text{ N}$

Explanation of Solution

Given data:

The load is $200\text{ N}$.

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:

$\begin{array}{l}{\displaystyle \sum F_x}=0\\ {\displaystyle \sum F_y}=0\end{array}$

Here, ${\displaystyle \sum F_x}$ is the sum of the forces in the x- direction or horizontal direction and ${\displaystyle \sum F_y}$ 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,

$\begin{array}{c}{T}_2=T_3\\ =T_4\end{array}$

Substitute $100\text{ N}$ for $T_2$

$T_4=100\text{ N}$

Conclusion:

The magnitude of the tension in the rope wound around the pulley is $100\text{ N}$.

To determine

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

Answer to Problem 16SP

Solution:

$100\text{ N}$

Explanation of Solution

Given data:

The load is $200\text{ N}$.

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:

$\begin{array}{l}{\displaystyle \sum F_x}=0\\ {\displaystyle \sum F_y}=0\end{array}$

Here, ${\displaystyle \sum F_x}$ is the sum of the forces in the x- direction or horizontal direction and ${\displaystyle \sum F_y}$ 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.

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 $T_4$.

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

${\displaystyle \sum F_y}=0$

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

$T_4-F=0$

Substitute $100\text{ N}$ for $T_4$

$\begin{array}{c}100\text{ N}-F=0\\ F=100\text{ N}\end{array}$

Conclusion:

The magnitude of the maximum force applied by the man is $100\text{ N}$.

To determine

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

Answer to Problem 16SP

Solution:

$300\text{ N}$

Explanation of Solution

Given data:

The load is $200\text{ N}$.

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:

$\begin{array}{l}{\displaystyle \sum F_x}=0\\ {\displaystyle \sum F_y}=0\end{array}$

Here, ${\displaystyle \sum F_x}$ is the sum of the forces in the x- direction or horizontal direction and ${\displaystyle \sum F_y}$ is the sum of the forces in the y-direction or vertical direction.

Explanation:

Draw the free body diagram of the upper pulley:

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

${\displaystyle \sum F_y}=0$

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

$T_5-T_4-T_3-T_2=0$

Substitute $100\text{ N}$ for $T_4$, $100\text{ N}$ for $T_2$, and $100\text{ N}$ for $T_3$

$\begin{array}{c}{T}_5-100\text{ N}-100\text{ N}-100\text{ N}=0\\ T_5=300\text{ N}\end{array}$

The net downward force acting on the uppermost pulleyhas the same magnitude as the tension of the rope, which is $T_5$.

Conclusion:

Therefore, the total downward force acting on the upper pulley is $300\text{ N}$.

To determine

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

Answer to Problem 16SP

Solution:

$300\text{ N}$

Explanation of Solution

Given data:

The load is $200\text{ N}$.

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:

$\begin{array}{l}{\displaystyle \sum F_x}=0\\ {\displaystyle \sum F_y}=0\end{array}$

Here, ${\displaystyle \sum F_x}$ is the sum of the forces in the x- direction or horizontal direction and ${\displaystyle \sum F_y}$ 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 $T_5$, which is equal to $300\text{ N}$.

Conclusion:

Therefore, themaximum force applied downward on the hook in the ceilingis $300\text{ N}$.