Chapter 4, Problem 17SP

(a) A 600-N load hangs motionlessly in Fig. 4-8. Assume the ropes to all be vertical and the pulleys to be weightless and frictionless. (a) What is the tension in the bottom hook attached, via a ring, to the load? (b) How many lengths of rope support the movable pulley? (c) What is the tension in the long rope? (d) How much force does the man apply? (e) How much force acts downward on the ceiling?

To determine

The magnitude of the tension in the bottom hook that is attached by a ring to the load in the Figure $4.8$.

Answer to Problem 17SP

Solution:

$600\text{ N}$

Explanation of Solution

Given data:

Refer to the figure 4-8.

Theload hanging motionlessly is $\text{600 N}$.

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:

Consider all pulleys and hook as a system and draw its free body diagram:

In the above diagram, $600\text{ N}$ is the load hanging at rest, $T_1$ is the tension in the bottom hook that is attached by a ring, $T_3$ and $T_4$ are the tensions in the ropeon pulley $2$, $T_2$ and $T_5$ are the tensions in the rope on pulley $3$, $F_2$ is the force applied by a man, and $F_1$ is the force acts downward on the ceiling.

Here, pulley $2$ and pulley $3$ are hinged at the same point.

Since thesame long rope is passing over the pulley $2$ and $3$, the tensions $T_5$, $T_4$, $T_3$, and $T_2$ will be thesame. Therefore,

$T_5=T_4=T_3=T_2$ …… $\left(1\right)$

Draw the free body diagram of the bottom hook, which is attached via a ring to the load.

In above the diagram, $T_1$ is the tension attached in a bottom hook.

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 $600\text{ N }$ for $W$

$\begin{array}{c}{T}_1-600\text{ N}=0\\ T_1=600\text{ N}\end{array}$

Conclusion:

The magnitude of the tension attached in a bottom hook is $600\text{ N}$.

To determine

The number of sections of the rope that supports the movable pulleyin the Figure $4.8$.

Answer to Problem 17SP

Solution:

$3$

Explanation of Solution

Given data:

Refer to the figure 4-8.

The load hanging at rest is $\text{600 N}$.

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 the schematic diagram from the first part (a).

To observed the free body diagram, there are only three ropes that support the movable pulley.

Conclusion:

A segment of the rope that supports the movable pulley is $3$.

To determine

The magnitude of the tension in the long ropein the Figure $4.8$.

Answer to Problem 17SP

Solution:

$200\text{ N}$

Explanation of Solution

Given data:

Refer to the figure 4-8.

The load hanging at rest is $\text{600 N}$.

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 pulley $1$:

In the above diagram, $T_1$ is the tension attached in a bottom hook in downward direction and tensions in the long rope are $T_4$, $T_3$, and $T_2$.

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,

$\left(T_2+T_3+T_4\right)-T_1=0$

Refer to the equation (1)

Substitute $600\text{ N}$ for $T_1$, $T_2$ for $T_3$, and $T_2$ for $T_4$

$\begin{array}{c}\left(T_2+T_2+T_2\right)-600\text{ N}=0\\ 3T_2-600\text{ N}=0\\ T_2=\frac{600\text{ N}}{3}\\ =200\text{ N}\end{array}$

Conclusion:

The magnitude of the tension in the long rope is $200\text{ N}$.

To determine

The maximum force applied by the man to pull the rope in the Figure $4.8$.

Answer to Problem 17SP

Solution:

$200\text{ N}$

Explanation of Solution

Given data:

Refer to the figure 4-8.

The load hanging motionlessly is $\text{600 N}$.

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 a man applied a pulling force on the rope:

To observed the above diagram, $F_2$ is the pulling force applied by a man on the rope.

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,

$T_5-F_2=0$

Since the same rope is passing over the pulley, the tension $T_5$ is equal to the $T_2$.

Substitute $200\text{ N}$ for $T_5$

$\begin{array}{c}200\text{ N}-F_2=0\\ F_2=200\text{ N}\end{array}$

Conclusion:

The magnitude of the force applied by the man is $200\text{ N}$.

To determine

The maximum force that is acting downward on the ceilingin the Figure 4.8.

Answer to Problem 17SP

Solution:

$\text{800 N}$

Explanation of Solution

Given data:

Refer to the figure 4-8.

The load hanging motionlessly is $\text{600 N}$.

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:

Consider both the pulley 3 and pulley 3 as a system and draw theirfree body diagram:

In the above diagram, $F_1$ is the force on the ceiling that pulls upward and tensions $T_5$, $T_4$, $T_3$, and $T_2$ in the rope pull downward.

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,

$F_1-\left(T_2+T_3+T_4+T_5\right)=0$ …… (2)

Refer to the equation (1)

$T_5=T_4=T_3=T_2$

Substitute $200\text{ N}$ for $T_2$, $200\text{ N}$ for $T_3$, $200\text{ N}$ for $T_4$, and $200\text{ N}$ for $T_5$

$\begin{array}{c}{F}_1-\left(200\text{ N}+200\text{ N}+200\text{ N}+200\text{ N}\right)=0\\ F_1=800\text{ N}\end{array}$

Since $F_1$ is the upward force on the ceiling, it has the same magnitude when a force acts downward on the ceiling, which is equal to $800\text{ N}$.

Conclusion:

The maximum force acting on the ceiling is $800\text{ N}$.