Q1DQ - University Physics with Modern Physics

Question

A man sits in a seat that is hanging from a rope. The rope passes over a pulley suspended from the ceiling, and the man holds the other end of the rope in his hands. What is the tension in the rope, and what force does the seat exert on him? Draw a free-body force diagram for the man.

Step-by-Step Solution

Answer

The tension in the rope is $\frac{\left(m_1+m_2\right)g}{2}$here $m_1$ is the mass of the man and $m_2$ is the mass of the chair.

The force the seat exerts on man is $\frac{m_{2}g-m_{1}g}{2}$.

Step 1: Significance of the tension.

Tension is the force that is being transmitted in an axial direction by using cable, chair or strings. The tension force mainly pulls a body from both ends in order to exert an action and also reaction force.

Step 2: Determination of the tension in the rope.

The diagram of the pulley is drawn below:

                                

 

 

In the diagram of the tension force, the tension is acting downwards and the rope is exerting an upward force on the pulley.

The acceleration of the whole system is zero as the system remains stationary.

The force diagram of the whole system is drawn below:

 

                                            

 

In the force diagram, the man and the chair are exerting downward force $\left(m_1+m_2\right)g$ here $m_1$ is the mass of the man and $m_2$ is the mass of the chair. Moreover, the upward tensions $T$ are also being experienced by the man and the chair.

According to the force diagram, the summation of the system of the forces in the $y$ direction is zero.

The equation of the summation of the system of the forces in the $y$ direction is expressed as:

 $$\begin{gathered} \sum F_y=0 \\ 2T-\left(m_1{m}_2\right)g=m{a}_3 \end{gathered}$$

Here, $\sum F_y$ is the summation of the forces in the $y$ direction, $T$ is the tension of the rope, $m$ is the mass of the system, $a_3$ is the acceleration of the system, $g$ is the acceleration due to gravity, $m_1$ is the mass of the man and $m_2$ is the mass of the chair.

Substitute $0$ for $a_3$ in the above equation.

$$\begin{gathered} 2T-\left(m_1+m_2\right)g=m\left(0\right) \\ 2T-\left(m_1+m_2\right)g=0 \\ 2T=\left(m_1+m_2\right)g \\ T=\frac{\left(m_1+m_2\right)g}{2} \end{gathered}$$

Thus, the tension in the rope is $\frac{\left(m_1+m_2\right)g}{2}$.

Step 3: Determination of the free-body diagram for the man

The free body diagram for the man is drawn below:

 

In the above diagram, the man is exerting a force $m_{1}g$ in the downwards direction. The rope is pulling the man at a tension $T$. The force of the chair on the man is $F_n$.

                                                 .

Step 4: Determination of the force exerted by the seat

The force diagram on the chair has been drawn below:

 

                                

 

In the above diagram, the chair is exerting a force $m_{2}g$ in the downwards direction and the rope is exerting a tension $T$ in the upwards direction. The force of the man on the chair is described as $F_n$.  

From the diagram of the man, the summation of the system of the forces in the $y$ direction is zero.

The equation of the summation of the system of the forces in the $y$ direction is expressed as:

 

 $$\begin{gathered} \sum F_y=0 \\ T+F_n-m_{1}g=m_1{a}_3 \end{gathered}$$

 

The above equation can be expressed as:

 $$\begin{gathered} T+F_n-m_{1}g=0 \\ T=m_{1}g-F_n \end{gathered}$$

 

The chair’s normal force on the man will be the same according to the Newton’s third law.

Hence, the equation of the normal force exerted by the chair is expressed as:

 $F_n=m_{2}g-T$

Substitute $\frac{\left(m_1+m_2\right)g}{2}$ for $T$ in the above equation.

$$\begin{gathered} F_n=m_{2}g-\frac{\left(m_1+m_2\right)g}{2} \\ =m_{2}g-\frac{1}{2}{m}_{2}g-\frac{1}{2}{m}_{1}g \\ =\frac{1}{2}{m}_{2}g-\frac{1}{2}{m}_{1}g \\ =\frac{m_{2}g-m_{1}g}{2} \end{gathered}$$

Thus, the force that seat exerts on him is $\frac{m_{2}g-m_{1}g}{2}$.