Question

Figure shows a man sitting in a bosun’s chair that dangles from a massless rope, which runs over a massless, frictionless pulley and back down to the man’s hand. The combined mass of man and chair is 95.0 kg. With what force magnitude must the man pull on the rope if he is to rise (a) with a constant velocity and (b) with an upward acceleration of $\mathbf{1}\mathbf{.}\mathbf{30}\mathbf{}\mathbf{m}\mathbf{/}{\mathbf{s}}^{\mathbf{2}}$? (Hint:A free-body diagram can really help.) If the rope on the right extends to the ground and is pulled by a coworker, with what force magnitude must the co-worker pull for the man to rise (c) with a constant velocity and (d) with an upward acceleration of $\mathbf{1}\mathbf{.}\mathbf{30}\mathbf{m}\mathbf{/}{\mathbf{s}}^{\mathbf{2}}$? What is the magnitude of the force on the ceiling from the pulley system in (e) part a, (f) part b, (g) part c, and (h) part d?

Short answer

1. The tension T, in the rope is, 466 N
2. The tension T, in the rope is, 527 N
3. The tension T, in the rope is, 932 N
4. The tension T, in the rope is,$1.05\times {10}^3\mathrm{N}$
5. The force on ceiling is, 931 N
6. The force on ceiling is,$1.05\times {10}^3\mathrm{N}$
7. The force on ceiling is, $1.86\times {10}^3\mathrm{N}$
8. The force on ceiling is,$2.11\times {10}^3\mathrm{N}$

Step-by-step solution

Given information

The combined mass of man and chair is, m = 95.0 kg

Understanding the concept

Newton’s second law states that the net force acting on the body is equal to the product of mass and the acceleration of the body. The net force is equal to the vector sum of all the forces acting on the body.

This problem is based on Newton’s laws, especially second law and third law will be beneficial for solving this problem.

Formula:

F = ma (i)

Here, F is the net force, is the mass of the body, and is acceleration.

(a) Calculate the magnitude of force with which the man should pull on the rope if he is to rise with a constant velocity

When the man is pulling the rope, the total upward force will be 2T. So, the equation of motion can be written as,

$2T-mg=ma$ (i)

As the man is rising with constant velocity, its acceleration is zero.

So, the equation for tension will become,

$$\begin{gathered} T=\frac{mg}{2} \\ =\frac{92.0\mathrm{kg}\times 9.80\mathrm{m}/{\mathrm{s}}^2}{2} \\ =466\mathrm{N} \end{gathered}$$

Therefore, the magnitude of the force with which man should pull on is 466N.

(b) Calculate the magnitude of force with which the man should pull on the rope if he is to rise with an upward acceleration of 1.30 m/s2

When,$\mathrm{a}=1.30\mathrm{m}/{\mathrm{s}}^2$, the equation (i) for tension will become,

$$\begin{gathered} 2T-mg=ma \\ T=\frac{ma+mg}{2} \\ =\frac{95\mathrm{kg}(1.30\mathrm{m}/{\mathrm{s}}^2+9.80\mathrm{m}/{\mathrm{s}}^2)}{2} \\ =527\mathrm{N} \end{gathered}$$

Therefore, the tension in the rope is 527 N.

(c) Calculate the magnitude of force with which the man should pull on the rope if he is to rise with a constant velocity

When the co-worker is pulling the rope, then only reaction force of gravitational force will act in the upward force. So, the equation of motion will become,

T - mg = ma

So, when a = 0, the tension in the rope will be

$$\begin{gathered} T-mg=ma \\ T=mg \\ =95.0\mathrm{kg}\times 9.80\mathrm{m}/{\mathrm{s}}^2 \\ =931\mathrm{N} \end{gathered}$$

Therefore, the tension in the rope is 931 N

(d) Calculate the magnitude of force with which the man should pull on the rope if he is to rise with an upward acceleration of 1.30m/s2

When,$a=1.30\mathrm{m}/{\mathrm{s}}^2$the tension will be,

$$\begin{gathered} T-mg=ma \\ T=ma+mg \\ =95.0\mathrm{kg}(1.30\mathrm{m}/{\mathrm{s}}^2+9.80\mathrm{m}/{\mathrm{s}}^2) \\ =1.05\times {10}^3\mathrm{N} \end{gathered}$$

Therefore, the tension in the rope is $1.05\times {10}^3\mathrm{N}$.

(e) Calculate the magnitude of the force on the ceiling from the pulley system in part (a)

The magnitude of force on the ceiling will be equal to the total downward force.

As the rope is downward from both sides of pulley the force on the ceiling will be 2T.

In part a, we have,

T = 466N

So,

F = 2T

= 931N

Therefore, the magnitude of the force is 931N.

(f) Calculate the magnitude of the force on the ceiling    from the pulley system in part (b)

For part b, we have, T = 527 N

So, the force on ceiling is,

F = 2T

=1054 N

Therefore, the force on the ceiling is$1.05\times {10}^3\mathrm{N}$

(g) Calculate the magnitude of the force on the ceiling   from the pulley system in part (c)

In part c, we have, T = 932N

So, force on ceiling is,

F = 2T

=1864 N

Therefore, the force on the ceiling is $1.86\times {10}^3\mathrm{N}$.

(h) Calculate the magnitude of the force on the ceiling   from the pulley system in part (d)

We have, $T=1.05\times {10}^3\mathrm{N}$in part d, so, the force on ceiling is,

$$\begin{gathered} F=2T \\ =2.11\times {10}^3\mathrm{N} \end{gathered}$$

Therefore, the force on the ceiling is $2.11\times {10}^3\mathrm{N}$.