91Ó°ÊÓ

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

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.

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,

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

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

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.

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

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

Therefore, the tension in the rope is 931 N

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

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

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

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.

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$1.05×{10}^3\mathrm{N}$

In part c, we have, T = 932N

So, force on ceiling is,

F = 2T

=1864 N

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

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

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

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