Question

A man of mass 60 kg is standing on a light horizontal plank AB that is supported by two light strings shown in the figure. Distance of the man from the left end of the plank is equal to one third of the length of the plank. The pulleys are ideal. What force should the man apply on the string to keep the system in equilibrium. Also, find the reaction force applied by the plank on the man.

Answer 1

Force applied by man on string = 240 N, Reaction force by plank on man = 360 N

Answer 2

To solve this problem, we will analyze the forces and torques acting on the man and the plank separately, and then ensure the system is in equilibrium.

Let:

• M = mass of the man = 60 kg
• g = acceleration due to gravity (take as 10 m/s² for calculations)
• L = length of the plank AB
• F = force applied by the man on the string (which is the tension in the string he is holding)
• N = reaction force applied by the plank on the man (normal force)
• T_R = tension in the right string

The plank is light, so its mass is negligible. The pulleys are ideal, meaning they are massless and frictionless, so the tension in a continuous string passing over them remains constant.

1. Free Body Diagram (FBD) of the Man:

The forces acting on the man are:

• His weight, Mg, acting downwards.
• The upward force F from the string he is pulling.
• The upward normal force N from the plank.

For the man to be in equilibrium: Sum of upward forces = Sum of downward forces F + N = Mg --- (Equation 1)

2. Free Body Diagram (FBD) of the Plank:

The forces acting on the plank are:

• The downward normal force N exerted by the man on the plank (by Newton's third law, equal in magnitude to the normal force N exerted by the plank on the man). This force acts at a distance L/3 from the left end A.
• The upward tension F from the left string. The string the man is holding goes over the left pulley and is attached to the left end A of the plank. So, the force exerted by this string on the plank at A is F.
• The upward tension T_R from the right string. This string is attached to the right end B of the plank.

For the plank to be in equilibrium:

a) Translational Equilibrium (Vertical Forces): Sum of upward forces = Sum of downward forces F + T_R = N --- (Equation 2)

b) Rotational Equilibrium (Torques): Let's take torques about point A (the left end of the plank) to eliminate the torque due to force F at A.

• Torque due to N (downwards) at L/3: N * (L/3) (clockwise)
• Torque due to T_R (upwards) at L: T_R * L (counter-clockwise)

For rotational equilibrium, sum of clockwise torques = sum of counter-clockwise torques: N * (L/3) = T_R * L T_R = N/3 --- (Equation 3)

3. Solving the System of Equations:

We have three equations:

1. F + N = Mg
2. F + T_R = N
3. T_R = N/3

Substitute Equation 3 into Equation 2: F + (N/3) = N F = N - N/3 F = 2N/3 --- (Equation 4)

Now substitute Equation 4 into Equation 1: (2N/3) + N = Mg 5N/3 = Mg N = (3/5)Mg

Now that we have N, we can find F and T_R. Given M = 60 kg and taking g = 10 m/s²: N = (3/5) * 60 kg * 10 m/s² N = 3 * 12 * 10 N = 360 N

This is the reaction force applied by the plank on the man.

Now, find the force F applied by the man on the string using Equation 4: F = (2/3)N F = (2/3) * 360 N F = 2 * 120 N F = 240 N

This is the force the man should apply on the string.

Summary of Results:

• Force applied by the man on the string (F) = 240 N
• Reaction force applied by the plank on the man (N) = 360 N