Question

In the given diagram, two rods of mass $m$, length $L$ are connected with each other at the top on which a man of mass 'M' is standing. If both the rods are connected by a string at their midpoint and the whole system is in equilibrium, then find the normal force acting on the rod from the ground & the tension force in the string connecting both the rods.

Answer 1

Normal force: $N = \frac{(M+2m)g}{2}$

Tension force: $T = \frac{(M+m)g}{3\sqrt{3}}$

Answer 2

The problem describes a system in equilibrium consisting of two rods, a man, and a string. We need to find the normal force on each rod from the ground and the tension in the string.

1. Normal Force (N) on each rod from the ground:

Consider the entire system (two rods + man). The total downward force is the sum of the weights: Total weight = Weight of man + Weight of two rods Total weight = $Mg + mg + mg = (M + 2m)g$

Since the system is in vertical equilibrium, the total upward normal force from the ground must balance the total downward weight. Let $N$ be the normal force on each rod. Due to symmetry, the normal force on both rods will be equal. Total upward force = $N + N = 2N$

Equating upward and downward forces: $2N = (M + 2m)g$ $N = \frac{(M + 2m)g}{2}$

2. Tension Force (T) in the string:

To find the tension, let's consider the equilibrium of one rod (say, the left rod). Draw a Free Body Diagram for the left rod. Let the base of the left rod be point P and the top joint be point Q. The string is attached at midpoint A. The angle of the rod with the ground is $\theta = 60^\circ$.

Forces acting on the left rod:

• Normal force (N): Acting vertically upwards at P.
• Weight of the rod (mg): Acting vertically downwards at its center of mass, which is at the midpoint of the rod (point A). The distance from P to A along the rod is $L/2$.
• Tension (T): Acting horizontally to the right at the midpoint A.
• Forces at the top joint Q: These are the forces exerted by the other rod and the man. Let these be $R_x$ (horizontal, to the left) and $R_y$ (vertical, downwards).

From the vertical equilibrium of the entire system, we found $N$. From the vertical equilibrium of a single rod: $N - mg - R_y = 0$ We know $N = \frac{(M+2m)g}{2}$. So, $R_y = N - mg = \frac{(M+2m)g}{2} - mg = \frac{Mg + 2mg - 2mg}{2} = \frac{Mg}{2}$. This confirms that each rod supports half the man's weight.

From the horizontal equilibrium of a single rod: $T - R_x = 0$ (assuming no friction at the ground, or that the horizontal force on the ground is zero due to symmetry for the whole system, so no horizontal external force on the rod from the ground). So, $R_x = T$. This means the horizontal force exerted by the other rod at the top joint is equal to the tension in the string.

Now, apply rotational equilibrium for the left rod by taking moments about the base point P. The normal force $N$ passes through P, so its torque is zero.

• Torque due to weight of the rod (mg): The force $mg$ acts at point A (midpoint of the rod). The horizontal distance from P to the line of action of $mg$ is $(L/2) \cos \theta$. Torque$_{mg}$ = $mg \times (L/2) \cos 60^\circ = mg \times (L/2) \times (1/2) = \frac{mgL}{4}$ (clockwise).

• Torque due to tension (T): The force $T$ acts horizontally at point A. The vertical distance from P to the line of action of $T$ is $(L/2) \sin \theta$. Torque$_{T}$ = $T \times (L/2) \sin 60^\circ = T \times (L/2) \times (\sqrt{3}/2) = \frac{T L\sqrt{3}}{4}$ (counter-clockwise).

• Torque due to vertical force $R_y$ at Q: The force $R_y = Mg/2$ acts vertically downwards at Q. The horizontal distance from P to the line of action of $R_y$ is $L \cos \theta$. Torque$_{R_y}$ = $R_y \times L \cos 60^\circ = \frac{Mg}{2} \times L \times (1/2) = \frac{MgL}{4}$ (clockwise).

• Torque due to horizontal force $R_x$ at Q: The force $R_x = T$ acts horizontally to the left at Q. The vertical distance from P to the line of action of $R_x$ is $L \sin \theta$. Torque$_{R_x}$ = $R_x \times L \sin 60^\circ = T \times L \times (\sqrt{3}/2) = \frac{T L\sqrt{3}}{2}$ (counter-clockwise).

For rotational equilibrium, the sum of clockwise torques must equal the sum of counter-clockwise torques: Torque$_{mg}$ + Torque$_{R_y}$ = Torque$_{T}$ + Torque$_{R_x}$ $\frac{mgL}{4} + \frac{MgL}{4} = \frac{T L\sqrt{3}}{4} + \frac{T L\sqrt{3}}{2}$

Multiply by $\frac{4}{L}$ to simplify: $mg + Mg = T\sqrt{3} + 2T\sqrt{3}$ $(M+m)g = 3T\sqrt{3}$ $T = \frac{(M+m)g}{3\sqrt{3}}$

The normal force acting on the rod from the ground is $N = \frac{(M + 2m)g}{2}$. The tension force in the string connecting both the rods is $T = \frac{(M+m)g}{3\sqrt{3}}$.

Explanation of the solution:

1. Normal Force: Consider the entire system (two rods + man). For vertical equilibrium, the total upward normal force (2N) must balance the total downward weight (Mg + 2mg). $2N = (M+2m)g \implies N = \frac{(M+2m)g}{2}$.
2. Tension Force: Consider one rod. Apply rotational equilibrium by taking moments about its base (ground contact point).
   • Forces causing clockwise torque about the base: Weight of the rod (mg) and the vertical component of the force from the man and the other rod at the top joint ($R_y = Mg/2$).
   • Forces causing counter-clockwise torque about the base: Tension (T) and the horizontal component of the force from the other rod at the top joint ($R_x = T$).
   • The perpendicular distances for torques are found using the rod's length (L) and the angle ($60^\circ$).
   • Equating clockwise and counter-clockwise torques yields the tension. $\frac{mgL}{4} + \frac{MgL}{4} = \frac{T L\sqrt{3}}{4} + \frac{T L\sqrt{3}}{2}$ $(M+m)g = 3T\sqrt{3} \implies T = \frac{(M+m)g}{3\sqrt{3}}$.