Question

A uniform bar of mass $m$ is supported by a pivot at its top about which the bar can swing like a pendulum. If a force $F$ is applied perpendicular to the lower end of the bar as shown in figure, what is the value of $F$ in order to hold the bar in equilibrium at an angle $\theta$ from the vertical

• A. $2mg \sin \theta$
• B. $mg \sin \theta$
• C. $mg \cos \theta$
• D. $\frac{mg}{2} \sin \theta$
• E. $\frac{mg}{2} \cos \theta$

Hint:

Always remember that for uniform objects, the gravitational force acts at the geometric center. This usually introduces a factor of $1/2$ when compared to forces applied at the edges.

Answer 1

The Correct Option is D

Concept:
For the bar to be in rotational equilibrium, the net torque about the pivot must be zero ($\sum \tau = 0$). [itemsep=4pt]
• Torque due to Gravity: Acts at the center of mass ($L/2$).
• Torque due to applied Force F: Acts at the end ($L$).

Step 1: Calculate the clockwise torque (Gravity).
The weight $mg$ acts vertically downwards at a distance $L/2$ from the pivot. The perpendicular distance from the pivot to the line of action of the weight is $(L/2) \sin \theta$. \[ \tau_{gravity} = mg \times \frac{L}{2} \sin \theta \]

Step 2: Calculate the counter-clockwise torque (Force F).
As seen in the figure, the force $F$ is applied perpendicular to the bar at its lower end ($L$). \[ \tau_{F} = F \times L \]

Step 3: Equate torques for equilibrium.
To hold the bar in equilibrium, the magnitude of the counter-clockwise torque must equal the magnitude of the clockwise torque: \[ F \times L = mg \times \frac{L}{2} \sin \theta \] Dividing both sides by $L$: \[ F = \frac{mg}{2} \sin \theta \]