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

Options

A. `2mg sin theta`
B. `mg sin theta`
C. `mg cos theta`
D. `(mg)/(2) sin theta`

Answer

D

Explanation

Let the length of bar is l, then according to the problem,

For equilibrium of the bar, `F_("ext")^(net")= 0 and tau_("ext")^(net")=0`
taking torque about pivot O, `OB xx Mg +OA xx F=0`
`rArr -OB Mg sin theta+ OA F sin 90^(@)= 0`
(clockwise) (anti-clockwise)
`therefore lF= (1)/(2) Mg sin theta`
`rArr F= (Mg sin theta)/(2)`