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 θ from the vertical

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 θ from the vertical, 2 m g sin ⁡ θ, m g sin ⁡ θ, m g 2 sin ⁡ θ, m g 2 cos ⁡ θ

MARKS App · Accepted Answer

Let the length of bar is l , then according to the problem. For equilibrium of the bar, F e x t n e t = 0 and τ e x t n e t = 0 taking torque about pivot O , O B × M g + O A × F = 0 ⇒ - O B M g s i n θ + O A . F s i n 90 o = 0 (clockwise) (anti-clockwise) ∴ l F = l 2 M g s i n θ ⇒ F = M g s i n θ 2

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