A square lamina OABC of side $10\ \text{cm}$ is pivoted at O. Forces act on the lamina as shown in the figure. If the lamina remains stationary, then the magnitude of $F$ is:
Answer: (C) $10\ \text{N}$
The pivot can supply any force, so only the net torque about O must vanish. Take O as origin with A $(l, 0)$, B $(l, l)$, C $(0, l)$, $l = 10\ \text{cm}$, and anticlockwise as positive.
- At A: $10\ \text{N}$ to the right passes through the line OA, so no torque; $10\ \text{N}$ downward gives $-10l$.
- At B: $10\ \text{N}$ upward gives $+10l$; $10\ \text{N}$ to the right gives $-10l$.
- At C: $10\ \text{N}$ upward passes through O's vertical line, so no torque; $F$ to the left gives $+Fl$.
$$\tau_O = -10l + 10l - 10l + Fl = 0 \Rightarrow F = 10\ \text{N}$$
Solution by Sreeraj P, M.Sc Physics