A light rod is free to rotate about a point OA force $\mathrm{F}=12 \mathrm{~N}$ is applied at the p

Question

A light rod is free to rotate about a point OA force $\mathrm{F}=12 \mathrm{~N}$ is applied at the point A perpendicular to OA . The torque about the point O is $(\mathrm{OA}=0.5 \mathrm{~m})$

Professor PC Thomas Classes · Answer

Torque $	au = r F \sin 	heta$. Here $	heta = 90^{\circ}$, so $	au = 0.5 	imes 12 = 6 \mathrm{N m}$.