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(B) Let the forces acting at $20 \, cm$ and $60 \, cm$ be directed downwards $(-F)$ and the forces at $40 \, cm$ and $80 \, cm$ be directed upwards $(

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experiences a torque

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(B) Let the forces acting at $20 \, cm$ and $60 \, cm$ be directed downwards $(-F)$ and the forces at $40 \, cm$ and $80 \, cm$ be directed upwards $(+F)$.Net force $F_{net} = (-F) + (+F) + (-F) + (+F) = 0$.Since the net force is zero,the rod does not experience any linear motion.Now,calculate the net torque about the end of the rod $(x = 0)$:$	au_{net} = \sum (F_i 	imes r_i) = (-F 	imes 20) + (F 	imes 40) + (-F 	imes 60) + (F 	imes 80)$$	au_{net} = F(-20 + 40 - 60 + 80) = F(40) = 40F 
eq 0$.Since the net torque is non-zero,the rod experiences a torque and will rotate.

Four equal and parallel forces are acting on a rod (as shown in the figure) at distances of $20 \, cm, 40 \, cm, 60 \, cm$,and $80 \, cm$ respectively from one end of the rod. Under the influence of these forces,the rod -

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