A uniform rod of length 200 , cm and mass 500 | vedclass

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(D) The rod is uniform,so its center of mass acts at the midpoint,i.e.,at $100 \, cm$ mark. The mass of the rod is $0.5 \, kg$.The wedge is at $40 \,

Vedclass · Accepted Answer

$\frac{1}{12} \, kg$

Vedclass · Answer

(D) The rod is uniform,so its center of mass acts at the midpoint,i.e.,at $100 \, cm$ mark. The mass of the rod is $0.5 \, kg$.The wedge is at $40 \, cm$. We take the torque about the pivot point (wedge) at $40 \, cm$.Clockwise torque due to the rod's weight $(0.5 \, kg)$ acting at $100 \, cm$:$	au_{cw} = (0.5 \, kg 	imes g) 	imes (100 \, cm - 40 \, cm) = 0.5 	imes g 	imes 60 \, cm$.Clockwise torque due to mass $'m'$ acting at $160 \, cm$:$	au_{cw}' = (m 	imes g) 	imes (160 \, cm - 40 \, cm) = m 	imes g 	imes 120 \, cm$.Counter-clockwise torque due to $2 \, kg$ mass acting at $20 \, cm$:$	au_{ccw} = (2 \, kg 	imes g) 	imes (40 \, cm - 20 \, cm) = 2 	imes g 	imes 20 \, cm$.For equilibrium,$	au_{ccw} = 	au_{cw} + 	au_{cw}'$:$2 	imes g 	imes 20 = 0.5 	imes g 	imes 60 + m 	imes g 	imes 120$.Dividing by $g$:$40 = 30 + 120m$.$10 = 120m$.$m = \frac{10}{120} \, kg = \frac{1}{12} \, kg$.

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