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  4. For equilibrium of the system, value of mass m should be

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Q. For equilibrium of the system, value of mass $m$ should bePhysics Question Image

System of Particles and Rotational Motion

Solution:

Net torque $=0$ for equilibrium
$12 l=m\left(\frac{l}{2}\right)+3\left(\frac{3 l}{2}\right)$
$12 l-4.5 l=\frac{m l}{2}$
$7.5 l=\frac{m l}{2}$
$m=15\, kg$