A uniform rod of length $200 ext{ cm}$ and mass $500 ext{ g}$ is balanced on a wedge placed at $40 ext{

Question

A uniform rod of length $200 	ext{ cm}$ and mass $500 	ext{ g}$ is balanced on a wedge placed at $40 	ext{ cm}$ mark. A mass of $2 	ext{ kg}$ is suspended from the rod at $20 	ext{ cm}$ and another unknown mass '$m$' is suspended from the rod at $160 	ext{ cm}$ mark as shown in the figure. Find the value of '$m$' such that the rod is in equilibrium. ($g = 10 	ext{ m/s}^2$)

Accepted Answer

The rod is in rotational equilibrium about the wedge at $40 	ext{ cm}$. The center of mass of the rod is at $100 	ext{ cm}$. Torque balance: $2 	ext{ kg} 	imes (40-20) 	ext{ cm} = 0.5 	ext{ kg} 	imes (100-40) 	ext{ cm} + m 	imes (160-40) 	ext{ cm}$. $2 	imes 20 = 0.5 	imes 60 + m 	imes 120$. $40 = 30 + 120m$. $10 = 120m \Rightarrow m = 10/120 = 1/12 	ext{ kg}$. (Note: The provided options might contain a typo based on the calculation, but following the torque balance principle).

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