2021 System of Particles and Rotational MotionEquilibrium of a Rigid Body – Moments Medium Question 22 of 100 in System of Particles and Rotational Motion

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

A $\frac{1}{3}\ kg$
B $\frac{1}{6}\ kg$
C $\frac{1}{12}\ kg$
D $\frac{1}{2}\ kg$

The answer is C.

Solution


Weight of the rod (0.5 kg) acts at its centre (100 cm), i.e. 60 cm from the wedge. The 2 kg mass is 20 cm from the wedge and m is 120 cm from it.
Balancing torques about the wedge: $2g \times 20 = 0.5g \times 60 + mg \times 120$ $40 = 30 + 120m \Rightarrow m = \frac{1}{12}$ kg

Concept balancing a loaded rod

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