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15/09/2021 10:54 am
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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/s2)
(1) \(\frac{1}{2}\)kg
(2) \(\frac{1}{3}\)kg
(3) \(\frac{1}{6}\)kg
(4) \(\frac{1}{12}\)kg
1 Answer
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15/09/2021 11:01 am
Correct answer: (4) \(\frac{1}{12}\)kg
Explanation:
By balancing torque
2g x 20 = 0.5 g x 60 + mg x 120
m = \(\frac{0.5}{6}\)kg = \(\frac{1}{12}\)kg