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) [figure]
- 112 kg
- 12 kg
- 13 kg
- 16 kg
Answer
(1) 112 kg
Solution
Take moments about the wedge at the 40 cm mark.
Anticlockwise moment: 2 kg at 20 cm (arm = 20 cm)
Clockwise moments: rod's weight (0.5 kg) at its centre, 100 cm mark (arm = 60 cm), and m at 160 cm (arm = 120 cm)
For equilibrium:
2g 20 = 0.5g 60 + mg 120
40 = 30 + 120m
m = 10120 = 112 kg
Answer: (1)