A uniform meter ruler is pivoted at its centre. A 20g mass is placed at 10cm and 50g at 40cm mark. What mark second 50g maa be placed for the system to be in equilibrium?

Answer 1

To find the equilibrium position of the second 50g mass, we need to calculate the total torque acting on the ruler with respect to the pivot point.

The torque due to the 20g mass at 10cm is given by:
Torque1 = (20g)(10cm)

The torque due to the 50g mass at 40cm is given by:
Torque2 = (50g)(40cm)

Let the position of the second 50g mass be x cm from the pivot point. The torque due to the second 50g mass is given by:
Torque3 = (50g)(x cm)

For the system to be in equilibrium, the total torque must be zero. Therefore:
Torque total = Torque1 + Torque2 + Torque3 = 0
(20g)(10cm) + (50g)(40cm) - (50g)(x cm) = 0

Solving for x:
200 + 2000 - 50x = 0
2200 - 50x = 0
50x = 2200
x = 44

Therefore, the second 50g mass should be placed at the 44cm mark for the system to be in equilibrium.