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Problem 7: A uniform rod 1 m long with weight 6 N can be supported in a horizontal position on a sharp edge with weights of 10 N and 15 N suspended from its ends. What is the position of point of balance?

Solution

Explanation: The rod is uniform. Its center of gravity would be at the center of the rod. When weights of unequal magnitudes are hung from its ends, the center of gravity doesn’t remain at the center of the rod. It shifts towards the end to which the larger weight is hung. Now the rod can be balanced from that new center of gravity and we have to locate that position.

Now consider the figure. By hanging the weights, the center of gravity (from where the rod can be balanced now) has shifted to the new position ‘c’. There are 3 forces giving rise to the torques. Weight of 10 N in the clockwise direction, weight of the rod 6 N in the clockwise direction and weight of 15 N in the counterclockwise direction. If the rod is balanced, the counterclockwise torque will be equal to the clockwise torque. Let’s calculate the torques.

Suppose, the new balancing position, c, is at distance r from the 15 N weight end. Then

Distance from 15 N end = r meter

Distance from the original center of gravity to the new balancing position, c = (½ – r) m

Distance from the 10 N end = (1 – r) m

Torque due to 15 N weight = 15 × r (counterclockwise)

Torque due to 6 N weight of the rod = (½ – r) × 6 (clockwise)

Torque due to 10 N weight = (1 – r) × 10 (clockwise)

Equating clockwise and counterclockwise torques,

Hence the new balancing position c, is 0.41 m from 15 N weight end.

3 Comments

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