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A fulcrum is placed at the 40 cm mark on a 50 g meter stick (center...

A fulcrum is placed at the 40 cm mark on a 50 g meter stick (center of mass at the 50 cm mark). A 35 g mass is placed at the 10 cm mark. What mass would have to be placed at the 75 cm mark to balance the meter stick?

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Answer #1

Considering equilibrium of torque

hanging mass * distance from hinge = mass of meter * distance of COM from hinge + balncing mass * distance feom hinge

35* (40 - 10) = 10 * 50 + 35* m

m = 15.714 g

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Comment before rate in case any doubt, will reply for sure.. Goodluck

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