Mechanics: Equilibrium of forces
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A meter stick is pivoted at its 50 cm mark but does not balance due to nonuniformities in its material that cause the center of gravity to be displaced from its geometric center. However when weights of 150 and 200g are placed at the 10 cm and 75 cm marks respectively balance is obtained. The weights are then interchanged and balance is obtained by shifting the pivot point to the 43 cm mark. Find the mass of the meter stick and the location of its centre of gravity.
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Solution Summary
The equilibrium of forces is examined for the mass of gravity. The solution provides a step by step solution provided.
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Let C.G. of the meter stick be shifted by d from its geometrical centre as shown. Also let mass of the meter stick be m. Hence, weight of the stick (mg) acts at the C.G. of the stick as shown in the fig. Weights 150g and 200g are placed as ...
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