2. Assume the center of gravity of the meter stick (the point at which the weigh

Question

2. Assume the center of gravity of the meter stick (the point at which the weight of the entire meter stick seems to be concentrated) is at the 0.500m mark. The meter stick is pivoted on a fulcrum at the 0.350m mark and balanced by placing a total mass of the 80 g at the 0.100m mark. What is the mass of the meter stick? Recall that according to the definition of the center of mass(gravity), the mass of the meter stick may be treated as though it is all at the center of mass. Make a sketch of the situation before making the calculation.

Explanation / Answer

Here ,

let the mass of meter stick is M

mass , m = 80 g

Now , for the meter stick to balance at 0.35 m mark

the net torque about the fulcrum must be zero

m *(0.35 - 0.10) * g - M *(0.50 - 0.35) * g = 0

80 * 0.25 - M * 0.15 = 0

M = 133.3 gm

the mass of meter stick is 133.3 gm

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