This document describes a system involving a meter stick with nickels attached at 0.120 m, a knife edge applying a force F at 0.455 m, and the force of gravity acting at the center of the stick at 0.500 m. Since the meter stick is in equilibrium, the sum of the torques about the knife edge at 0.455 m must be zero. This results in an equation relating the mass and positions of the objects.

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The x axis is along the meter stick, with the origin F
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at the zero position on the scale. The forces acting .
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on it are shown on the diagram to the right. The .
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nickels are at x = x1 = 0.120 m, and m is their .
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total mass. The knife edge is at x = x2 = 0.455 m .
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and exerts force F . The mass of the meter stick is .
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M , and the force of gravity acts at the center of x1 .
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x3
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the stick, x = x3 = 0.500 m. Since the meter stick
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x2 .
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is in equilibrium, the sum of the torques about x2 .
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must vanish: M g(x3 − x2 ) − mg(x2 − x1 ) = 0. mg .
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Thus, Mg
x2 − x1 0.455 m − 0.120 m
M= m= (10.0 g) = 74 g .
x3 − x2 0.500 m − 0.455 m