A centimeter ruler, balanced at its center point, has two coins placed on it, as

Question

A centimeter ruler, balanced at its center point, has two coins placed on it, as shown in the figure. (Figure 1) One coin, of mass M1=10g, is placed at the zero mark; the other, of unknown mass M2, is placed at the 4.7 cmmark. The center of the ruler is at the 3.0 cm mark. The ruler is in equilibrium; it is perfectly balanced.

Does the pivot point (i.e., the triangle in the diagram upon which the ruler balances) exert a force on the ruler? Does it exert a nonzero torque about the pivot?

find the mass?

Explanation / Answer

First draw your ruler with all the forces on it. You have two weights pointing down, and the reaction force pointing up. There's also the weight of the ruler itself, going through the same point as the reaction force in the opposite direction.

If we sum the moments (the force times the distance to the point) about the point of balance, we don't need to find the weight of the ruler or the reaction force. We'll call a counterclockwise moment positive:

SM = 0 = (3.0cm)(0.010kg)(g) - (1.7cm)(M2)(g)
M2 = (3.0cm)(0.010kg)(g) / (1.7cm)(g)
M2 = 0.0176 kg
M2 = 17.6 g   

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