As illustrated, a weight is attached to an arm that rests on the left plate of a balance scale. A beaker of water sits on the right hand plate. The scale is initially balanced. If the hanging weight is submerged in the water, the right hand plate moves downward and it is shown that an additional weight equal to twice the buoyant force is needed to rebalance the scale.

When the hanging weight is submerged, a buoyant force F pushes up on the weight. At the same time (see 2B - 9) a reaction force equal to F pushes down on the right hand plate. Thus a weight equal to 2F must be added to the left side to restore balance. For this apparatus, the hanging weight has a volume of 100 cm3, so the “weight” on the left pan is reduced by 100 g when the weight is completely submerged. At the same time the “weight” on the right hand side is increased by 100 g (see discussion above regarding the reaction force). Therefore, to restore equilibrium 200 g must be added to the left pan.

Directions: Be sure that the system is balanced as shown. Swing the weight arm over the beaker, allowing the weight to be submerged. The 200-g weight will rebalance the system. (Note: The lack of precision of the balance may result in the compensating weight being approximately, but not exactly, 200 g. The point is that the weight is close to 200 g, not 100 g, as might be expected.)

Suggestions for Presentation: Although the explanation is fairly straightforward, the system is a real puzzle for most students (and perhaps faculty!). The set-up has the appearance of being an internal system of forces, suggesting that action-reaction forces will cancel each other and nothing will happen. Or possibly that 100 g will rebalance the system, since that much was "lost" in the immersion process. Ask the students if anything will happen to the balance if the block is submerged. If so, what has to be done to restore equilibrium? To which side must weights be added? Why?

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Last Updated: Nov 30, 2023 11:25 AM

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