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Bouyant Force on a Scaleby shyboyswin
Tags: buoyancy 
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#1
Jul1609, 08:05 PM

P: 9

This is not homework, but the question is really bugging me. [Link to Yahoo Answers Question]
Here is the problem: On a scale, you place a glass of water and a ball on each side of the scale. The scale is balanced at this point. If you take the ball on side B of the scale, and place it into the glass of water on side B and the ball floats, will the scale become unbalanced? Although the posters on Yahoo Answers think that the scale will remain balanced, I think not. The scale only detects the forces acted on each side. Because the ball floats, there is a buoyant force from the displaced water that is preventing the ball to sink to the bottom, thus the ball's force of gravity is not acting on the scale. For side A: [tex]\Sigma F_A = F_{G}_{water} + F_{G}_{ball}[/tex] For side B: [tex]\Sigma F_B = F_{G}_{water} + (F_{G}_{ball}  F_{buoy}) = F_{G}_{water}[/tex] Can someone please confirm/correct. 


#2
Jul1609, 08:53 PM

Sci Advisor
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PF Gold
P: 2,907

Hi shyboy. I'm sorry but the Yahoo guy has it right. Just to make sure there's a clear understanding of the question, I'll copy it here:
In other words, the force downwards is due to water pressure integrated over an area, and that pressure increases by an amount that is equal to the weight of the floating ball divided by the cylinder's area. Note that the volume of water displaced by the object has a weight which is equal to the weight of the object. So this volume of water must rise up in the cylinder exactly the same amount that it would rise if water of this volume were added. We could drop either a ping pong ball in of weight W or add water of weight W to the cylinder, and both of these would cause the water to rise inside the cylinder an equal amount. It can get very complicated if we talk about a noncylindrical water container, like an old fashion Coke bottle, but the conclusion is the same. The weight of the object in the water increases the downward force on the scale when the ball is put in by an amount equal to the weight of the ball. Therefore, the scale remains ballanced. 


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