I've asked different versions of this question in the past, but I always find it interesting. How about you?
You have a glass of water sitting on a perfectly balanced scale. You put your finger into the glass and into the water. Your finger does not touch the glass, it only is submerged in the water. It makes the water go higher up the sides of the glass but it does not overflow.
What happens to the scale?
I think the weight of the scale will not change.
ReplyDeleteThe finger will displace some of the water, hence the water rises. But I think since the weight of the finger is still being borne by the hand attached to the finger, its weight won't register on the scale.
I think it is comparable to holding a pencil and sticking it in the water. The weight of the pencil is being borne by the holder, not the cup.
The fact that the water rises has no bearing on the reading on the scale. The amount of water in the cup hasn't changed, only its position.
The finger in the cup isn't pushing down on the water and therefore not going to be registered on the scale.
I agree with USAF Pilot, but I also think that buoyancy of water will play its role.
ReplyDeleteWhen we immerse the finger in the water, water level rises, at the same time buoyancy of water is trying to push the finger up.
Thus weight of the finger will not be registered on the scale and scale will not change.
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I agree that the finger in the water would have no effect on the scale, but only after it reaches steady state. The initial thrust of the finger into the water would be met with resistance and would impart a small force onto the scale.
ReplyDeleteAs an example of this principle consider when objects fly past your face. If it gets close enough and it moving fast enough you can feel the air that is displaced. In the same way that the air imparts a small and temporary force on your skin, the water would impart a small and temporary force because of the water displaced from the finger.
Doesn't the bouyancy of the water push your finger up? Isn't there an upward force acting on your finger? And then by Newton's (third?) law, doesn't that mean that your finger is exerting downward force on the water?
ReplyDeleteSo by my reckoning, the scale will register extra weight. How much, I don't know... Perhaps weight equal to the weight of the submerged part of your finger?
The upward force acting on your finger is equal to the weigth of the displaced water. Thus the scale will register a force of the same magnitude in the opposite direction.
ReplyDeleteNo, I don't think the buoyancy of your finger will cause a reaction to the scale because your finger is not buoyant in this situation.
ReplyDeleteThe upward force acting against your finger is equal to its displaced water weight, however that weight is not being borne by buoyant forces, it is being borne by the hand attached to the finger.
Instead of your finger, think of a piece of dense steel. Won't ever float. The amount of water it displaces is way less than its weight. If you held a rod of steel and put a portion of it into the water, there are some buoyant forces in effect here but very small forces. However, the weight of the steel rod is still being borne by your hand and not the buoyant forces. Hence it doesn't sink, because you hold it and don't let it sink. It won't register on the scale because similarly, it's weight is being borne by your hand.
I believe (not having a scale or bucket of water to work with to prove any of this) that the scale will move down (the water side). The force you're exerting to displace the water will push the scale down.
ReplyDeleteMike,
ReplyDeleteJust because we're displacing water doesn't mean there will be an increase in weight.
What about an object in the water that is neutrally buoyant? Doesn't sink or float to the top. It just floats half way down the glass? It displaces water but I doubt it causes the scale to weigh more.
Alright, maybe I should explain my reasoning a little bit more.
ReplyDeletePut a boat in a pond and it will float. Why does it float? The force coming up from the water. The water in the pond will be displaced which, if you could see the shore line and measure it very accurately, you would see it rise some. But the force from the boat is still there. That force pushes down. Put the pond on a balanced scale and you would see it drop a little.
At least, that's my thought process.
Mike, sorry, I've been preparing reports and stuff for the end of the semester, I didn't come to your defense.
ReplyDeleteYou're definitely right. Any displaced fluid will exert a "buoyancy force" on a submerged object (proportional to the ratio of the densities and the submerged volume). The scale will, in the end, and with whatever method you choose to use to understand it, have to bear that force, so the force will increase on the scale. The easiest way is probably thinking about the object compressing the fluid, and that increased pressure eventually pushing a bit harder on the scale or something...
The comment about the weight of the finger being still borne by the hand is misleading. What will happen is that your arm will slightly lessen the amount of force it provides to accommodate the new force. Think about how much force you exert when holding a child in your arms, then compare that to the amount of force when you're holding them in the water...same principle here. If you exerted the same amount of force (the full weight of the object) you'd pull your finger (or the child) up out of the water. You have to exert less force for an equilibrium case, and the buoyancy force (transmitted to the scale) picks up the slack.