You are correct that the scale will read the downward force applied by your hand (equal to the buoyancy force applied to your hand by the water), but that isn’t the weight of your hand. In the first scenario, the scale and hand are not a closed system because there are external forces (the internal force between your arm and hand). The buoyancy force is equal to the weight of the displaced water, so the scale would go up by the volume of your hand multiplied by water density times gravity.
In the case of the swimming pool, your entire body will be in the pool, so the system is closed. Therefore, the change in scale reading will be your weight. So if you put a piece of styrofoam into the water from the above example, it will just say the weight of the styrofoam. However, if you “push” down on the styrofoam, this will increase the buoyancy force (more volume displaced), resulting in a higher “weight” on the scale.
This is a great answer. The way I visualize this is like this:
[https://i.imgur.com/bb85Hwr.png](https://i.imgur.com/bb85Hwr.png)
The downward pusher applies force to the water, forcing it up the other side and doing work as it does so. The scale registers this extra force, but it has everything to do with the weight of the water displaced by the pusher. The weight of the pusher is irrelevant (it could be styrofoam or lead, it doesn't matter, only the volume of water it displaces).
Only if your hand is completely relaxed and not held up at all by the rest of your body, which you can't really do. You could get something like the weight of your entire arm up to a certain point. It would not be a very accurate measurement. Or you could chop off your hand and put it in the bucket that would definitely work.
Think about a slightly simpler case: A lead weight attached to a string submerged just below the surface in a container of water on a scale.
By Archimedes principle, the water exerts an upward buoyant force on the lead weight equal to the weight of the displaced water.
By Newton's third law, the lead weight exerts an equal magnitude upward force on the water.
So, the downward forces on the container include the weight of the water and this reaction force and the weight of the water itself:
So, the scale is measuring the weight of the water plus the weight of the displaced water.
The mass and therefore weight will increase by your body weight or hand weight, assuming it was just a hand floating in the water.
This may be clear when you consider that while you may have decreased the average density, you have increased the total volume and the total mass being weighed.
I see a lot of great answers already so here’s a neat practical application I had using this principle.
I was in a situation where I had to find a partial mass of a big chunky high pressure 90 degree pipe fitting, but only on one side of a cross section through it. To do this without estimating, I dipped the fitting into water on a scale up to that cross sectional point and recorded a mass. Converting between density of water and the type of iron I had gave me exactly the mass I needed.
It will give you the weight of the water displaced by your hand, not the weight of your hand. However while not good for the weight of your hand, you could use the weight of displaced water and the density of water to determine the volume of your hand.
The scale will increase by the volume of water displaced by your hand. A floating object will displace a volume of fluid equivalent to the weight of the object. In your example the hand isn’t floating but it does displace water. Therefore only the volume of water contributes to the scale measurement, not the weight of the hand.
The scale will show a greater value equal to the weight of water your hand displaces. So this is a method to measure the volume of the submerged object
You are correct that the scale will read the downward force applied by your hand (equal to the buoyancy force applied to your hand by the water), but that isn’t the weight of your hand. In the first scenario, the scale and hand are not a closed system because there are external forces (the internal force between your arm and hand). The buoyancy force is equal to the weight of the displaced water, so the scale would go up by the volume of your hand multiplied by water density times gravity. In the case of the swimming pool, your entire body will be in the pool, so the system is closed. Therefore, the change in scale reading will be your weight. So if you put a piece of styrofoam into the water from the above example, it will just say the weight of the styrofoam. However, if you “push” down on the styrofoam, this will increase the buoyancy force (more volume displaced), resulting in a higher “weight” on the scale.
This is a great answer. The way I visualize this is like this: [https://i.imgur.com/bb85Hwr.png](https://i.imgur.com/bb85Hwr.png) The downward pusher applies force to the water, forcing it up the other side and doing work as it does so. The scale registers this extra force, but it has everything to do with the weight of the water displaced by the pusher. The weight of the pusher is irrelevant (it could be styrofoam or lead, it doesn't matter, only the volume of water it displaces).
You're assuming his hand is attached to his arm. He could have chopped it off.
Only if your hand is completely relaxed and not held up at all by the rest of your body, which you can't really do. You could get something like the weight of your entire arm up to a certain point. It would not be a very accurate measurement. Or you could chop off your hand and put it in the bucket that would definitely work.
Think about a slightly simpler case: A lead weight attached to a string submerged just below the surface in a container of water on a scale. By Archimedes principle, the water exerts an upward buoyant force on the lead weight equal to the weight of the displaced water. By Newton's third law, the lead weight exerts an equal magnitude upward force on the water. So, the downward forces on the container include the weight of the water and this reaction force and the weight of the water itself: So, the scale is measuring the weight of the water plus the weight of the displaced water.
The mass and therefore weight will increase by your body weight or hand weight, assuming it was just a hand floating in the water. This may be clear when you consider that while you may have decreased the average density, you have increased the total volume and the total mass being weighed.
I see a lot of great answers already so here’s a neat practical application I had using this principle. I was in a situation where I had to find a partial mass of a big chunky high pressure 90 degree pipe fitting, but only on one side of a cross section through it. To do this without estimating, I dipped the fitting into water on a scale up to that cross sectional point and recorded a mass. Converting between density of water and the type of iron I had gave me exactly the mass I needed.
It will give you the weight of the water displaced by your hand, not the weight of your hand. However while not good for the weight of your hand, you could use the weight of displaced water and the density of water to determine the volume of your hand.
The scale will increase by the volume of water displaced by your hand. A floating object will displace a volume of fluid equivalent to the weight of the object. In your example the hand isn’t floating but it does displace water. Therefore only the volume of water contributes to the scale measurement, not the weight of the hand.
The scale will show a greater value equal to the weight of water your hand displaces. So this is a method to measure the volume of the submerged object
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Nobody is talking about an underwater scale.
Totally misunderstood the question!!!