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According to Newton's third law of motion, if there is a beaker with 100g water, the normal force exerted by the bottom of the beaker would be equal to the force by the weight of the water in it..

What would happen if a wooden block is now half submerged into the beaker? Is the new normal force going to be the sum of the weight of the beaker and the weight of the wooden block? Will upthrust affect the normal force in causing it to be less than the sum of these two masses? I think that it would, but am really just confused and unable to explain it. An explanation of what actually happens during these two situations would be greatly appreciated, thanks!

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If the block is floating then the upthrust on the block due to the water is exactly equal in magnitude but opposite in direction to the weight of the block.

The block must therefore exert a force on the water equal but opposite to the upthrust.

That force causes the water in the container to rise and hence the pressure on the bottom of the container increases as now the water is deeper and so the force on the bottom of the container also increases by an amount equal to the upthrust that the water exerted on the block.

So the normal force on the water due to the bottom of the container also increases by an amount equal to the upthrust.

So that normal force is equal to the normal force before the block was added plus the weight of the block.

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  • $\begingroup$ Oh.. But what about when only half the block is submerged? Since upthrust is equal to the weight of fluid displaced by the block, then is it right to say that the normal force is equal in magnitude and opposite in direction to half the weight of the block and the weight of water? Or does the water actually still support all of the block's weight somehow? Would the normal force vary with the depth at which an object sinks to? Thank you! $\endgroup$
    – y.han
    Sep 28 '16 at 0:16
  • $\begingroup$ If the block is in static equilibrium the net force on it is zero. If you held the block so that it is half submerged then the upthrust plus the force you exert on the block must equal the weight of the weight of the block. The force that the water feels is equal to the upthrust it exerts on the block. If the block floats a quarter submerged and you push the block down so that it is half submerged the upthrust will double as the upthrust is the weight of water displaced. $\endgroup$
    – Farcher
    Sep 28 '16 at 5:13
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The bottom of the beaker supports the whole weight of the water, plus the weight of anything floating in the water or resting entirely on the bottom (ie not touching the sides of the beaker). If you push down on the block, this is equivalent to making the block heavier.

Upthrust is an internal action-reaction force between the water and the block. The block pushes down on the water, the water pushes up on the block. If your 'system' is the contents of the beaker (ie water & block), you do not need to consider internal forces, because they are all action-reaction pairs of forces. The only external forces on the 'system' are gravity, atmospheric pressure, and the normal reaction from the bottom of the beaker.

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