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Author Topic: Weight of an immersed object in liquid  (Read 26712 times)
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Fu-Kwun Hwang
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on: March 10, 2007, 01:39:18 pm »

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The buoyant force on an immersed object equals the weight of displaced liquid.
Use the top/right slider (mouse drag the center of scale) to move the object up/down.
When part of the object is immersed in the liquid, the weight indicated on the top scale is reduced.
However, the same amount is increased in the lower scale (support the whole system).
You can change the density of the object or the liquid.



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desert knight
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Reply #1 on: November 25, 2007, 07:16:44 am »

thats very nice, but i have a question plz :

how can a body floating on a liquid increases the total weight of the liquid , isn't the buoyancy of the liquid cancels the weight of the body ?

i have studied that in school but i dont know why.

could u plz answer me .thnx
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Fu-Kwun Hwang
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Reply #2 on: November 25, 2007, 08:12:44 am »

Assume we have object A(mass ma), liquid and container as B(mass mb)

If we just put the liquid and the container on top of the measurement device, due to gravity g
B will have  mb*g download force acting on the device, that is how the device get the reading mb*g.

If we put object A into the container, the total force acting download will be ma*g + mb*g

If the object is floating on the liquid, the weight of the object is support by the liquid. i.e. the liquid provide upward force on the object (ma*g up) to support the gravitation force for object (ma*g down).
So the same time, the object give the liquid ma*g downward force. Now the total force acting on the liquid is ma*g + mb*g (extern force from object and gravitation force due to it's own mass). So the measurement device need to provide ma*g+mb*g upward force to support the liquid. That is why the reading will be ma*g+mb*g
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desert knight
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Reply #3 on: November 25, 2007, 09:12:35 am »

thnx for responding, i just want to understand somthing:

when the  body (A) float, it affects on the liquid(b) by force (ma*g down) then the liquid affects it by an upward force (ma*g up) so they cancel each other so the remaining force which acts on measuring device will be the weight of the liquid (mb*g down ),

is that right or wrong? 

so how will it be (ma*g)+(mb*g) ??

 thnx alot
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Fu-Kwun Hwang
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Reply #4 on: November 25, 2007, 04:36:11 pm »

Action force and re-action force would not cancel each other. Those two forces are acting on different objects.

There are two forces acting on your object A:
 gravitation force (ms*g down) and the liquid try to push it up with buoyancy (ma*g up).

Because the liquid give your object A an upward force , so your object A also affects on the liquid ma*g down.
So the total force acting on the liquid is it's own gravitation force mb*g + ma*g (from object A).
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desert knight
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Reply #5 on: November 25, 2007, 04:49:21 pm »

thnx alot prof. for ur help and ur time
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