You have a cup, it has a mass of 1 Kg.

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You have a cup, it has a mass of 1 Kg.
In the cup is a liquid, with a mass of 1.5 Kg.
the Cup (and liquid) are on a weight scale (call it X).
We have another object, with known volume (call it V), attached to a spring-scale from upwards (call it Y).
The object was put inside the liquid (not floating), X reads 7.5*g N, Y reads 2.5*g N, what is the density of the liquid.
Now, what most people i know did is that they said :
The weight of cup and liquid = (1+1.5)*g = 2.5*g N
And since scale X reads 7.5*g N, then the real (not apparent) weight of the object must be 5*g N (and they start to solve from there).
What i have made is that i drew a free body diagram, and i reached the conclusion that the real weight of the object is 7.5*g N.
Who do u think is right ?
Thanks in advance.
 
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  • #2
HallsofIvy
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Now, what most people i know did is that they said :
The weight of cup and liquid = (1+1.5)*g = 2.5*g N
And since scale X reads 7.5*g N, then the real (not apparent) weight of the object must be 5*g N (and they start to solve from there).
Not quite: the "Y" scale is supporting SOME of the weight.
The total weight of all parts: cup, fluid and object is the sum of the readings of X and Y: (7.5+ 2.5)g N= 10g N. Since we know that the cup and fluid together weigh 2.5g N it follows that weight of the object is, as you say, 10gN- 2.5gN= 7.5g Newtons.

The difference between that and the 2.5g N reading of the Y scale (i.e. 5.0g N) is the bouyancy of the fluid. It is equal to the volume of fluid displaced (and since it is not floating, that is the volume of the object) times the density of the fluid. You have the equation dV= 5 so that d, the density of the fluid is given by 5/V
 
  • #3
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Just as i expected.
Thanks !
 

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