Pressure, Force question

  • Thread starter nemzy
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  • #1
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Here is the problem. A swimming pool has dimensions 30 m x 12 m and a flat bottom. What is the force caused by the water on each end? (the ends are 12m and 30m)

I have no clue on how to do this problem

i know that p1=p2 , but how would u find the pressure? thanks
 

Answers and Replies

  • #2
Tide
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How much force would it take to support a column of water of a given depth? How much force is the atmosphere above exerting on that column?
 
  • #3
hemmul
The total force applied to each of the sides of a pool is:
F=P*S+ro*g*H/2;
where the height of the swimming pool is H (do i need to proove that the pressure of a water column is proportional to its height? at least this is the reason of 1/2 in the answer)
P - is atmospheric pressure, at the "surface-level". According to Pascal's law it makes its contribution to the total pressure.
S - is the area of a side of the pool. That is either 12*H or 30*H - depending on which side you choose.
ro - the density of the water...

This was in case you needed the force applied to each of the sides. But the fact that the word "end" is used in your post makes me think that you are interested in the forces that act the surface-part-of-the-side-faces. In this case you should use the surfcae tension formula: the total force on each of the sides (neglecting the distortions in the corners) is given simply by
F=sigma*L
where sigma - is a surface tension, and L - the length of the "end" (12 or 30 m).
If you need to take into the account the non-planar type of water surface (wave-peaks etc) then you just have to integrate dF=sigma*dl along each face...

does it make sense? ;)
 
  • #4
Clausius2
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hemmul said:
The total force applied to each of the sides of a pool is:
F=P*S+ro*g*H/2;
where the height of the swimming pool is H (do i need to proove that the pressure of a water column is proportional to its height? at least this is the reason of 1/2 in the answer)
P - is atmospheric pressure, at the "surface-level". According to Pascal's law it makes its contribution to the total pressure.
S - is the area of a side of the pool. That is either 12*H or 30*H - depending on which side you choose.
ro - the density of the water...

Your "F" hasn't got units of force. [tex]\rho g H/2[/tex] is a pressure.
 
  • #5
hemmul
Clausius2 said:
Your "F" hasn't got units of force. [tex]\rho g H/2[/tex] is a pressure.
oh, sure... i have forgotten to specify the multiplication by S along with P... :approve:
 

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