- #1

- 125

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I have no clue on how to do this problem

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

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- Thread starter nemzy
- Start date

- #1

- 125

- 0

I have no clue on how to do this problem

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

- #2

Tide

Science Advisor

Homework Helper

- 3,076

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- #3

hemmul

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

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

Science Advisor

Gold Member

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hemmul said:Thetotalforce 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

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

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