Calculating Force on Swimming Pool Bottom and Short Sides | Fluid Statics

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SUMMARY

The discussion focuses on calculating the force exerted by water on the bottom and short sides of a swimming pool measuring 23.0 m by 8.0 m with a depth of 3.0 m. The force on the bottom, calculated using the weight of the water and gravity, is confirmed to be 5.41E+06 N. For the short sides, participants suggest using the formula Pressure = F/A, but clarification is needed on how to apply it correctly in conjunction with the formula F = ρgAd, where ρ is the water density, g is the acceleration due to gravity, A is the area, and d is the depth.

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A swimming pool has dimension 23.0 m by 8.0 m with a depth of 3.0 m. When it is filled with water, what is the force (resulting from the water alone) on the bottom?

When it is filled with water, what is the force (resulting from the water alone) on each short side?

i figured out part a already by finding the weight of the water in the swimming pool and multiplying it by gravity (which gave me 5.41E+06 N) but i cannot figure out part b... am i suppose to use the formula
Pressure = F/A?? (i tried with serveral different numbers and still got the wrong answer)
 
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[tex]F=\rho g A d[/tex]
 

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