What is the total force on the face of a small dam with a hemisphere shape?

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SUMMARY

The total force on the face of a small dam shaped as a hemisphere with a diameter of 40 meters can be calculated using the equation for a semicircle. The radius of the hemisphere is 20 meters, and the equation for the semicircle is x² + y² = 400. The length of a line across the face at a depth y is determined to be 2√(400 - y²). This approach eliminates the need for similar triangles, focusing instead on the geometric properties of the semicircle.

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Homework Statement



The following figures show the shape and dimensions of small dmas.
Assuming the water level is at the top of the dam, find the total force on that face of the dam.

There's a figure of a hemisphere whose diameter is 40 meters

Homework Equations


The Attempt at a Solution



So it's radius is 20 meters. I'm struggling with this problem with trying to determine the function for the radius in terms of y so that way I can find the total force. Sense this is a hemisphere shape I can't use similar triangles but is the ratio of height, y, to length of the diameter, still 1 to 2? I just wanted to make sure I could before I started solving this problem. I know what to do if I can make this assumption I just wanted to make sure it was a valid assumption.
 
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There is no figure so I don't know exactly what you mean. But it seems very strange that a dam face would be a three dimensional figure! Are you sure the dam face is not a semicircle?

If it is a semicircle, of radius 20, with the straight diameter at the top, then, taking the origin at the center of that diameter, x^2+ y^2= 400, so that x= \pm\sqrt{400- y^2}. So the length of a line across the face at depth y is 2\sqrt{400- y^2}.
 
My bad. Yes it is and thanks. I thought I had to apply some similar shape thing as that's what I have been doing but I forgot about x^2 + y^2 = r^2 thanks
 

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