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I have semi-circle with its diameter facing the sky and it's sticking out of the water by 1 meter.

I need to use integration to figure out the pressure, but I'm getting the formula wrong. I'm not quite sure how to approach it since it's sticking out of the water.

What I did was density * gravity * depth * width = 1000(9.8)(5-x)(2x)

I know my formula is wrong, I just can't figure it out why.

I need to calculate the pressure on one surface of the semicircle.

1 Answers 1

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What is your semicircle made of, and why is it sticking out of the water? Where are you measuring the pressure?

The pressure (in pascal = newtons per square meter) in a fluid at rest is the product of the depth below the free surface of the fluid, the density and the gravitational acceleration. No integration is involved. The shape is irrelevant.

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    My semi-circle is just a shape. It says to express the hydrostatic force as an integral. The problem simply states that there is a partially submerged semicircle with a diameter of 6 meters facing up. It's submerged up to 5ft into the water. What I did was say my depth was (5-x) and the width of a layer of the semi-circle was 2x. I figured out 2x by using the similar triangle method. (width/x) = (5 over 5), hence width = 5x/5 Then I integrated. I have 6-7 problems like these, and I can't seem to get any of the solutions. – 2017-02-05
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    Well, then the problem is completely different from what you stated. You asked for the pressure, but you want the total force. – 2017-02-05
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    The semi-circle is out of the water by 1 meter and submerged up to 5ft into the water. With this information we have volume and applying https://en.wikipedia.org/wiki/Archimedes'_principle we can calculate density. Now with volume and density we can obtain the requested vertical hydrostatic force. What is not clear is if you want lateral hydrostatic pressure, vertical hydrostatic pressure, vertical hydrostatic force. Your question needs to be rewritten in an appropriate manner. – 2017-02-06