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## Homework Statement

Submerged (vertically) right triangle 12x9x15(hypotenuse) with 12 m leg parallel to water surface. Top of triangle is 3m below surface. Find force on triangle.[/B]

## Homework Equations

I know mass of water is 840 k/m

^{3}. I think I should also multiply times 9.8 for gravity? One formula I found has to get Pressure=w*g*d for water mass, gravity and depth below surface respectively. Then to take pressure times area to get force. F=PA. But does this only work when submerged horizontally?

## The Attempt at a Solution

I feel like I'm not "getting" how to decide what domain to integrate over and when to rotate axes. I let the surface of water be x-axis.

Assuming the short leg to be set on y-axis, I found the slope of hyp to be 3/4 so the x coordinate of dy to be L(x)=4/3y

So my integral looked like:

840* INT[-12 to -3] -y(4/3y)dy

840*[(-4y

^{3})/9] |eval over(-12 to -3)

840*[12-768] Now this gives me a negative number, so I should just reverse their order? Is this otherwise correct? I get so confused on the origin placement. If I reverse them it becomes 840[768-12]=635,040

Then since this is metric, I express it in Newtons? 6.35*10

^{5}N Should this have been multiplied times the 9.8 gravity thing or no?

Thanks for any input or suggestions!