Calculating Fluid Force of Triangle Submerged in Water

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SUMMARY

The discussion focuses on calculating the fluid force exerted on a triangle submerged in water, specifically a triangle with a base of 6 ft and a height of 4 ft, submerged to a depth of 3 ft. The user correctly set up the problem using the equation W ∫ h(y) L(y), where W is the weight density of water at 62.4 lbs/ft³. The integral was evaluated from -7 to -3, resulting in a calculated fluid force of 3244.8 lbs. The only noted concern was the unconventional use of negative y-values, which could lead to confusion in hydrostatic calculations.

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Homework Statement



Find the fluid force of a triangle with base of 6 ft and height of 4 ft submerged vertically into a body of water,vertex down, to a depth of 3 feet.


Homework Equations



from c to d W ∫ h(y) L(y)

The Attempt at a Solution



I set up the triangle on an x and y axis, splitting it down the middle. I made h(y) = -y and
L(y)= 2 (3/4(y+7)) (I found the equation of the line for the side of the triangle and multiplied it by 2). W is equal to 62.4

I set the problem up like this -6/4 x 62.4 times the integral from -7 to -3 of y(y+7).
When I integrated this I got 3244.8 lbs. I just want to know if I did this correctly. If not, please explain where I went wrong.
 
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Looks like you got it to me.

The only thing unconventional to me is using -ve y in your set up. Usually in hydrostatic problems y is just taken as +ve in the down direction. Using your convention the force should have come out -ve mathematically. But as long as you are consistent it didn't matter.
 

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