- #1
McAfee
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Before I begin I would like to say hello to everyone. This is my first post. I will follow and abide by all rules here.
A rectangular plate measuring 2m by 5m is suspended vertically in a liquid with density 500 kg per cubic meter in such a way that one of its longer sides is on top and is 3 meters below the surface of the liquid. Find the hydrostatic force on the plate.
The force(F) equals = Area*depth*density*acceleration
F= A * depth * density * acceleration
The density and acceleration are 500 * 9.8 = 4900, so that goes outside the ∫
The bounds of the integral are 3 to 5 since the plate is 3 meters deep.
The Area I think is equal to 5 * x and the depth I think is (3-x)
So my final integral would be 4900 ∫5x (3-x) dx from 3 to 5
Not sure if I'm right or not
Homework Statement
A rectangular plate measuring 2m by 5m is suspended vertically in a liquid with density 500 kg per cubic meter in such a way that one of its longer sides is on top and is 3 meters below the surface of the liquid. Find the hydrostatic force on the plate.
Homework Equations
The force(F) equals = Area*depth*density*acceleration
The Attempt at a Solution
F= A * depth * density * acceleration
The density and acceleration are 500 * 9.8 = 4900, so that goes outside the ∫
The bounds of the integral are 3 to 5 since the plate is 3 meters deep.
The Area I think is equal to 5 * x and the depth I think is (3-x)
So my final integral would be 4900 ∫5x (3-x) dx from 3 to 5
Not sure if I'm right or not