Variation of Pressure with Depth problem

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SUMMARY

The discussion focuses on calculating the torque exerted by water on a hinged hatch in a rectangular tank filled with water 2 meters deep. The force on the hatch is derived using the equation P = pg(2-y), leading to a calculated force of approximately 2.943 x 10^-4 N. The correct approach to find torque involves integrating the varying force across the hatch's height, ultimately yielding a torque value of 16.3 kN. Key insights include the necessity of correctly identifying the distance from the hinge to the force application point, which is not simply (2-y).

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Homework Statement



A rectangular tank is filled with water 2 m deep. At the bottom of one side wall is a rectangular hatch 1 m high and 2 m wide that is hinged at the top of the hatch. a) Determine the force the water causes on the hatch. b) Find the torque caused by the water about the hinges.

I only need help with part (b)

Homework Equations



T = Fd

The Attempt at a Solution



For part a)
P = pg(2-y) ; dA = w*dy
dF = PdA = (integral from 0 to 1) pg(2-y)(wdy)
Plugging in numbers, (integral from 0 to 1) (1000)(9.81)(2-y)(2)dy
=1000(9.81)(2) [2y - y^2/2]
= 2.943 * 10^-4 N

I found out how to get force. With this answer I need to find the torque about the hinges. I know T = Fd but I'm confused as to how to approach this problem because the force varies with depth. What is the value of d?

dT = dF*r = pg(2-y)*2dy*2

This was my attempt but it's definitely way off. The answer is 16.3 kN.

Thanks!
 
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you have to integrate the distance to the top of the hatch multiplied by the force on a horizontal strip of the hatch. If your y-coordinates start with 0 at the bottom of the tank and positivie is up, than the distance to the top of the hatch is NOT (2-y).
I also don't understand the dy^2 in your expression for dT.
 

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