# Fluid mechanics and torque

Fluid mechanics and torque....

## Homework Statement

A square gate of size 1m*1m is hinged at its mid point. A fliud of density rho fills the space to the left of the gate. A force F is applied at the bottom of the gate towards right. Find value of F such that the gate is stationary....

## The Attempt at a Solution

First i calculated the clockwise torque( which is due to the top half of the water)....

If i take an element dh at a depth h from the top then excess pressure on this part due to the water is rho*g*h so force is rho*g*h*dh*1 and toque due to this element is ... rho*g*h*(0.5 - h)dh...Integrating this expression with h varying from 0 to 0.5 m we get net torqe(due to upper half) as
rho*g*0.53/6.....

Now calculating the anticlockwise torque(due to lower half) we get the expression for torque as rho*g*h*dh*(h-0.5).......integrating this with limits of h from 0.5 to 1 we get net torque due to this half as rho*g*0.375/6.......
So net torque (which is anticlockwise) is (anticlock wise - clock wise torque) which come out to be equal to rho*g*.25/6......

This torque is equal to torque due to the force F.....So F*.5= rho*g*.25/6......From here we get F=rho*g/12.....but answer is rho *g/6.....I tried it many times but i am still wrong...Please help.....

Please attach a diagram. Cannot figure it out correctly.

Heres the diagram.....the left part is water...and the arrow gives the point of application of force.....

#### Attachments

• fi.bmp
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My answer came rho *g/6. The only thing I did differently was that I did not separate the upper and lower part. I used torque = rho*gh(h-0.5)dh