Fluid mechanics and torque: Finding the equilibrium force on a hinged gate

[Abhishekdas]

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...

Homework Equations

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...

 

[ashishsinghal]

Please attach a diagram. Cannot figure it out correctly.

 

[Abhishekdas]

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

 

[ashishsinghal]

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

 

[ashishsinghal]

Now, I did it using your method still, I got the correct answer. I'll advise you to check your calculations

 

[ashishsinghal]

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...

Your answer here is 3*rho*g/48 while the answer should be 5*rho*g/48

 

[Abhishekdas]

Ya i made a mistake there...i checked it again with a calci and i got it now...