- #1
Kaushik
- 282
- 17
- Homework Statement
- A rectangular gate, ##3m * 1m## , stands vertical with the water on one side of it, hinged at the middle. What is the force F required to be applied at the bottom to keep the gate in equilibrium?
- Relevant Equations
- .
##\int \!dτ = \int \!P.dA.x ##
Here, I am taking torque of the part above the axis##(τ_1)## ,which is clockwise, and the torque of the part below the axis ##(τ_2)##, which is anticlockwise, separately.
Now, let me consider a thin strip ##dx## at a height ##x## from the axis.
##τ_1 = \int_0^{\frac{1}{2}} \!ρg(x-\frac{1}{2})dA.3x## ##\tag{1}##
Considering, a thin strip ##dx## at a depth ##x## from the axis.
##τ_2 = \int_0^{-\frac{1}{2}} \!ρg(x+\frac{1}{2})dA.3x## ##\tag{2}##
Net torque due to water is given by,
##τ_{water} = τ_1 - τ_2 ## (considering the anticlockwise torque to be negative direction)
Then,
##τ_{water} = \frac{F}{2}##
But this does not yield the answer. It seems like taking ##0## to ##-\frac{1}{2}## as the limit is the reason why I am getting an incorrect answer. But why is it so? What is wrong when I take ##-\frac{1}{2}##? I took ##-##ve as the points below the axis are negative while the points above the axis are ##+##ve, considering the axis as the origin?
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