Calculating torque for a small horizontal strip

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The discussion focuses on calculating the torque exerted by water on a gate at a dam, which is hinged at its center. The user initially integrates the torque from the top to the center of the gate and then from the center to the bottom, but arrives at an incorrect net torque. Clarifications reveal that the force on a thin horizontal strip should be expressed as dF = 4ρgy dy, emphasizing that pressure varies with depth. The correct approach involves integrating the torque contributions from both halves of the gate while considering the pressure at each depth. Ultimately, the user successfully arrives at the correct torque value after addressing the misunderstanding about pressure and force calculations.
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The uppper edge af a gate in a dam runs along the water surface. The gate is 2.00 m high and 4.00 m wide and is hinged along a horizontal line trough its center. I have to calculate the torque about the hinge arising from the force due to the water. What I have done so far is to consider the top of the gate as point y=0 , the center as y=1 and the bottom y=2.
Then I calculate the torque of a thin horizontal strip at a depth y and integrate that over the gate.

Total torque from the top to the center would be

Torque=intergrate(r*g* dy *dA ) or 4*r*g*integrate(y^3)
(where r=density of water or 1.00*10^3 kg/m^3)

from this I get torque to be 9800 N*m

Then I do the same for the gate from height y=1 to y=2
and get -147000

The net torque I get is not the right answer, which is by the way2.61*10^4 N*m

Could someone please give me a hint to what I am doing wrong
 
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It's not clear to me what you are integrating. The force on a thin horizontal strip is dF = 4 \rho g y dy. For the top half of the gate, the torque on that piece (about the center) is d\tau = (1-y) 4 \rho g y dy. Now integrate from y = 0 to 1 to find the net torque on the top half.

Write a similar expression for the torque on the bottom half.
 
Did what you told me Doc Al, and got the right answer. Just one thing I'm not understanding. You say that the force on a thin horizontal strip is dF=4*rho*g dy

The way I see the force is dF=dp*dA
since dp=rho*g dy and dA=4*dy
shouldn't the integration contain y^2
 
Swatch said:
Just one thing I'm not understanding. You say that the force on a thin horizontal strip is dF=4*rho*g dy
No, I said: dF = 4 \rho g y dy (Don't forget the y.)


The way I see the force is dF=dp*dA
since dp=rho*g dy and dA=4*dy
shouldn't the integration contain y^2
dF = P dA, not dP dA. You need the actual pressure at the depth y: P = \rho g y, not \rho g dy.
 
O.K.

So at this little horizontal strip or dA I consider the pressure as constant. Is that the correct way to think of it?
 
Swatch said:
So at this little horizontal strip or dA I consider the pressure as constant. Is that the correct way to think of it?
Right. For a small horizontal strip you can consider the pressure to be uniform, just like you can consider it to be at a single depth.

(FYI: Realize that the pressure varies from P to P + dP across the thickness of the strip. So, the average pressure is P + dP/2. So the average force would be (P + dp/2)(dA) = PdA + dp dA/2. Speaking loosely, that last term is a higher order infinitesimal and can be ignored compared to PdA.)
 
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