# A dam, torque problem

1. Jul 27, 2005

### Swatch

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

2. Jul 27, 2005

### Staff: Mentor

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.

3. Jul 27, 2005

### Swatch

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

4. Jul 27, 2005

### Staff: Mentor

No, I said: $dF = 4 \rho g y dy$ (Don't forget the y.)

$dF = P dA$, not $dP dA$. You need the actual pressure at the depth y: $P = \rho g y$, not $\rho g dy$.

5. Jul 27, 2005

### Swatch

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?

6. Jul 27, 2005

### Staff: Mentor

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.)