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Homework Help: A dam, torque problem

  1. Jul 27, 2005 #1
    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. jcsd
  3. Jul 27, 2005 #2

    Doc Al

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    Staff: Mentor

    It's not clear to me what you are integrating. The force on a thin horizontal strip is [itex]dF = 4 \rho g y dy[/itex]. For the top half of the gate, the torque on that piece (about the center) is [itex]d\tau = (1-y) 4 \rho g y dy[/itex]. 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.
     
  4. Jul 27, 2005 #3
    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
     
  5. Jul 27, 2005 #4

    Doc Al

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    Staff: Mentor

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


    [itex]dF = P dA[/itex], not [itex]dP dA[/itex]. You need the actual pressure at the depth y: [itex]P = \rho g y[/itex], not [itex]\rho g dy[/itex].
     
  6. Jul 27, 2005 #5
    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?
     
  7. Jul 27, 2005 #6

    Doc Al

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    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.)
     
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