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Homework Help: Torque and water pressure

  1. Dec 9, 2004 #1
    There is a gate in a dam. The upper edge of the gate runs along the water surface. The gate is 2 m high and 4 m wide and is hinged along a horizontal line through its center. I have to calculate the torque about the hinge arising from the force due to the water. I know how to calculate torque and water pressure with depth but I have no clue about this problem, I got a clue that tells me to calculate the torque of a thin horizontal strip at a depth h and integrate this over the gate, how should I do that. If someone has a clue, please help me.
  2. jcsd
  3. Dec 9, 2004 #2


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    Calculate the hydrodynamical force with which the water acts upon the dam's gate.Use particular geometry of the problem (meaning the perpendicularity between the force and the distance between the point of application and the the center of rotation) to find the torque.

  4. Dec 9, 2004 #3
    There is in fact no problem finding the force on, lets say the bottom of the gate, but the force is not constant on the whole gate because deeper you go into the water, more pressure there is, isn't that correct. So I think my biggest problem is finding a equation for the variables and integrating it, and that's where I'm stuck
  5. Dec 9, 2004 #4


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    It seems like a pretty straightforward problem. Let's take a look at a thin layer of water at height [itex]h[/itex]. Then we can say that the water pressure at that height is [itex]p(h)[/itex]. (We can assume that water pressure at a particular depth is constant.)

    Now, the force exerted by water is equal to the pressure multiplied by the area. The area of a thin strip of the gate is going to be [tex]w \times dh[/tex]. So the force at a particular depth will be [tex]f(h)=p(h)\times w \times dh[/tex]. The torque due to the force is going to be [tex](h-h_{pivot}) \times f(h)[/tex], so all you have to do is integrate that from the top of the gate to the bottom:
    [tex]\int_{h_{min}}^{h_{max}} (h-h_{pivot}) p(h) w dh[/tex]

    Remeber that the sign depends on how your coordinates are set up.
  6. Dec 10, 2004 #5
    OK. Let's see If I got this right.
    Lets say that the pivot has height 0 meters, the top of the gate then is 1 and the bottom is -1

    The equation that I came up with looks like this:

    Torque(h)= (Po + density*g* -h)-4h

    What I don't know is how to integrate this equation

    I guess I have to integrate once with the limits 0 to 1 and then once with 0 to -1
    that way I get the torque on either side of the pivot
    Last edited: Dec 10, 2004
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