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- Thread starter gunnar
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gunnar said:

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.

Daniel.

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NateTG

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gunnar said:

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.

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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

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

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