Calculating Torque on a Dam: How Does Changing the Height Affect the Lever Arm?

[Fisicks]

Homework Statement

Water stands at a height h behind a vertical dam of uniform width b.
Show that the torque about the base of the dam due to this force can be considered to act with a lever arm equal to h/3.

Homework Equations

F=PA
T=Fr(perpendicular)

The Attempt at a Solution

Im picturing a giant dam in my head and I just do not see how reducing the height to 1/3, will have the same torque. I also calculated the entire force on the dam which is pg(h^2)b/2.

Another part of the question is For a freestanding concrete dam of uniform thickness t and height h, what minimum thickness is needed to prevent overturning? and nothing is coming to mind.

Thanks for your help,
Fisicks

 

[Delphi51]

You have a great start with the F = pg(h^2)b/2.
You must have integrated dF over the height of the dam.
Repeat that for integral of dT = h*dF to get the torque.
Having Torque and Force, you can use T = F*d to solve for d. It works out to h/3.

The second part must be to think of the dam as a rectangular block of concrete with a force acting 1/3 of the way up. If it is too thin, the force will tilt it up on the outer edge like pushing a domino over. I'm thinking you would do torques about this edge - the force of the water torque vs the weight of the concrete torque. The first will have 1/3 the height of the dam in it and the second will have half the thickness of the dam.

 

[Fisicks]

Thanks! I appreciate it.

 

[Delphi51]

Most welcome! Interesting problem.

 

[Fisicks]

Ok. the second part is giving me a little trouble.

The weight of the dam must equal to force caused by the torque at 1/3h.

Thus (tbh)(density of concrete)g=(density of water)g((h/3)^2)b/2 which is incorrect.

Also, i replaced the right side with the value i got for total force and it was still wrong.

 

[Delphi51]

There is something wrong with that equation. Back up a bit.
The sum of the torques on the bottom left (outer) of the concrete = 0
That is, mgt/2 - F*h/3 = 0
since the weight acts at the center of the concrete and the water at height h/3.