Pressure on a Dam: Calculating Force & Torque

[Kalie]

As the reservoir behind a dam is filled with water, the pressure that the water exerts on the dam increases. Eventually, the force on the dam becomes substantial, and it could cause the dam to collapse. There are two significant issues to be considered: First, the base of the dam should be able to withstand the pressure rho*g*h, where rho is the density of the water behind the dam, h is its depth, and x is the magnitude of the acceleration due to gravity. This means that the material of which the dam is made needs to be strong enough so that it doesn't crack (compressive strength).

http://session.masteringphysics.com/problemAsset/1011118/16/SFL_pl_5.jpg

Consider a horizontal layer of the dam wall of thickness located a distance above the reservoir floor. What is the magnitude dF of the force on this layer due to the water in the reservoir?

The force of the water produces a torque on the dam. In a simple model, if the torque due to the water were enough to cause the dam to break free from its foundation, the dam would pivot about its base (point P). What is the magnitude tau of the torque about the point P due to the water in the reservoir?

Well on part A I found the pressure, vertical height above the floor of the reservoir
p(x)=pg(h-x) but I don't know what to do from there. There are no values so it is just letter variables

 

[OlderDan]

It will probably help you to assume the dam has a width L so that you can calculate the area of a strip at a given depth. The pressure at that depth exerts a force on the strip that is the pressure times the area of the strip. The torque that you calculate will actually be the torque about a line through point P, and it will be proportional to L. The wider the dam, the greater the force on the dam and the greater the torque. As you probably already realize, this is an integral calculus problem.