Question

The force of the added 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 τ of the torque about the point P due to the water in the reservoir?

Answer 1

Answer:

The appropriate response is "$\tau=\frac{1}{6} PgLh^3$". A further explanation is described below.

Explanation:

The torque ($\tau$) produced by the force on the dam will be:

⇒  $d \tau=XdF$

On applying integration both sides, we get

⇒  $\tau = \int_{0}^{a}x pgL(h-x)dx$

⇒     $= pgL\int_{0}^{h}(h-x)dx$

⇒     $=pgL[\frac{h^3}{2} -\frac{h^3}{3} ]$

⇒     $=\frac{1}{6} PgLh^3$