Question

a circular opening 1.2m in diameter in the vertical side of a reservoir is closed be a disc which just fits the opening and is pivoted on a shaft passing trough its horizontal diameter. show that, if the water level in the reservoir is above the top of the disc, the turning moment of the shaft required to hold yhe disc vertical is independent of the head of water. calculate the amount of this moment

Answer 1

Showing that it is constant is fairly trivial.
The area above and below the shaft is the same shape.
Pressure - rho g h - i.e. it increses linearly with height.

Now you can consider the sum of the pressure on the disc (= force)

For each point above the pivot there is a corresponding point below the pivot. The pressure difference between them is rho g (delta h) (where delta h is the depth difference between corresponding points)
Thus when you sum all those pressure differences , above and below th epivot, the sum depends only on delta h , and NOT on the overall H (i.e. Not on the depth of the reservoir.

Answer 2

I THINK that to calculate the magnitude you need integrate



The Area is a function of h (possibly r sin theta dh)

SO this is not a simple integration

I don't have time (and I'm not sure I have the skill) to do this right now

Quite interesting problem though