Fluids at Rest — Question 5

Question 5

Tilting a gate changes depth without changing area

A plane rectangular gate of width $b$ and length $L$ lies at angle $\theta$ above horizontal, with its upper edge at the liquid free surface. Its underside is wet and the opposite side has the same surface air pressure. Let distance $s$ run down the gate; its vertical depth is $s\sin\theta$.

1. Find the resultant net force and its distance along the gate from the upper edge.

2. Compare the force and hinge moment for $\theta=30^\circ$ with those for a vertical gate of the same dimensions. What happens as $\theta\to0$ in this geometry?

Question 5 - Solution

Tilting a gate changes depth without changing area

(a) Derivation and calculation

$$h=s\sin\theta,\qquad dF=\rho gb\,s\sin\theta\,ds.$$$$\boxed{F=\tfrac12\rho gbL^2\sin\theta},\qquad M=\tfrac13\rho gbL^3\sin\theta.$$$$\boxed{s_R=M/F=2L/3}\quad(\theta>0).$$

(b) Result and interpretation

$$\frac{F_{30^\circ}}{F_{90^\circ}}=\frac{M_{30^\circ}}{M_{90^\circ}}=\sin30^\circ=\boxed{\tfrac12}.$$$$\theta\to0:\quad h(s)\to0,\quad F\to0,\quad M\to0.$$The entire gate approaches the free surface in this geometry. A horizontal gate at finite depth is a different limit.

Physical check

The force is normal to the gate, so the along-gate distance is the correct moment arm.