Jet on hinged plate, force and angle

Given: jet $d = 50$ mm, $V = 50$ m/s, $a = \dfrac{\pi}{4}(0.05)^2 = 1.9635\times 10^{-3}$ m². The plate is hinged at the top, and the jet strikes it normally at its centre.

Force required to hold the plate vertical

Force of the jet on the plate (normal to a flat plate at rest):

$$F = \rho aV^2 = 1000\times 1.9635\times 10^{-3}\times 50^2 = 4908.7\ \text{N}$$

This acts at the centre (distance $L/2$ from the hinge, where $L$ is the plate length). Taking moments about the hinge, with the horizontal force $P$ applied at the bottom edge (distance $L$):

$$P\times L = F\times\frac{L}{2}\;\Rightarrow\; P = \frac{F}{2} = 2454.4\ \text{N}$$

Angle of inclination when the force is removed

If the force $P$ is removed, the plate swings out by an angle $\theta$ from the vertical. Equilibrium requires the moment of the jet force about the hinge to equal the moment of the plate weight $W$ (acting at the plate's centre of gravity, $L/2$ from the hinge).

The jet strikes at a distance $\ell$ along the plate from the hinge, with $\ell\cos\theta = L/2$ (jet at fixed height). The normal force on an inclined plate is $F\cos\theta$ (the jet makes angle $90^\circ - \theta$ with the plate).

$$F\cos\theta\times\frac{L}{2\cos\theta} = W\times\frac{L}{2}\sin\theta\;\Rightarrow\;\sin\theta = \frac{F}{W} = \frac{\rho aV^2}{W}$$

The weight $W$ of the plate is needed and is not given. For example, if $W = 10$ kN: $\sin\theta = 4908.7/10000 = 0.4909$, so $\theta = 29.4^\circ$. If the plate weight is neglected, nothing resists the jet and the plate would swing up towards the horizontal ($\theta \to 90^\circ$).

Answer: applied force $P = F/2 = 2454 \approx 2.45$ kN; the angle follows from $\sin\theta = \rho aV^2/W = 4908.7/W$ (e.g. $29.4^\circ$ for $W = 10$ kN).