Question 18.15: A flat plate, weighing 80 N is suspended vertically by a hin......

Given data:
Weight of plate                            W = 80 N
Diameter of jet                           d = 30 mm = 0.03 m
Velocity of jet                             V = 6 m/s

Distance of weight W from hinge             = 120 mm = 0.12 m

Distance of the axis of jet from hinge     = 200 mm = 0.2 m

Area of jet is              a=\frac{\pi}{4} d^2=\frac{\pi}{4}(0.03)^2=0.000707 \mathrm{~m}^2

Let P be the horizontal force applied at the centre of gravity of the plate to keep it vertical as shown in Fig. 18.12(a).

The force exerted by the jet on the plate is given by Eq. (18.1) as

F=\rho a V^2

=1000 \times 0.000707 \times 6^2=25.452 \mathrm{~N}

Taking moments about the hinge O, we have

P \times 0.12-F \times 0.2=0

or                                  P=\frac{F \times 0.2}{0.12}=\frac{25.452 \times 0.2}{0.12}=42.42 \mathrm{~N}

Let V be the velocity of jet when the plate is deflected through an angle of 30° as shown in Fig. 18.12(b).
The forces acting on the plate as shown in Fig. 18.12(b) are as follows:

(a) Weight of the plate, W acting at centre of gravity (G) at a distance 0.12 m from the hinge 0.
(b) Horizontal force, P acting at G
(c) Force, F_n due to jet of water acting normally at A.

The force exerted by the jet on the plate is given by Eq. (18.3) as

F_{x}=\rho a V^{2}\sin^{2}\theta      (18.3)

F_n=\rho a V^2 \sin \theta

=1000 \times 0.000707 \times V^2 \sin 30^{\circ}=0.3535 V^2 \mathrm{~N}

Taking moments about the hinge O, we have

F_n \times O A-P \times O C-W \times C G=0

or                  F_n \times \frac{0.2}{\cos 30^{\circ}}-P \times 0.12 \cos 30^{\circ}-W \times 0.12 \sin 30^{\circ}=0

or              0.3535 \mathrm{~V}^2 \times \frac{0.2}{\cos 30^{\circ}}=42.42 \times 0.12 \cos 30^{\circ}+80 \times 0.12 \sin 30^{\circ}

or              0.0816 V^2=4.408+4.8=9.208

or            V^2=112.84

or            V=\sqrt{112.84}=10.62 \mathrm{~m} / \mathrm{s}