A plate is acted upon at its centre by a jet of water of diameter 20 mm with a velocity of 20 m/s. The plate is hinged and is deflected through an angle of 15°. Find the weight of the plate. If the plate is not allowed to swing, what will be the force required at the lower edge of the plate to keep the plate in vertical position.

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Analysis of a Jet on a Hinged Plate

Problem Statement

A plate is acted upon at its centre by a jet of water of diameter 20 mm with a velocity of 20 m/s. The plate is hinged and is deflected through an angle of 15°. Find the weight of the plate. If the plate is not allowed to swing, what will be the force required at the lower edge of the plate to keep the plate in vertical position.

Given Data & Constants

  • Diameter of jet, \(d = 20 \, \text{mm} = 0.02 \, \text{m}\)
  • Velocity of jet, \(V = 20 \, \text{m/s}\)
  • Deflection angle, \(\theta = 15^\circ\)
  • Density of water, \(\rho = 1000 \, \text{kg/m}^3\)

Solution

1. Calculate the Normal Force of the Jet (\(F\))

First, we calculate the force the jet would exert if it were striking the plate perpendicularly.

$$ \text{Area of jet, } A = \frac{\pi}{4} d^2 = \frac{\pi}{4} (0.02)^2 \approx 0.00031416 \, \text{m}^2 $$ $$ F = \rho A V^2 = 1000 \times 0.00031416 \times (20)^2 $$ $$ F = 1000 \times 0.00031416 \times 400 \approx 125.66 \, \text{N} $$

Part 1: Find the Weight of the Plate (W)

The plate is in equilibrium when the moment from the jet's force equals the restoring moment from the plate's weight. Let \(L\) be the side length of the plate.

$$ (F \cos\theta) \times \frac{L}{2} = W \times \left(\frac{L}{2} \sin\theta\right) $$ $$ F \cos\theta = W \sin\theta \implies W = \frac{F \cos\theta}{\sin\theta} = \frac{F}{\tan\theta} $$ $$ W = \frac{125.66}{\tan(15^\circ)} = \frac{125.66}{0.2679} \approx 469.05 \, \text{N} $$

Part 2: Force Required to Keep Plate Vertical (\(F_k\))

If the plate is held vertical, the jet strikes normally. The moment from the jet's force must be balanced by the moment from the required force (\(F_k\)) applied at the lower edge (distance L from the hinge).

$$ \text{Moment from jet} = \text{Moment from required force} $$ $$ F \times \frac{L}{2} = F_k \times L $$ $$ F_k = \frac{F}{2} = \frac{125.66}{2} \approx 62.83 \, \text{N} $$
Final Results:

The weight of the plate is approximately \(469.05 \, \text{N}\).

The force required at the lower edge is approximately \(62.83 \, \text{N}\).

Explanation of the Equilibrium Conditions

This problem analyzes two different scenarios of static equilibrium by balancing moments around the hinge.

  • Swinging Plate: The jet's force pushes the plate away, creating a turning moment. As the plate swings, its own weight creates an opposing, restoring moment. The plate settles at the angle (15°) where these two moments are equal. By equating them, we can solve for the unknown weight of the plate.
  • Vertical Plate: When the plate is held vertically, its weight acts directly through the hinge and creates no moment. The jet's force, acting at the center, creates a turning moment. To prevent the plate from swinging, an opposing force must be applied. Since this force is applied at the lower edge (twice the distance from the hinge as the jet's force), it only needs to be half the magnitude of the jet's force to create an equal and opposite moment.
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