A jet of water of 30 mm diameter, moving with a velocity of 15 m/s, strikes a hinged square plate of weight 245.25 N at the centre of the plate. The plate is of uniform thickness. Find the angle through which the plate will swing.

Force of a Jet on a Hinged Plate

Problem Statement

Given Data & Constants

Diameter of jet, $d = 30 \, \text{mm} = 0.03 \, \text{m}$
Velocity of jet, $V = 15 \, \text{m/s}$
Weight of plate, $W = 245.25 \, \text{N}$
Density of water, $\rho = 1000 \, \text{kg/m}^3$

Solution

1. Calculate the Force of the Jet ($F$)

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

$$ \text{Area of jet, } A = \frac{\pi}{4} d^2 = \frac{\pi}{4} (0.03)^2 \approx 0.00070686 \, \text{m}^2 $$ $$ F = \rho A V^2 = 1000 \times 0.00070686 \times (15)^2 $$ $$ F = 1000 \times 0.00070686 \times 225 \approx 159.04 \, \text{N} $$

2. Set up the Moment Balance Equation

Let $\theta$ be the angle the plate swings from the vertical. The plate is in equilibrium when the moment caused by the jet's force equals the restoring moment caused by the plate's weight. Let $L$ be the side length of the square plate. The hinge is at the top edge, and the jet strikes at the center ($L/2$ from the hinge).

$$ \text{Moment from jet} = \text{Restoring moment from weight} $$ $$ (F \cos\theta) \times \frac{L}{2} = W \times \left(\frac{L}{2} \sin\theta\right) $$

The term $L/2$ cancels from both sides, simplifying the equation.

$$ F \cos\theta = W \sin\theta $$ $$ \frac{\sin\theta}{\cos\theta} = \tan\theta = \frac{F}{W} $$

3. Calculate the Swing Angle ($\theta$)

Now we substitute the known values for the force and weight.

$$ \tan\theta = \frac{159.04 \, \text{N}}{245.25 \, \text{N}} \approx 0.6485 $$ $$ \theta = \arctan(0.6485) \approx 32.96^\circ $$

Final Result:

The plate will swing through an angle of approximately $33^\circ$.

Explanation of the Equilibrium

The solution is based on the principle of moments. For the plate to be held in a stable, angled position, the turning forces (moments) around the hinge must be perfectly balanced.

Jet's Moment: The water jet pushes the plate away. As the plate swings to an angle $\theta$, the effective force component acting perpendicular to the plate is $F \cos\theta$. This force acts at the center of the plate, creating a turning moment.
Weight's Moment: Gravity pulls the plate's weight straight down. As the plate swings, its center of gravity moves horizontally away from the hinge. This creates a restoring moment that tries to pull the plate back to its vertical position.

The plate settles at the exact angle where these two moments are equal and opposite, achieving a state of static equilibrium.

A jet of water of 30 mm diameter, moving with a velocity of 15 m/s, strikes a hinged square plate of weight 245.25 N at the centre of the plate. The plate is of uniform thickness. Find the angle through which the plate will swing.

Problem Statement

Given Data & Constants

Solution

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

2. Set up the Moment Balance Equation

3. Calculate the Swing Angle (\(\theta\))

Explanation of the Equilibrium

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A jet of water of 30 mm diameter, moving with a velocity of 15 m/s, strikes a hinged square plate of weight 245.25 N at the centre of the plate. The plate is of uniform thickness. Find the angle through which the plate will swing.

Problem Statement

Given Data & Constants

Solution

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

2. Set up the Moment Balance Equation

3. Calculate the Swing Angle (\(\theta\))

Explanation of the Equilibrium

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