---
title: "Jet on hinged plate, force and angle"
subject: "Fluid Mechanics"
chapter: 7
key: "semester_3/fluid_mechanics:jet-on-hinged-plate-force-angle"
source: https://www.bctengineer.com/q/semester_3/fluid_mechanics/jet-on-hinged-plate-force-angle
site: BCT Engineering
---

# Jet on hinged plate, force and angle

Fluid Mechanics · Chapter 7: Momentum principle and flow analysis

- Asked in: 2079 Baisakh, 3+3 marks
- Video: [Force Exerted by a Jet on a Hinged Plate](https://www.youtube.com/watch?v=gA7kDBVgD-c) — TutorialsPoint
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/fluid_mechanics/jet-on-hinged-plate-force-angle
- Whole chapter: https://www.bctengineer.com/md/semester_3/fluid_mechanics/chapter-7

## Question

A jet of the water, 50 mm in diameter, is striking normally with velocity of 50 m/s at the center of the plate which is hinged at its top edge and a horizontal external force is applied at the bottom edge to keep it vertical. What should be the amount of the applied force? If the force is removed what will be the angle of inclination of the plate with vertical for equilibrium condition?

## Short answer

Applied force at bottom to hold vertical plate is P = F/2 = 2454 N (about 2.45 kN). If force removed, equilibrium angle satisfies sinθ = ρaV²/W = 4908.7/W; e.g., 29.4° for a 10 kN plate weight.

## Answer

**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).**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. What is the force of a jet on a stationary normal plate?
   A. 2 rho a V^2
   B. rho a V^2
   C. rho a V
   D. rho a V^2 / 2
   Answer: B. All horizontal momentum is destroyed: F = rho a V^2 = 4908.7 N. (Key point: Jet force)

2. Why is the applied force P = F/2?
   A. P acts at L/2
   B. F acts at the hinge
   C. P acts at distance L from the hinge while F acts at L/2
   D. P equals the weight
   Answer: C. Moments about the hinge give P x L = F x L/2. (Key point: Moment balance)

3. What condition gives the equilibrium angle once P is removed?
   A. tan(theta) = F x W
   B. sin(theta) = rho a V^2 / W
   C. cos(theta) = W / F
   D. theta = 45 degrees
   Answer: B. Jet moment equals plate weight moment, giving sin(theta) = F/W. (Key point: Equilibrium angle)

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