---
title: "Water jet on fixed vane, 100 mm2"
subject: "Applied Mechanics (Dynamics)"
chapter: 3
key: "semester_3/applied_mechanics_dynamics:nozzle-fixed-vane-force-100mm2"
source: https://www.bctengineer.com/q/semester_3/applied_mechanics_dynamics/nozzle-fixed-vane-force-100mm2
site: BCT Engineering
---

# Water jet on fixed vane, 100 mm2

Applied Mechanics (Dynamics) · Chapter 3: System of particles

- Asked in: 2079 Bhadra, 2+6 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/applied_mechanics_dynamics/nozzle-fixed-vane-force-100mm2
- Whole chapter: https://www.bctengineer.com/md/semester_3/applied_mechanics_dynamics/chapter-3

## Question

Define linear and angular momentum of system of particles. A nozzle discharges a stream of water of cross-sectional area $A = 100$ mm$^2$ with a speed of $v = 60$ m/s and the stream is deflected by a fixed vane as shown in figure. The mass density of water $\rho = 1000$ kg/m$^3$. Determine the resultant force $\vec F$ exerted on the stream by fixed vane. [Figure: stream arriving horizontally at $V_B$ and leaving the vane at C at $60^\circ$ to the horizontal with $V_C$.]

## Short answer

Linear momentum of a system is L = m v̄, angular momentum H_O = Σ r_i × m_i v_i. For the fixed vane, mass flow rate ṁ = ρ A v = 6 kg/s. Force on stream: F = ṁ (v_C – v_B) = (-180 i + 311.8 j) N, magnitude 360 N at 120° to the incoming flow.

## Answer

#### Linear and angular momentum of a system of particles
- **Linear momentum:** $\vec L=\sum m_i\vec v_i=m\vec{\bar v}$, where $m$ is the total mass and $\vec{\bar v}$ the velocity of the mass centre. Newton's law: $\sum\vec F=\dot{\vec L}$.
- **Angular momentum about O:** $\vec H_O=\sum\vec r_i\times m_i\vec v_i$, with $\sum\vec M_O=\dot{\vec H}_O$.

#### Force of a fixed vane on a steady stream
**Assumption:** the stream arrives horizontally (to the right) and leaves at C at $60^\circ$ above the horizontal, in the forward direction. Since the vane is fixed and friction is neglected, the speed is unchanged: $v_B=v_C=60$ m/s.

```text
   V_B --->   ______
              \      \  V_C (60 deg above horizontal)
               \______\
                 vane
```

**Mass flow rate:**
$$
\dot m=\rho Av=1000(100\times10^{-6})(60)=6.0\ \text{kg/s}
$$

**Momentum equation** (force on the stream by the vane):

$$
\vec F=\dot m(\vec v_C-\vec v_B)
$$

$$
\begin{aligned}
F_x&=6\left(60\cos60^\circ-60\right)=-180\ \text{N}\\
F_y&=6\left(60\sin60^\circ-0\right)=311.8\ \text{N}\\
F&=\sqrt{180^{2}+311.8^{2}}=360\ \text{N}
\end{aligned}
$$

The direction is $\tan^{-1}(311.8/180)=60^\circ$ above the negative x-axis (that is $120^\circ$ from +x). Check: $F=2\dot mv\sin(\tfrac{\alpha}{2})=2(6)(60)\sin30^\circ=360$ N, where $\alpha=60^\circ$ is the deflection angle.

**Answer: the vane exerts $\vec F=(-180\hat i+311.8\hat j)$ N on the stream, magnitude $360$ N at $120^\circ$ to the original flow (the stream pushes on the vane with an equal and opposite force).**

## Check yourself

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

1. What is the linear momentum of a system of particles?
   A. Sum of r_i x m_i v_i
   B. m times the speed of the fastest particle
   C. m v_bar, the total mass times the mass-centre velocity
   D. Sum of (1/2) m_i v_i^2
   Answer: C. L = sum m_i v_i = m v_bar, and Sum F = dL/dt. (Key point: Linear momentum)

2. What is the mass flow rate for A = 100 mm^2, v = 60 m/s?
   A. 60 kg/s
   B. 0.6 kg/s
   C. 6.0 kg/s
   D. 600 kg/s
   Answer: C. m dot = rho A v = 1000(100e-6)(60) = 6.0 kg/s. (Key point: Mass flow rate)

3. Which equation gives the force on the stream by the vane?
   A. F = m dot (v_C + v_B)
   B. F = m dot (v_C - v_B)
   C. F = m dot v_C
   D. F = m dot (v_B - v_C)
   Answer: B. Force equals the rate of change of momentum, m dot times (exit velocity minus entry velocity). (Key point: Force on stream)

4. For deflection angle 60 degrees and unchanged speed, what is the force magnitude?
   A. 2 m dot v sin(30) = 360 N
   B. m dot v = 360 N for any angle
   C. m dot v cos(60) = 180 N
   D. 2 m dot v = 720 N
   Answer: A. F = 2 m dot v sin(alpha/2) = 2(6)(60)(0.5) = 360 N. (Key point: Magnitude check)

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