# Force required to hold a fire hose

1. Sep 6, 2014

### toothpaste666

1. The problem statement, all variables and given/known data

A fire hose exerts a force on the person holding it. This is because the water accelerates as it goes from the hose through the nozzle.

How much force is required to hold a 6.7-cm-diameter hose delivering 490L/min through a 0.80-cm-diameter nozzle?

2. Relevant equations
A1V1 = A2V2

P1 + .5ρV1^2 + ρgY1 = P2 + .5ρV2^2 + ρgY2

3. The attempt at a solution

First I converted the diameters into meters, divided it by 2 to get the radius and then found the areas of the hose and nozzle. A1 (the hose) = .0036 m^2 A2 (the nozzle) = 5.0x10^-5 m^2. Then i converted V1 (the velocity in the hose) to m^3/s. V1 = .0082 m^3/s.
Next I must find V2.

A1V1 = A2V2

V2 = A1V1/A2 = (.0036)(.0082)/(5.0x10^-5) = .59 m^3/s

Then I must use bernoulli's equation to find the pressure in the hose and multiply by the area to find the force. the part of the hose that connects to the nozzle and the nozzle are at the same height so Y1 and Y2 = 0. So I am left with

P1 + .5ρV1^2 = P2 + .5ρV2^2

This is where I am stuck. I know the density of water is around 1000 kg/m^3 and I know both of the velocities , but i still have two unknowns (P1 and P2). since the hose/nozzle is a confined space, can I say that P1=P2 or does that only work for non moving fluid?

anyway after that I would have to find P1 (pressure in hose) and multiply by A1 to find the force the hose puts on the person, which would be equal in magnitude to the force needed to hold it. How does my solution look so far?

2. Sep 7, 2014

### SteamKing

Staff Emeritus
This is a round-about way to do this: For a circle,

$A = \pi r^{2} = \pi D^{2}/4$

What units are used to measure velocity?

Why must you use Bernoulli's equation? It doesn't seem particularly applicable to your problem.

Think about what is happening in the nozzle. You are flowing a certain amount of water thru the hose, which means it is travelling at constant velocity. When the flow reaches the nozzle, what happens to the velocity?

Hint: Try re-writing Newton's Second Law, F = m a, for the nozzle.

3. Sep 7, 2014

### toothpaste666

ok I see my first mistake. V1 = .0082/A1 = .0082/.0036 = 2.3 m/s. V2 = .0082/(5.0x10^-5) = 164m/s. When it gets to the nozzle the area gets smaller so the velocity has to go up which is an acceleration. When there is acceleration there is a force so F = ma = m(Δv/Δt)

Δv = V2-V1 = 164 - 2.3 = 162 m/s

I will choose a 1 sec interval so Δt = 1

a = (162 m/s)/s = 162m/s^2

to find m multiply the volume by the density

m = (.0082)(1000) = 8.2 kg

F = ma = (162m/s^2)(8.2 kg) = 1300 kgm/s^2 = 1300 N

4. Apr 10, 2015

### Andy Barrette

Bernoulli's equation relates pressure (F/A) and flow velocity. You can solve for A and flow velocity, and the goal is to find F. Why do you say that Bernoulli's equation is not applicable here? It is misleading.

Last edited: Apr 10, 2015