How Much Force Must a Fireman Exert to Hold a Steadily Ejecting Hose?

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Homework Help Overview

The discussion revolves around a physics problem involving a fireman's hose, specifically focusing on the forces involved as water is ejected from the nozzle. The problem includes parameters such as the cross-sectional areas of the hose and nozzle, the speed of the water, and the rate of water flow.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants explore different methods to calculate the force exerted by the fireman, with one participant attempting to use kinematic equations and another suggesting the conservation of momentum. Questions arise regarding the validity of the initial calculations and the assumptions made about mass flow rate.

Discussion Status

The discussion is active, with participants questioning the original approach and seeking clarification on the calculations related to mass flow rate and momentum. Some guidance has been offered regarding the use of conservation of momentum, but there is no explicit consensus on the correct method or interpretation of the problem.

Contextual Notes

Participants are grappling with the implications of varying velocities within the hose and the relationship between mass flow rate and cross-sectional area. There is an ongoing debate about the assumptions made in the calculations and the relevance of different speeds at various points in the hose.

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The question states: A firemans hose has a nozzle of area 5cm^2 and the hose has a cross sectional area of 360cm^2. The hose ejects water at 25m/s and at a rate of 12L/s What is the force that the fireman must exert to keep the hose steady?

My attempt:

What I have done so far is to calculate the speed of water as it travels through the parts of the hose.

In the nozzle it travels at 24m/s (12000/500) and in the hose it is traveling at (12000/36000) = 0.33ms^-1, and as it is ejected, it is traveling at 25m/s.

So it goes from traveling at 0.33m/s to 24m/s to 25m/s. Hence, the force is
m(v-u)/t = 12kg/s * (25-0.33) = 296N​

So the fireman exerts a force of 296N in the direction of the flow of water.

Is this correct? It seems to simplistic to ignore the 24m/s in the nozzle, yet it seems to work.
 
Last edited:
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This question is to do with momentum. Try and use the conservation of momentum to find the force the firefighter must exert on the hose to steady himself. You know the volume of water being ejected so its easy to find the mass. Also think of the time dependence of the mass.
 
Sorry, but I don't understand what you're getting at. What was wrong with the original approach?
 
zaguar said:
Sorry, but I don't understand what you're getting at. What was wrong with the original approach?

How is m=12kg/s?
The mass flow rate will depend on the cross section of the hose and the velocity. So, it won't be constant.

More specifically \dot{m} = \rho A v, where \dot{m} is the mass flow rate.

What Kurdt says, is to find the difference in the momentum entering and leaving the hose. You can find the force from that.
 
Last edited:
Isn't that what I did?

Isn't the flow rate always going at 12L/s and the velocity changing, so that there is always 12 litres of water going through the pipe per second, but the speed is different in different parts of the pipe?

So the water is traveling at 0.33m/s in the hose, and exiting at 25m/s. If it is doing that, F=(mv-mu)/t=12*(25-0.33)=296N
 

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