Force and Volume Relationship in Fire Hose Operation

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Homework Help Overview

The problem involves the relationship between force and volume in the context of a fire hose operation, specifically examining how changes in water flow affect the force exerted by a fireman holding the hosepipe. The subject area includes concepts from fluid dynamics and Newton's laws of motion.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the relationship between force, velocity, and mass flow rate. There are attempts to derive the force based on changes in water flow and velocity, with some questioning the implications of doubling the flow rate on the force required.

Discussion Status

The discussion is active, with participants exploring different interpretations of the problem. Some have suggested qualitative reasoning regarding momentum and mass flow, while others are seeking clarification on the mechanics involved. There is no explicit consensus on the final answer yet.

Contextual Notes

Participants are working under a homework deadline, which may influence the urgency of their inquiries. There is an ongoing exploration of assumptions regarding the relationship between flow rate and velocity in a fixed cross-sectional area.

Ipos Manger
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Need help on this problem:

10. A fireman is holding a hosepipe so that water leaves the pipe horizontally. The hosepipe has a constant cross-sectional area. The magnitude of the force that the fireman exerts to hold the hosepipe stationary is F.
The volume of water delivered by the hose per second doubles, the force that the fireman must now exert is
A. sqrt2F.
B. 2F.
C. 4F.
D. 8F.

I only get 2F... Need help!
Thank you guys!
 
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First, try to find the relationship between the force and the velocity of the water. The generalized version of Newton's second law helps: F=dp/dt
 
Gotcha, but i only get 2F as an answer..
 
Anyone got any idea? My HW is for Saturday. :S
 
Momentum is p = m*v. Force, as ideasrule stated, is dp/dt.

When the flow rate doubles, what happens to the exit velocity of the water? What happens to the rate of mass ejected?

Suppose you differentiated the momentum, p = m*v. How would that look for a given fixed exit velocity v?
 
The mass would then double, and therefore, velocity would double also?

Giving a value of 4F.

Is this the correct way to solve the problem?
 
Ipos Manger said:
The mass would then double, and therefore, velocity would double also?

Giving a value of 4F.

Is this the correct way to solve the problem?

Yes, the RATE at which mass is ejected doubles, so the velocity had to double to accomplish this. Both contribute to the rate that momentum is changing. That's the qualitative approach.

Whether this qualitative argument will hold water with whomever is marking your answers is a question only you can answer :smile:. To be sure, you should do the differentiation of the momentum as suggested and show that if v is doubled and the rate of mass ejection doubles, then dp/dt = F quadruples.
 
Gotcha, it does, thank you!

Last question:
Why should the velocity of the water change?
 
Ipos Manger said:
Gotcha, it does, thank you!

Last question:
Why should the velocity of the water change?

How else can you double the flow rate through a fixed cross sectional area?
 
  • #10
Damn what an idiot! Thank you!
 

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