Force and Volume Relationship in Fire Hose Operation

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The discussion revolves around the relationship between force and volume in fire hose operation, specifically when the flow rate of water doubles. Participants analyze how this increase affects the exit velocity of the water and the force exerted by the fireman. It is concluded that doubling the flow rate necessitates a doubling of the exit velocity, resulting in a quadrupling of the force required to maintain the hose's position. The correct answer to the problem posed is determined to be 4F, based on the principles of momentum and the differentiation of force. The conversation emphasizes the importance of understanding the relationship between mass flow rate, velocity, and force in fluid dynamics.
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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?
 
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Damn what an idiot! Thank you!
 
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