How Much Additional Force Is Needed to Hold a Water Hose Stationary?

[djlightsout06]

I really just need help getting started...

What additional force is necessary to hold a water hose stationary after the water flow is turned on, if the discharge rate is .6 kg/sec with a speed of 25 m/s?

Thanks in advance..

 

[KillaMarcilla]

I'm not sure, but until someone better at physics responds, this'll probably tide you over:

When the water is going forward, I guess it's pushing the hose/person holding it backward

the momentum going forward is .6kg*25m/s, and you get the force by taking the change in momentum over the change in time, 0.6kg/s * 25m/s => 15kg*m/s^2 = 15N

I don't know if my underlying assumption is correct there, tho

 

[djlightsout06]

thanks, that's exactly the info that i needed to get going.

 

[hi_rudra]

confused: can anyone tell how can i ask my own problems?/creat threads?

 

[Phantom]

I think you'd also need to know H20 pressure, or at least, I guess, differential pressure, because if you think about the idea of a culvert through which a river runs, there is no force because the water just flows through the bottom part. So if you had a hose and you could get flow rate of .6kg/s through it without the hose being full, there's little or no pressure drop as the water leaves the hose. does that make sense?

All you are measuring with flow rate and mass is the force the water can exert on an object it hits, which doesn't help you that much in answering your question.

 

[kuenmao]

Phantom...you don't need to know that. The question only involves Newton's third law. The force the water can exert on an object, in fact, can't be known unless you know how much speed the water recoils with.

 

[Phantom]

But you DO know the speed. I just misread the question. The answer was correct. The momentum equation works fine. Zero momentum before you turn on the hose and 15 kg-m/s2 afterwards. Got it.

 

[mee]

new threads

confused: can anyone tell how can i ask my own problems?/creat threads?

Just click on the "new thread" button at the top of the listed page.

 

[ConceptuallyInept]

(ignore this message)

Hi people
I'm just testing to see if this thread gets posted. This is the first time I've used theis site.

 

[nx01rules]

I'm not sure, but until someone better at physics responds, this'll probably tide you over:

When the water is going forward, I guess it's pushing the hose/person holding it backward

the momentum going forward is .6kg*25m/s, and you get the force by taking the change in momentum over the change in time, 0.6kg/s * 25m/s => 15kg*m/s^2 = 15N

I don't know if my underlying assumption is correct there, tho

Yeah KillaMarcilla, you are precicely right. As water has a density of 1.00 kg/L you don't need to account for it's density change. So, to hold the hose steady, you would have to apply a force of 15N onto the hose.

 

[Gokul43201]

Killamarcilla is correct.

Anyways, a useful trick to help you get started on any problem is dimensional analysis. For example, in this problem you are given 2 quantities whose dimensions are kg/s and m/s. You are asked to find a quantity whose dimensions are Newtons or kg.m/s^2. Clearly the product of the dimensions of the given quantities gives you the dimensions of the required answer. This suggests that multiplying the 2 numbers may be the way to go.

However, let this not be the only thing you do to solve a problem. It does not consider dimensionless constants or involve an understanding of the underlying physics. It's just a useful trick for checking solutions.