# Two questions on momentum.

1. Aug 25, 2006

### MathematicalPhysicist 1) water shoots out of a fire hydrant having nozzle diameter D with nozzle speed V0. what is the reaction force on the hydrant?
2) An inverted garbage can of weight W is suspended in air by water from a geyser. the water shoots up from the ground with a speed v0 , at a constant rate dm/dt. the problem is to find the maximum height at which the garbage can rides. what assumption must be fulfiled for the maximum height to be reached?

about the second question, i think that the assumption is that W equlas the force exerted from the water, but i dont know how to calculate the height.

2. 3. Aug 25, 2006

### Hawknc I'll pay you the obvious equation you'll need: F = dp/dt, where p = momentum. Work out your control volume for each question and then work out what the change in momentum across that control volume is.

4. Aug 26, 2006

### MathematicalPhysicist you didn't help me much, i know already this equation my problem is how to apply it in here.
in the first question im only given the diameter of the nozzle and the speed, in order to calculate the change in momentum i need the mass of the water, from what i m given i know how to calculate the area of the nozzle which is piD^2/4, and i think i need to use here hydrostatic pressure but im not sure.
in the second question im quite clueless as to the change of momentum in here.
here are the equation i got (it's ofcourse incorrect):
(W/g+dm/dt*t)v0=(W-F)*t
t=h/v0 and F=(dm/dt*v0)t^2=dm/dt(h^2/v0)

i need some help to clear me out what exactly is the change in momemntum in this question.

5. Aug 26, 2006

### Andrew Mason The question assumes that you know that the density of water is 1g/cm^3. For question 1) you can determine the volume of water, and, knowing the density, the mass of the water, passing through the nozzle per second: dm/dt. Use F = dp/dt = vdm/dt.

For the second question, you have to assume that all the water is impacting the garbage can. The force is again vdm/dt. Does v of the water change with height? How does v change with height? (think energy and gravity).

AM

6. Aug 27, 2006

### MathematicalPhysicist is this approach correct:
i assume for the first question that the shape of the nozzle is cylinder than:
h=V0*t
V=pi*(D^2/4)*h
m=p*V
dm/dt=pi(D^2/4)V0*p
F=dm/dt*V0=pi(D^2/4)V0^2*p
where p is the density of water.
for the second question, do you know approach that doesnt employ energy, cause in the text the questions are in the momentum section, i havent yet delved into energy.

7. Aug 27, 2006

### Andrew Mason Ok for the first question.

For the second, you can think of the geyser at the ground supporting a column of water of height h plus the garbage can. What is the weight of that? That is the force (= vdm/dt) at the ground.

AM

8. Aug 27, 2006

### MathematicalPhysicist for the second question, what force do you refer to, i think the net force is W-F where W is the weight of the garbage can and F is the force of the water.
about v and height i know of the equation:
v^2-v0^2/2g=h but im not sure it will work here.

9. Aug 27, 2006

### Astronuc Staff Emeritus
One way to look at F = dp/dt = Vdm/dt

is that dm/dt = $\rho$VA, where $\rho$ is fluid density, V is mean fluid velocity, and A is the cross-sectional area of the fluid.

then F = V *$\rho$VA = $\rho$V2A.

In the second problem, any mass traveling upwards must be decelerating due to gravity. If one observes water flowing from a hose in the vertical direction, one observes that it reaches some height where the water stops traveling upward then starts falling.

So to suspend a garbage can, one has to figure the velocity of the water at whatever height, and the force of the water must balance the weight of the garbage can.

The right idea, but be careful with the signs. If one applies the conservation of energy, then the initial kinetic energy (with zero relative grav. potential energy) must be equal to the kinetic energy mV2/2 + mgh, or the change in kinetic energy must equal the change in gravitational potential energy. Note that V2/2 is specific kinetic energy.

Last edited: Aug 27, 2006
10. Aug 27, 2006

### MathematicalPhysicist the weight is W+Mg where M is the mass of the water column, but my question is how do i calculate M, if im not given the diameter of the stream of water?

11. Aug 27, 2006

### Andrew Mason You have to do a little thinking on your own. The force is vdm/dt. You are given dm/dt. So you can determine the volume per second that is coming out of the geyser opening. You are also given v so you can determine the area (just use the formula you developed for the first question).

Determine the weight of a column of water of height h and area A, add the weight of the garbage can and set that equal to the force at the geyser opening.

AM

Last edited: Aug 27, 2006
12. Aug 27, 2006

### civil_dude Something you should try to do is personally try to provide the required force to change the flowing water's direction. Imagine pushing against the momentum of the water, and how the volume of water impacts tremendously the force required to change the momentum of the water flow. That is why we civils have thrust block typical details for construction of water mains. They are constructed with concrete blocks thrust against insutu soils positioned normal to the change in direction.

13. Aug 29, 2006

### MathematicalPhysicist here what i did, but i reackon it's incorrect:
A=(dm/dt)*(1/vp)
(p is the density of water).
V=Ah
A=V/h=(dm/dt)(1/vp)
V=(hdm/dt)(1/vp)
Vgp+W=vdm/dt
but obviously it's wrong, i would some one point me where am i wrong here?

14. Aug 29, 2006

### Andrew Mason I am not sure why you say it is obviously wrong!

$$A = \frac{dm}{dt}\frac{1}{\rho v}$$ - that is right

$$V = Ah$$ - that is right

$$F = mg + W = \rho Vg + W = vdm/dt$$ - that is right

So:

$$\rho Ahg + W = vdm/dt$$ - that is right

$$h = \frac{{v\frac{dm}{dt} - W}}{\rho Ag} = \frac{{v\frac{dm}{dt} - W}}{\rho (\frac{dm}{dt}\frac{1}{\rho v})g} = \frac{v^2}{g} - \frac{Wv}{\frac{dm}{dt}g}$$

AM

Last edited: Aug 29, 2006
15. Aug 30, 2006

### MathematicalPhysicist the problem is that if i plug in:
v=20 W=10 dm/dt=0.5 iget with this equation h=0, and according to my text book, h=15, perhaps it's a mistake in the text?

16. Aug 30, 2006

### Andrew Mason Oops. I see it is not quite as simple as I suggested. Sorry to have misled you. The water column that is supported is the height that it would go if the garbage can was not there, so the equation I gave you is wrong. You need to find the height of that column first.

The best way to solve this problem is to use energy. The relationship between the speed at the garbage can and the speed at the ground is:

$$v_{can}^2 = v_0^2 - 2gh$$

But if you want to avoid using energy, you can do it by working out the height of the column of water without the can using

$$\rho AHg = vdm/dt$$

[Note H is the height of the column without the can] Then work out the speed of the water in that column as a function of vertical distance (the speed at that vertical distance x dm/dt has to equal the weight of the water above that point. When you put the can there, you are replacing the weight of the can with the weight of the water that would otherwise be above that point). The height at which v = W/(dm/dt) will give you the answer you are looking for.

In other words, you have to find the height of the column of water without the can and then find the length, y, of the portion of that column of water that is equal to the weight of the can and subtract the two:

$$W = \rho Ay$$

$$h = H - y$$ where h is the height of the can

AM

Last edited: Aug 30, 2006
17. Aug 31, 2006

### MathematicalPhysicist let me see if i understand you:
pAHg=vdm/dt where v is the initial speed, i.e v0.
and W=pAy where A is the area is the same area as the above equation (the area which i have already got from before) right?
if so, i tried plugging the numbers and i still got 0.

18. Aug 31, 2006

### Andrew Mason I think the problem is with the numbers. You can see this by just determining the area. A = .5/20000 = 2.5x10^-5 m^2. The weight of the column of water (without the can) is less than the weight of the can.

$$H = vdm/dt/\rho Ag$$

$$y = W/\rho A$$ and

$$A = dm/dt/\rho v$$

So :

$$H = v^2/g$$

$$y = Wv/(dm/dt)$$

$$h = H-y = v^2/g - Wv/(dm/dt)$$

Plugging in your numbers (v = 20 m/sec, W = 10 N, dm/dt = .5 kg/sec g = 9.8 m/sec^2) h = 40.8 - 400 . The can is too heavy to be supported.

AM

19. Aug 31, 2006

### MathematicalPhysicist so you're telling me that kleppner has a mistake in his book, im surpsrised.
anyway thanks for your help.

20. Aug 31, 2006

### Andrew Mason Not necessarily a mistake in the book. Check the numbers and units. With the numbers you have given us, assuming MKS units, you can see for yourself:

The force required to support the can is W = vdm/dt where v is the speed of the water where it impacts the can. The force is in Newtons.

If dm/dt is .5 kg/sec and v is 20 m/sec at the ground, you need to be at ground level to provide a force of 10 N.

AM

21. Sep 1, 2006

### MathematicalPhysicist the numbers and units are the same as you have written.
but according to the book h equals 15 meters, why isnt this a mistake?
havent we got a different answer here?