# Garbage can suspended in the air

1. Sep 5, 2013

### Saitama

1. The problem statement, all variables and given/known data
An inverted garbage can of weight W is suspended in air by water from a geyser. The water shoots up the ground with speed $v_0$, at a constant rate $dm/dt$. The problem is to find the maximum height at which garbage can rides. What assumption must be fulfilled for the maximum height to be reached.

Ans. clue. If $v_0=20\, m/s$, W=10kg, dm/dt=0.5 kg/s, then $h_{max} \approx 17 \, m$

2. Relevant equations

3. The attempt at a solution
I don't know how to begin with this problem. I think that I have to find the force due to water but I can't see how to form the equations here. I need a few hints to begin with.

Any help is appreciated. Thanks!

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2. Sep 6, 2013

### ehild

Does the momentum of the water change when it interacts with the can? Can you solve the problem if there were balls shot up instead of water?

ehild

3. Sep 6, 2013

### rude man

Think F = dp/dt with p = p(height).

BTW I got a very different answer. (I assume W = 10*9.81 = 98.1 N).
@ehild?

4. Sep 6, 2013

### Staff: Mentor

The geyser alone reaches 20.4m

Last edited: Sep 6, 2013
5. Sep 6, 2013

### ehild

I also think that the example solution is wrong.

ehild

6. Sep 6, 2013

### Saitama

Yes, the momentum of water changes. If the base of garbage can is at a height h, the velocity with which water strikes is $\sqrt{v_0^2-2gh}$. The problem is how do I find the change in momentum of mass of water striking the can. The question doesn't state the type of collision between water and can.

7. Sep 6, 2013

### Enigman

So under what conditions does it?

8. Sep 6, 2013

### ehild

Yes, when is it maximum? At elastic or inelastic collision?

ehild

9. Sep 6, 2013

### Saitama

Elastic.
Does it mean the water bounces back with $\sqrt{v_0^2-2gh}$? I guess that would be wrong because the garbage can is also moving. Correct?

10. Sep 6, 2013

### Enigman

Nope...
Yep, so what are you missing? What does the water do?

11. Sep 6, 2013

### ehild

What does maximum height mean? Does the can raise higher from that?

ehild

12. Sep 6, 2013

### D H

Staff Emeritus
Same here. I think the correct answer for a 10 kg garbage can (and calling this "weight" in a physics text is not good) is zero. 20 m/s * 0.5 kg/s = 10 newtons. Double that for a pure elastic collision and you get 20 newtons, which is not enough to lift a 10 kg object off the ground.

13. Sep 6, 2013

### voko

The problem says "of weight W". So I'd think it should be W = 10 N, rather than 10 kg.

14. Sep 6, 2013

### ehild

And it also says that W=10 kg. The text of the problem is confusing.

ehild

15. Sep 6, 2013

### Saitama

Yes, giving weight as 10 kg is incorrect but the problem is from a very good book, errors may creep in despite the best efforts by the authors. :)

The source of the problem is "An Introduction to Mechanics by Daniel Kleppner and Robert Kolenkow".

Change in momentum of water at maximum height is $2\Delta m\sqrt{v_0^2-2gh}$. The force applied on can in time $\Delta t$ is $2(\Delta m/\Delta t)\sqrt{v_0^2-2gh}$. Taking the limit $\Delta t\rightarrow 0$, the force acting on can is $2(dm/dt)\sqrt{v_0^2-2gh}$. This is balanced by the weight of can. Hence,
$$2\frac{dm}{dt}\sqrt{v_0^2-2gh}=W$$
From here I can solve for h.

Looks correct?

16. Sep 6, 2013

### Staff: Mentor

The method looks correct, yes. But as for the answer you'll get .....

17. Sep 6, 2013

### voko

In the 2010 edition, the answer clue is $v_0 = 20 \ \text{m/s}, \ W = 8.2 \ \text{N}, \ dm/dt = 0.5 \ \text{kg/s}, \ h_{\text{max}} \approx 15 \ \text{m}$.

18. Sep 6, 2013

### Saitama

Yes, the given clue doesn't satisfy the final result as said by the other posters.

19. Sep 6, 2013

### Saitama

Mine is a low priced Indian edition 2009.

20. Sep 6, 2013

### voko

The text as you gave it is an exact copy from the original 1973 edition.