Garbage can suspended in the air

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Homework Statement


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##


Homework Equations





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
ehild
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Homework Statement


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##


Homework Equations





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!

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
rude man
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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
NascentOxygen
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The geyser alone reaches 20.4m
 
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  • #5
ehild
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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?
I also think that the example solution is wrong.

ehild
 
  • #6
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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
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
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1. The problem is to find the maximum height at which garbage can rides.

So under what conditions does it? :wink:
 
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  • #8
ehild
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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.
Yes, when is it maximum? At elastic or inelastic collision?

ehild
 
  • #9
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Yes, when is it maximum? At elastic or inelastic collision?

ehild
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
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Does it mean the water bounces back with ##\sqrt{v_0^2-2gh}##? ?
Nope...
I guess that would be wrong because the garbage can is also moving. Correct?
Yep, so what are you missing? What does the water do?
 
  • #11
ehild
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What does maximum height mean? Does the can raise higher from that?


ehild
 
  • #12
D H
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I also think that the example solution is wrong.
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
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The problem says "of weight W". So I'd think it should be W = 10 N, rather than 10 kg.
 
  • #14
ehild
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The problem says "of weight W". So I'd think it should be W = 10 N, rather than 10 kg.
And it also says that W=10 kg. The text of the problem is confusing.


ehild
 
  • #15
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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.
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".

What does maximum height mean? Does the can raise higher from that?
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
NascentOxygen
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Looks correct?
The method looks correct, yes. But as for the answer you'll get .....
 
  • #17
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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
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The method looks correct, yes. But as for the answer you'll get .....
Yes, the given clue doesn't satisfy the final result as said by the other posters.
 
  • #19
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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} ##.
Mine is a low priced Indian edition 2009.
 
  • #20
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The text as you gave it is an exact copy from the original 1973 edition.
 
  • #22
ehild
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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?
Yes. :smile:

ehild
 
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  • #23
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Yes. :smile:

ehild
Thanks for the help ehild! :)
 
  • #24
ehild
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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} ##.
It is 17 m with these data.
 
  • #25
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It is 17 m with these data.
And if I use 10 N instead of 10 kg, it comes out to be 15 m, lol.
 

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