How High Can a Garbage Can Be Lifted by a Geyser?

[geoffrey159]

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

The forces acting on the garbage can are its own weight downward, and the force of water upward.

I have no proof that the force is strictly upward, but it seems natural to me, so I will make this assumption.

The force of water on the garbage can is the opposite of the force of garbage can on water. This force is determined by taking the time derivative of the momentum difference of water before and after collision:

$F_{g\rightarrow w} = (\frac{dP}{dt})_f - (\frac{dP}{dt})_i$

To be consistent with the assumption that the water push on the garbage can is upward, the rebound direction of the geyser must go downward, otherwise it would change the direction of the force.

I get that,

$\triangle P_f = - (\triangle m) v_f$
$\triangle P_i = (\triangle m) v_i$

So the force of water on garbage can is :

$F_{w\rightarrow g} = - F_{g\rightarrow w}= (v_i + v_f) \frac{dm}{dt}$

But I don't know $v_i$ nor $v_f$ and don't see any way of knowing it, so I assume they are identical so that $v_w = v_i = v_f$.

So the motion of the garbage can be described by :

$\frac{W}{g} \frac{dv_{g}}{dt} = - W + 2v_w \frac{dm}{dt}$

If I integrate through time:

$\frac{W}{g} v_g = -W t + 2 y_w \frac{dm}{dt}$

And the garbage can reaches maximum height when $v_g = 0$.
I don't think this will lead me anywhere, because I don't have any expression of water height vs time.
Could you please give me a hint on how to find maximum height ? Thanks

 

[mfb]

I don't see how the garbage can could accelerate water downwards (away from the can that does not move). It just can stop the incoming water.

v_i will depend on the height and you have to find the point where the can does not accelerate (there is just one stable position).

So the motion of the garbage can be described by :

That looks like a mixture of masses and forces.

 

[Doc Al]

W=10kg

A typo? Did you mean W = 10 N? (M = 1 kg)

 

[haruspex]

An inverted garbage can of weight W is suspended in air by water from a geyser. The water shoots up the ground with speed v₀, at a constant rate dm/dt. The problem is to find the maximum height at which garbage can rides.

I'm not sure what is being asked. "is suspended" and "rides" imply steady state, so you just need to find the height where the forces are in balance. But "maximum" suggests the can is being shot up past the equilibrium position from some lower position, presumably the height of the water source.

$\frac{W}{g} \frac{dv_{g}}{dt} = - W + 2v_w \frac{dm}{dt}$
If I integrate through time:
$\frac{W}{g} v_g = -W t + 2 y_w \frac{dm}{dt}$

That step doesn't work. You cannot integrate $v_w \frac{dm}{dt}$ like that, and it doesn't make sense to be integrating v_w anyway. That would give you the distance moved by a particular atom of water. I would take it as steady state, meaning the height is constant and so v_w is constant.

 

[geoffrey159]

A typo? Did you mean W = 10 N? (M = 1 kg)

Yes it is a typo, but not mine :) It is stated that W is a weight, so I would say 10N, but I can't be sure.

 

[Bystander]

Just for calibrating "intuition," the geyser has a mass flow rate and velocity comparable to an ordinary garden hose. I might be able to "levitate" a ping-pong ball to a height of 17 m (the answer clue) on a good day, if the neighbors aren't watering, doing dishes, or washing their hands, but a garbage can?

 

[haruspex]

Ans. clue. If v0=20m/s, W=10kg, dm/dt=0.5 kg/s, then hmax≈17m

Seems wrong to me, regardless of whether the weight is 10N or 100N.

 

[geoffrey159]

Just for calibrating "intuition," the geyser has a mass flow rate and velocity comparable to an ordinary garden hose. I might be able to "levitate" a ping-pong ball to a height of 17 m (the answer clue) on a good day, if the neighbors aren't watering, doing dishes, or washing their hands, but a garbage can?

:-) Thanks ! I need intuition calibration !

You cannot integrate $v_w \frac{dm}{dt}$ like that, and it doesn't make sense to be integrating $v_w$ anyway. That would give you the distance moved by a particular atom of water. I would take it as steady state, meaning the height is constant and so v_w is constant.

What should be done according to you ? $\frac{dm}{dt}$ is assumed to be a constant, this is why I have integrated that way, but honestly, I don't know what to do next.

 

[haruspex]

:) Thanks ! I need intuition calibration !
What should be done according to you ? $\frac{dm}{dt}$ is assumed to be a constant, this is why I have integrated that way, but honestly, I don't know what to do next.

Assume it's steady state, so the can is hovering at a constant height, h.
In terms of h, at what speed does the water hit the can?
As mfb posted, the can cannot push the water back down, but it will stop its upward flow. What force does that give you on the can?
What other force(s) act on the can? What is the condition for equilibrium?

 

[geoffrey159]

Assume it's steady state, so the can is hovering at a constant height, h.
In terms of h, at what speed does the water hit the can?
As mfb posted, the can cannot push the water back down, but it will stop its upward flow. What force does that give you on the can?
What other force(s) act on the can? What is the condition for equilibrium?

Hello,

I will make an attempt but I'm unfamiliar with problems involving liquids, be lenient with me :-)

1 - The speed at which water hits the can, in terms of h
Before hitting the garbage can, any water molecule is under its own weight, so its height is

$y(t) = tv_0 - \frac{g}{2}t^2$,

With this equation, we can determine the time at which $y(t) = h$, and its expression is

$t_{i} = \frac{v_0 - \sqrt{v_0^2+2gh}}{g}$.

Therefore, the speed at which water hits the can is

$v(t_i) = v_0 - gt_i = \sqrt{v_0^2+2gh}$

2 - The forces on the garbage can

If the garbage can just stops water, there is no outgoing speed, and the force of water on the garbage can is

$F_{w\rightarrow g} = \frac{dm}{dt} v(t_i) =\sqrt{v_0^2+2gh} \frac{dm}{dt}$.

Furthermore, the garbage can is under its own weight.

3 - Equilibrium condition

If the height of the garbage can is constant, it does not move vertically, so it does not accelerate, so

$0 = -W +F_{w\rightarrow g} \Rightarrow h = \frac{1}{2g} ( ({\frac{W}{\frac{dm}{dt}}})^2 - v_0^2 )$

It must be wrong because I get 0 with the given values, and not 17 meters !

 

[mfb]

Therefore, the speed at which water hits the can is

$v(t_i) = v_0 - gt_i = \sqrt{v_0^2+2gh}$

That would mean the speed increases with increasing height.
There is a sign error.

2 - The forces on the garbage can

If the garbage can just stops water, there is no outgoing speed, and the force of water on the garbage can is

$F_{w\rightarrow g} = \frac{dm}{dt} v(t_i) =\sqrt{v_0^2+2gh} \frac{dm}{dt}$.

Furthermore, the garbage can is under its own weight.

Okay (apart from the sign error from above).

3 - Equilibrium condition

If the height of the garbage can is constant, it does not move vertically, so it does not accelerate, so

$0 = -W +F_{w\rightarrow g} \Rightarrow h = \frac{1}{2g} ( ({\frac{W}{\frac{dm}{dt}}})^2 - v_0^2 )$

It must be wrong because I get 0 with the given values, and not 17 meters !

Doesn't look so bad, with the fixed sign error it could be right.

 

[geoffrey159]

Hello,

Thank you for looking at it !
I see the sign error: $v(t_i) = \sqrt{v_0^2 - 2gh }$.

So $h = \frac{1}{2g} ( v_0^2 - ({\frac{W}{\frac{dm}{dt}}})^2 )$

But still, I get 0 with the given values !

 

[mfb]

Now the formula should be correct, and I agree. Looks like the answer clue is wrong.

 

[BvU]

Seems there is an error in Kleppner. Water flow should be 6 kg/s
See this thread (which is also in the list below)