How Does the Monkey's Movement Affect the Box in a Pulley System?

[kaspis245]

Homework Statement

A monkey which has a mass of $m$ is hanging on a rope as the picture shows. Then the monkey starts to move upwards with speed $u$. Find how the box which has a mass of $2m$ will move. Neglect friction between the rope and pulleys. Assume that pulleys are massless.

Homework Equations

Newton's laws.

The Attempt at a Solution

I don't understand why $2m$ has to move in the first place. The motion of the monkey affects only the box with mass $m$. The downward net force in pulley A remains the same. The monkey is even hanging on entirely separate rope. I don't see how $2m$ can be affected. Please help!

 

[CWatters]

Are you sure you need help :-) In many sports they have an expression "back yourself to win".

 

[insightful]

So, how would mass m move and why?

 

[Kyouran]

Are you sure the upward motion of the monkey is only relative to the second pulley? The problem says it moves upward with speed u, but it doesn't say anything about what combination of motions of the pulleys is used to achieve this upward velocity.

 

[insightful]

I would say the only reasonable assumption is that u is relative to the rope the monkey is on. I guess we also have to assume that the monkey has mass m.

 

[CWatters]

The mass m moves up at the same speed as the monkey. That's one of the Lewis Carroll problems..

http://www.numericana.com/answer/physics.htm#monkey

If they move up at constant velocity then there is no acceleration (other than gravity) acting on them and their pulley.

 

[insightful]

So, as the center of inertia of the monkey and mass m move up, what happens to the center of inertia of the monkey, mass m, and mass 2m?

 

[kaspis245]

The mass $m$ will move with the same upward speed as the monkey, so I suppose their center of mass will move upwards with the same speed they are moving. Then, $2m$ should move upwards with speed $u$ in order to maintain the same distance between $2m$ and A.

 

[theodoros.mihos]

The mass m will move with the same upward speed because both moves with the same force so the same acceleration and initial conditions are the same.

 

[haruspex]

The mass $m$ will move with the same upward speed as the monkey, so I suppose their center of mass will move upwards with the same speed they are moving. Then, $2m$ should move upwards with speed $u$ in order to maintain the same distance between $2m$ and A.

That works if the monkey's speed u is relative to the ground. It might be relative to the rope... it isn't clear.

 

[kaspis245]

That works if the monkey's speed u is relative to the ground. It might be relative to the rope... it isn't clear.

The monkey begins to move relative to the rope. Is there a difference?

 

[theodoros.mihos]

For constant speed of monkey, there is no acceleration so there is no additional force against to 2m. If u of monkey comes from the long past the 2m must be unmoving. If we have a problem "before-after" then we have a plastic collision. I think the 2m moves up with the same u as monkey, on land frame. If u is the speed on the rope frame then u--> u/2.

 

[haruspex]

The monkey begins to move relative to the rope. Is there a difference?

You are asked about the movement of the 2m box. You show that relative to the ground it will be the same upward speed as the monkey's, but that is not the same as the monkey's speed relative to the rope. How will those last two speeds be related?

 

[theodoros.mihos]

For the same time Δt if monkey take l rope moves up l/2 so the relation between two speeds is 1:2.

 

[haruspex]

For constant speed of monkey, there is no acceleration so there is no additional force against to 2m. If u of monkey comes from the long past the 2m must be unmoving. If we have a problem "before-after" then we have a plastic collision. I think the 2m moves up with the same u as monkey, on land frame. If u is the speed on the rope frame then u--> u/2.

No, the problem is subtler than that. For the monkey to move at all, it must slightly increase the rope tension. The 1m mass will experience the same increased rope tension and so accelerate in exactly the same way as the monkey. Similarly on slowing down. Thus the monkey and the 1m mass are always at the same vertical speed.
Now apply that argument to the monkey+1m mass as a combination, against the 2m mass.
But I agree with your conclusion in the land frame. Not sure about the rope frame yet.

 

[theodoros.mihos]

Right. Even acceleration is non constant, the 2m and the monkey moves both with the same acceleration every time. With the same initial conditions (i.e. v(t=0)=0) they have the same speed every time.

 

[CWatters]

+1 to what Theo said

The monkey starts off stationary and accelerates to velocity u. During that acceleration phase the mass m also accelerates to velocity u.

During the acceleration (a) there is a excess force on their pulley = 2ma

That force/tension acts on the mass 2m. So mass 2m accelerates upwards at 2ma/2m = a.

Once the monkey reaches velocity u it stops accelerating so mass 2m stops moving.

The problem doesn't ask us what happens when the monkey decelerates but mass 2m will descend again.

 

[insightful]

The monkey starts off stationary and accelerates to velocity u. During that acceleration phase the mass m also accelerates to velocity u.

Relative to what? Can we agree (or decide) that u in the original problem and henceforward is the monkey's velocity relative to the rope?

 

[kaspis245]

Relative to what? Can we agree (or decide) that u in the original problem and henceforward is the monkey's velocity relative to the rope?

Yes. The original problem says that monkey's velocity is relative to the rope, but I didn't know it matter so I skipped it.

By the way, I know that this problem can be solved using Lagrangian mechanics, but since it is not included in the course I am studying I can't use it.

Conservation of momentum must be applied here, but I haven't found a way how.

 

[theodoros.mihos]

The speed of 2m is the same with the speed of monkey relative to what you like.
Consevation of momentum used to skip kinematics for variations in very small time by large and variable forces. After monkey (and 2m) moves, there is no changes to use this principle. For the acceleration phase (collision) external forces acting tell us that momentum is not conserved.

 

[haruspex]

The speed of 2m is the same with the speed of monkey relative to what you like.
Consevation of momentum used to skip kinematics for variations in very small time by large and variable forces. After monkey (and 2m) moves, there is no changes to use this principle. For the acceleration phase (collision) external forces acting tell us that momentum is not conserved.

We all agree that the monkey and the two masses all rise at the same speed relative to any given reference. But it would be strange if the question were asking for the speed of the 2m mass relative to the bit of rope the monkey is holding. So knowing now that the speed u is relative to the rope, the speed of the 2m mass (relative to the ground, I suggest) is something else.
As you noted in post #14, the monkey and the 1m mass each rise at rate u/2 relative to the lower pulley. We can then apply the same logic to find how the 2m mass moves.

Kaspis, you ask about conservation of momentum. That doesn't really apply here because the tension in the rope hplding up the top pulley is an external force, and it's not always equal to 4mg. However, the arguments advanced in both #15 and #17 are akin to conservation of momentum. They note that the forces are the same on the monkey and the 1m mass, so integrating these over time leads to the momenta being the same. Conservation of momentum can be derived from action and reaction being equal and opposite, and integrating those over time. That leads to momenta being equal and opposite.

 

[insightful]

Conservation of momentum must be applied here, but I haven't found a way how.

Then try conservation of energy. Throw some numbers in, like a 10 kg monkey that climbs up the rope at 0.1 m/s a distance of 10 m. Compare the potential energy involved vs. the kinetic energy. Consider that the only input of energy to the system is the monkey.

 

[haruspex]

Then try conservation of energy. Throw some numbers in, like a 10 kg monkey that climbs up the rope at 0.1 m/s a distance of 10 m. Compare the potential energy involved vs. the kinetic energy. Consider that the only input of energy to the system is the monkey.

I don't see how that is going to work.

 

[insightful]

I don't see how that is going to work.

Are you saying energy is not conserved in this system?

 

[haruspex]

Are you saying energy is not conserved in this system?

I'm saying I don't see how you will get a useful equation out of it, but feel free to prove me wrong.

 

[insightful]

Consider an example with a 10 kg monkey, 10 kg mass m, and 20 kg mass 2m.

First consider the monkey climbing a rope attached to the ceiling.
Say at t = 0, he starts at u = 0.1 m/s, so using h = vt, at t = 100 s, he has climbed 10 m.
He has gained potential energy, PE = mgh, so at 10 m, he has gained (10)(9.8)(10) = 980 J.
He has gained kinetic energy, KE = 0.5mv^2 = (0.5)(10)(0.1)^2 = 0.05 J.
In this example, KE is not significant and can be ignored.
So, by conservation of energy, the monkey climbing 10 m expends 980 J energy.

Now consider a one-pulley system attached to the ceiling.
From previous arguments, we agree the monkey and mass m rise at the same velocity.
At t = 0, he again starts up relative to the rope at u = 0.1 m/s.
At t = 100 s, with 980 J available, the total 20 kg can be raised 980/[(20)(9.8) = 5 m.
Relative to the ceiling, their velocity is 5/100 = 0.05 m/s or u/2. (No news here.)

Now consider the two-pulley system attached to the ceiling.
We know the velocity of the monkey and mass m relative to pulley A is 0.05 m/s.
Therefore, the velocity of the monkey and mass m relative to the rope from pulley B to pulley A is also 0.05 m/s.

*[This is equivalent to the monkey with mass m strapped to his back climbing the rope from pulley B at u/2 relative to that rope.]*

At t = 0, our (now) 20 kg monkey starts up relative to the rope from pulley B at 0.05 m/s.
From the above, it's clear the monkey is now lifting himself, mass m, and mass 2m all at the same velocity.
At t = 100 s, with 980 J available, the total 40 kg can be raised 980/[(40)(9.8)] = 2.5 m.
Relative to the ceiling, their velocity is 2.5/100 = 0.025 m/s or u/4.

Final answer: Mass 2m travels upward with a velocity of u/4.

If KE is significant, i.e., a fast-climbing monkey, the final answer is the same.

Of course, given *[ ]* above, energy considerations are not needed for solution but can give a start if stumped.

 

[SammyS]

Final answer: Mass 2m travels upward with a velocity of u/4.

If KE is significant, i.e., a fast-climbing monkey, the final answer is the same.

Of course, given *[ ]* above, energy considerations are not needed for solution but can give a start if stumped.

If u is the monkey's velocity with respect to the rope, (and OP has confirmed that this detail was omitted from the original problem statement) then I agree with this result.

Of course, we can also arrive at this by considering the analysis in Post#15:

Let's say the monkey's velocity is $v$ in the inertial reference frame in which all the the objects are initially at rest.

Then the monkey, the 1m mass and the 2m mass all have velocity, $\ v\,,\$ upward.

Pulley A, will then have velocity, $\ -v\,,\$, i.e. Pulley A moves downward at speed, $\ v\ .$

The 1m mass is moving upward with velocity, $\ v\,,\$ so that the length of rope on the right side of Pulley A is increasing at rate $\ 2v\ .$ Combine that with the fact that Pulley A moves downward at $\ v\$ gives the rope the monkey holds onto is moving at speed $\ 3v\$ downward, So the monkey's velocity w.r.t the rope is $\ 4v\$ upward.

I.E. $\ 4v=u\ .$

 

[haruspex]

From previous arguments, we agree the monkey and mass m rise at the same velocity.

But that's the whole thing right there. Applying energy conservation on top gains nothing. You might as well apply conservation of charge.
Indeed, you can assume work is not conserved, and still solve the problem, which proves you have not actually used work conservation.

 

[insightful]

Indeed, you can assume work is not conserved, and still solve the problem, which proves you have not actually used work conservation.

How would you propose that work is not conserved?

 

[haruspex]

How would you propose that work is not conserved?

E.g., make the rope a little slippery. Or have the monkey bound up the rope, with monkey and 1m mass descending slightly just before the monkey gets a grip on the rope again. Work is not conserved, yet the result still stands.

 

[insightful]

You might as well apply conservation of charge.

Please apply the equations of conservation of charge as they relate to this problem.