Homework Help: Newton's laws Pulley Problem

1. May 23, 2015

kaspis245

1. The problem statement, all variables and given/known data
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.

2. Relevant equations
Newton's laws.

3. 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!

2. May 23, 2015

CWatters

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

3. May 23, 2015

insightful

So, how would mass m move and why?

4. May 23, 2015

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.

5. May 23, 2015

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.

6. May 23, 2015

CWatters

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

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

7. May 23, 2015

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?

8. May 24, 2015

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.

9. May 24, 2015

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.

10. May 24, 2015

haruspex

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

11. May 24, 2015

kaspis245

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

12. May 24, 2015

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.

Last edited: May 24, 2015
13. May 24, 2015

haruspex

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?

14. May 24, 2015

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.

15. May 24, 2015

haruspex

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.

16. May 24, 2015

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.

17. May 24, 2015

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.

18. May 24, 2015

insightful

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?

19. May 24, 2015

kaspis245

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.

20. May 24, 2015

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.

21. May 24, 2015

haruspex

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.

22. May 24, 2015

insightful

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.

23. May 24, 2015

haruspex

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

24. May 24, 2015

insightful

Are you saying energy is not conserved in this system?

25. May 24, 2015

haruspex

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