2 pulleys 3 masses problems

[phy_magic]

Homework Statement

3 equal weights are attached to the ends and in the middle of the rope. We raise the middle weight so that the rope is horizontal and in the middle between the pulleys, and let it go, after which the middle weight is lowered and the weights at the ends are raised.
How far will the middle weight travel before it starts to rise?

Relevant Equations

3 weights -m
2 pulleys
I don't understand if we let go of the middle mass it will move down the y-axis and the two edge masses up. So the tensions will have to be greater than mg for the edge masses to start moving but then when (if) the system stops then the tension will have to be equal to mg. Is that possible? Will the middle weight move up after some time? So, i came to this thoughts.

Is tension force in rope same all the time , does acceleration of middle weight change during time and why would middle weight go up (rise)?
I saw questions like this , but I am little confused about this questions?

 

[kuruman]

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• micromass
• Homework students
• Replies: 2
• Forum: Introductory Physics Homework Help

• 

 

[haruspex]

First, think about the equilibrium position. Can you draw that?
Clearly the centre mass will keep descending as long as it is above that position, but when it reaches it the system components have momentum, so will continue on some way. What do you think will happen thereafter?

 

[phy_magic]

First, think about the equilibrium position. Can you draw that?
Clearly the centre mass will keep descending as long as it is above that position, but when it reaches it the system components have momentum, so will continue on some way. What do you think will happen thereafter?

In equillibrium x components shoud cancel and y component of resultant tension should be equal to mg.
If i understand good, two end masse will keep going up and then on middle mass will start acting resultant force up. But is tension force constant at the magnitude ?

 

[haruspex]

In equillibrium x components shoud cancel and y component of resultant tension should be equal to mg.

The tension is at different angles in different places. Where will its y component be mg?

If i understand good, two end masse will keep going up and then on middle mass will start acting resultant force up. But is tension force constant at the magnitude ?

No, the tension will keep changing.

Often one can sidestep a lot of details by applying a conservation law. Can you think of one you can use here?

 

[phy_magic]

The tension is at different angles in different places. Where will its y component be mg?

No, the tension will keep changing.

Often one can sidestep a lot of details by applying a conservation law. Can you think of one you can use here?

Momentum conservation
2mv-mV=0 if up is positive direction.
And conservation energy ( x and y are displacenemts of end masses and middle mass)
2gx+gy=v^2+ 0.5V^2
In equilibrium position tension will be equal to mg and from that i found angle, but them tension will contiunue to decrease but the sume of 2 tension forces will be greater than mg(middle mass) and it will decrease velocitiy and end masses will decrease velocities ( it is belov equilibrium point) and then when they reach zero velociti they will star to accelerate in oposite direction (middle mass up ).
Is this way good way of understanding problem ?

 

[jbriggs444]

Momentum conservation

Is momentum conserved here? We have the external forces from the two pulleys and the external force from gravity. Do those always sum to zero?

Or, think about the position of the center of mass. Does it change between the initial setup and the equilibrium configuration? Is that possible if momentum is conserved?

 

[Lnewqban]

In equilibrium x components should cancel and y component of resultant tension should be equal to mg.

Welcome!
Consider that the y component acting on the middle mass is formed by the addition of two forces.
Each end mass should produce one of those vertical forces.
The angle that the rope takes at the equilibrium position must satisfy those values.

 

[haruspex]

And conservation energy ( x and y are displacenemts of end masses and middle mass)
2gx+gy=v^2+ 0.5V^2

when they reach zero velociti they will star to accelerate in oposite direction

How far will the middle weight travel before it starts to rise?

Think about the connection between those observations.