2:1 Pulley Question

[MarcusThatsMe]

TL;DR Summary

The math when reversing a pulley doesn't make sense, hoping someone can clarify.

I have Mass A being pulled vertically. I have Mass B on an incline that is pulling Mass A. There is a 2:1 pulley between them.

The math I'm using is:

FA = MA / 2 = ? t-force
MB * SIN(of the incline degree) = ?
If MB is greater then FA, it pulls FA up as MB moves down the incline.

BUT... If I reverse the 2:1 pulley. Then the math changes to...

FA = MA * 2 = ? t-force
MB * SIN(of the incline degree) = ?
If FA is greater then MB, it pulls MB up the incline as FA moves down.

It's confusing because I would have figured that MA would remain constant and MB * SIN would be divided by 2? Like the first equation...

 

[.Scott]

What exactly is meant by "reversing the pulley"? Is it simply reversing the direction that the pulley will be turning? Or is it changing the setup so that the mass B is being supported with a 2:1 force advantage (mass B connected to the center of the pulley) instead of a 2:1 distance advantage (mass A, on the ramp, connected to the pulley center)?

If the setup is changing, the MA/MB force formula will change. If it is only the direction of movement, that formula will not change.

 

[Lnewqban]

It's confusing because I would have figured that MA would remain constant and MB * SIN would be divided by 2? Like the first equation...

In the first case, mass B moves at double the speed of mass A.
In the second case, mass B moves at half the speed of mass A.

 

[MarcusThatsMe]

Here is my math and a sketch of the second case... Maybe if someone can just confirm that I'm doing this right. From what I think should happen A should move down 304 meters, and B and C should move 608 meters...

 

[MarcusThatsMe]

In the first case, mass B moves at double the speed of mass A.
In the second case, mass B moves at half the speed of mass A.

So A moves half the speed because A has two active rope legs sharing the load and it's where the rope is anchored and where the rope returns to. This lets A move 304 m and B/C each move 608 m...

A gets to have it's weight "doubled" because it is the anchor point of the 2:1 pulley system?

I just feel like I'm missing something. I've added a sketch above of my math... Just feels that the downward force is excessive even for this setup.

 

[MarcusThatsMe]

What exactly is meant by "reversing the pulley"? Is it simply reversing the direction that the pulley will be turning? Or is it changing the setup so that the mass B is being supported with a 2:1 force advantage (mass B connected to the center of the pulley) instead of a 2:1 distance advantage (mass A, on the ramp, connected to the pulley center)?

If the setup is changing, the MA/MB force formula will change. If it is only the direction of movement, that formula will not change.

"Reversing the direction" means the anchor point changes. But it just feels like I'm missing something in my math with the second part where B moves up and A moves down. I included a picture of what I'm doing and my math, maybe I am missing something?!?

 

[sophiecentaur]

"Reversing the direction" means the anchor point changes.

You have posted a diagram (which is always a good thing) but it doesn't contain the rope and the pulley. I'm not sure what's really going on. Your hand drawing is absolutely fine and you could make it much easier to read if you tinker with (edit) the brightness and contrast of the image. Black ink might help too.

If this problem has a practical application then your theoretical calculations should have an extra factor of the Friction forces. That would give an indication of the actual Mechanical Advantage. Without extra forces, your answers will involve Velocity Ratio and the answer from that can be very optimistic. Your pulley mass may be negligible and so could the friction - but you would only know from doing the calculations.