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.