2:1 Pulley Question

  • Thread starter Thread starter MarcusThatsMe
  • Start date Start date
Click For Summary
The discussion revolves around the mechanics of a 2:1 pulley system involving two masses, A and B, where A is pulled vertically and B is on an incline. The user is confused about the implications of "reversing the pulley," questioning whether it refers to changing the direction of movement or altering the setup to change the force dynamics. They present calculations for both scenarios, noting that mass B moves at different speeds relative to mass A depending on the configuration. Additionally, there is a suggestion to consider friction forces in practical applications, as they can significantly affect the mechanical advantage. Clarification on the sketch provided is requested to better understand the setup and calculations.
MarcusThatsMe
Messages
22
Reaction score
8
TL;DR
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...
 
Engineering news on Phys.org
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.
 
MarcusThatsMe said:
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.
 
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...

1000005619.webp
 
Lnewqban said:
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.
 
.Scott said:
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?!?
 
MarcusThatsMe said:
"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.
 
MarcusThatsMe said:
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.
Sorry, I am unable to understand your sketch above.
Could you show us a more realistic diagram, please?
 
Lnewqban said:
Could you show us a more realistic diagram, please?
Yes; one with circles for pulleys and lines for string etc.. As simple as you like.
 

Similar threads

Replies
3
Views
247
  • · Replies 21 ·
Replies
21
Views
4K
  • · Replies 3 ·
Replies
3
Views
11K
Replies
3
Views
4K
  • · Replies 9 ·
Replies
9
Views
13K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
Replies
2
Views
1K