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15+ Year Member
May 30, 2002
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can someone help me out on pulleys, specifically multiple pulleys. i don't understand how theres two tension forces pointing up per pulley if the force at the end of the string is going down. if you trace the force all throughout the string, one force would be going down while the other could be going up.

(o) (o) (o)
| | | | | |
| a b c d e
| | | | | |
F | | | | |
v (o) (o) Box

please use your imagination on this very rudimentary drawing. So A,B,C,D,E are all tension forces going up. But if you follow the tension force all throughout the rope starting from the box, A,C,E would all point up and BD would point down. Can anyone please explain this to me.

Sorry if you can't understand the pic... i'll try to surf the net and look for an actualy image


7+ Year Member
15+ Year Member
Oct 9, 2003
Resident [Any Field]
The first thing to keep in mind is that the force doesn't 'wrap around' the pulley the way it sounds like you're describing it. Think of it like a rubber band. If you stretch a rubber band, you're putting tension on it, and the rubber band pulls your hands toward each other, i.e. there's a force at each end pointing toward each other. At the same time, your hands are putting a force on each end of the rubber band, and they're pointing away from each other.
Now for the pulley's, the best way to see the force balance is to draw a free body diagram, which would look something like this:

...........^ Fa
.... ____|___
(sorry bout the dots, I'm too lazy to figure out the HTML tabs for whitespace)
Where Fa is the force of the anchor that's holding the pulley against the ceiling (or whatever), F is the force being applied to the end of the rope, and T is the force from the tension in the section you called a. You MUST remember to include the force of that anchor! And remember from the rubber band example, since that rope a is in tension, the force on the pulley must be acting down. The force section a would exert on the bottom pulley would be pointing up. Anyway, this is the free body diagram for the pulley.
From Newton's Law you know that:
Sum of Forces = Mass * Acceleration
Now that pulley isn't going to be moving (ignoring rotation), so it's not gonna be accelerating either. If Acceleration is zero, the sum of forces must be zero also. Note that this analysis only works if it's a frictionless pulley, otherwise the problem becomes complicated... If the sum of the forces equals zero, and you define up to be a positive direction and down to be negative, you get
Fa - F - T = 0, or
Fa = F + T
Because the pulley is frictionless, F and T are going to be equal in magnitude (trust me on this, I've made this post is way too long already). Finally, because you know that T equals F, you know the force section a is exerting on the bottom pulley is also T, and thus F. So if you trace it all the way thru the system, the force pulling up on the box is equal to force T. If this makes no sense whatsoever, let me know and I'll try to clear it up.



Lanius examinatianus
10+ Year Member
Apr 23, 2004
too far south
While the above looks correct to me (though honestly, I didn't make it all the way through), I think it might be helpful to look at it the quick way.

On the MCAT:

1. Ropes, cables and strings are just ways to apply forces.

2. The force on a rope etc. is always applied along the line of the rope, and it is the same in both directions (Matt explains this, above).

3. The amount of force is the same throughout the rope.​

This is why the "count the ropes pulling up" method works for multiple pulleys. In your diagram, the rope on the rope doesn't count because neither end touches something that can move (turning a pulley doesn't count), but a, b, c, d, and e each do, and each rope applies the same force, up and down, as the amount by which you pull.


10+ Year Member
5+ Year Member
Jan 4, 2005
I still don't understand the answer to the OP's question. All I know is that you don't sum forces on each end to determine the tension in the rope.