# Tension in a string over a pulley

1. Sep 24, 2009

### free-node-5

I'm stumped. I was working on a problem in a computerized system that tells you whether you got it right or not and I successfully found everything including the force acting on each end of the string, which in this case is suspended over a (frictionless) pulley.

The force on one end is 19.6N and the force on the other is 67.2N. (opposing each other across the pulley)
So what I don't get is, how do you find the tension in the string?

I figured it was the sum of the forces on each end but it said that was wrong.
So then since it is a moving system, I tried subtracting the lesser force from the larger, and it said that was wrong too. :-/

Thanks

2. Sep 24, 2009

### kuruman

Do you mean to say that a weight of 19.6 N is hanging on end and 67.2 N on the other?

3. Sep 25, 2009

### free-node-5

yeah pretty much

that's after some calculations but I used those values for the rest of the questions and they should be accurate

4. Sep 25, 2009

### kuruman

"Pretty much" is not very informative. What is the exact wording of the problem? Perhaps your interpretation of it is faulty.

5. Sep 25, 2009

### free-node-5

Two objects are connected by a light string that passes over a frictionless pulley. The incline is frictionless, m1 = 2.00 kg, m2 = 8.00 kg, and θ = 59.0°.

Find the tension in the string.

It also asked for the acceleration but it said I got that part right. (could have been a coincidence I suppose)

In the past I've spat out numbers like 86.8, 47.6, 67.2, all of which seem to be wrong. :-(

Last edited by a moderator: Apr 24, 2017
6. Sep 25, 2009

### kuruman

If you have the acceleration correctly figured out, the rest is easy. SInce you don't know the tension, call it T. Look at the hanging mass. If it accelerates down, there must be a net force acting on it that is not zero. Can you write an expression for this net force in terms of the tension T and the weight mg?

7. Sep 25, 2009

### free-node-5

T=(m1+m2)a ???

8. Sep 25, 2009

### kuruman

We are talking about the hanging mass, m2. How many forces act on it?

9. Sep 25, 2009

### free-node-5

it's own weight excluding the amount that the other mass, m1 is pulling back on it so...
2

10. Sep 25, 2009

### kuruman

The other mass is not pulling on m2. The other mass is not even in contact with m2. The rope is pulling on m2. The two forces acting on m2 are tension T (up) and weight mg (down). So what is the net force on m2?

11. Sep 25, 2009

### free-node-5

I feel so stupid

well then I'd say F=(m2)(9.8)-T

oh, so do I set up the one for m1 too and then solve for T?
edit:
on second thought, that may not make any sense

12. Sep 25, 2009

### kuruman

You're almost there.

You have assumed that "down" is positive and "up" is negative. That's OK. What you have above is the net force (=the sum of all the forces). According to Newton's Second Law, what is the net force equal to?

13. Sep 25, 2009

### free-node-5

well...
all I know is F=ma
is that what you mean?

14. Sep 25, 2009

### kuruman

That's exactly what I mean. You know m=m2 and you know the acceleration from before. So now you can find T by putting together Newton's Second Law.

15. Sep 25, 2009

### free-node-5

like (m2)(4.76019)=(m2)(9.8)-T ?

if so, how did we know to look at the hanging mass as opposed to the one on the slope?

16. Sep 25, 2009

### kuruman

Exactly like that. The mss on the slope has a different free body diagrams and different forces. However, if you did that one instead and did it right, you would get the same answer for the tension. I asked you to do the hanging mass because it is simpler to explain and understand. You can try doing the other mass if you wish. It is good practice.

17. Sep 26, 2009

### free-node-5

a = 4.76019

T = (m1)(9.8)+(m1)(a) = 29.1204
T = (m2)(9.8*sin59)-(m2)(a) = 29.1204

yay

when I started typing this post I was getting inconsistent results but then after poking it a while I figured out that not only did we have m1 and m2 backwards but that I had to add for m1 instead of subtract because of the direction of the acceleration

so, yay

thanks kuruman

18. Oct 15, 2009

### kn0where

Thanks for this thread. I had much the same problem, except the first mass was on a slope.

a = F/m
a = (m2g - m1g*sin(theta))/(m1+m2)

solve for tension using the second mass
T = Fg - ma
T = m2g - m2(gm2 - gm1*sin(theta))/(m1+m2)