# Spring & pulley

1. May 30, 2007

### f(x)

1. The problem statement, all variables and given/known data

Consider a pulley, which has two masses of 2kg each attached with a massless rope followed by a spring (on one side only). Consider another system of a pulley, same as the previous except that the mass connected to the rope with the spring is of 1 kg. Find the ratio of extensions of the two springs, assuming the spring constants to be equal in both cases.

2.What i think

I tried making a freebody diagram of the spring...it has a force equal to tension (2g in first case) acting upwards, and another tension force acting down. Both the forces try to stretch the spring, so total force on it is 4g in first case and 3g in second. I assume system to be in equilibrium in both cases.
Thus extension should be 4:3.
But this is not the right answer. Rather i feel i m very much off the track on this one. Kindly suggest how do i solve this.
Any help is appreciated

2. May 30, 2007

### Staff: Mentor

The spring extension will be proportional to the tension in the rope/spring. In the first case, the tension in the rope equals mg, since the system is in equilibrium. The tension equals mg, not 2mg! To resist the downward pull of the hanging mass (weight = mg) the tension must equal mg; to resist the upward pull, via the pulley, of the other mass (weight = mg) the tension must equal mg.

The spring is pulled from both sides by a force equal to the tension--the two forces don't "add up".

For the second case you first have to solve for the tension in the rope/spring by using Newton's 2nd law. In this case, the system is not in equilibrium.

3. May 30, 2007

### f(x)

Yeah thats what i want to know, why dont they add up ?

4. May 30, 2007

### Staff: Mentor

(1) Imagine a rope attached to a wall. You pull on it with a force of 100N. What's the tension in the rope? What force does the wall exert on the rope?

(2) Now imagine that you and a friend grab a rope at each end and pull with a force of 100N. What's the tension in the rope?

Is there a difference between these two cases?

5. May 30, 2007

### f(x)

Tension is 100 N in both cases

6. May 30, 2007

### Staff: Mentor

Exactly. Now apply that same thinking to your problem.

7. May 30, 2007

### f(x)

OK, i understand that the spring is being pulled by equal force, each of which is mg. But how to i get the extension of the spring ? Is it equal to mg/k in each direction, adding up to give 2mg/k totally ?

8. May 30, 2007

### Staff: Mentor

No. If you pull on the spring with a force of 100N, it must pull back on you with the same force. The force the spring exerts is related to how much it stretches, via Hooke's law. Note that the spring pulls just as hard--100N--at both ends, but the tension is only 100N.

Part of the point is that you can't just pull a spring (or rope) at one end--you have to pull at both ends to maintain a tension.

Last edited: May 30, 2007
9. May 30, 2007

### f(x)

So is the extension of the spring mg/k ?

And in part 2, does the presence of the spring make any difference , since it pulls both the ends with the same tension ?

10. May 30, 2007

### Staff: Mentor

Yes, in the first case, but not in the second. To find the extension in the second case, first find the tension.

I'm not sure what you mean by part 2 (do you mean case 2?), but if I understand you correctly: No, the spring doesn't effect the tension, it just responds to it. (Lets ignore any oscillitory effects.)

11. May 30, 2007

### f(x)

For the 2nd case,
2g-T=2a
T-g=a
Putting this in eqn 2, T=4g/3
So, the force on the spring is 4g/3, hence extension is 4g/3k
Extension in case 1 was 2g/k, so the ratio of extensions case 1:case 2::3:2
Is this correct ?
Would the ratio remain same if i interchange the 2kg and 1kg mass ?

12. May 30, 2007

### Staff: Mentor

All good.
Yes. The only difference will be the direction of the acceleration, but the tension in the rope/spring will be the same.

13. May 30, 2007

### f(x)

Yay ! Thanks a million Doc Al , specially for the quick replies.
I'll rework the problem and ask if I still have any doubts .
Thx once again