# Trying to understand the tension in a spring with two weights attached [duplicate]

Possible Duplicate:
Why I think tension should be twice the force in a tug of war

For (A) in this image, I would think the tension should be should be 2mg.

However, my book says the weight in (A) should be only mg. I don't understand why the tension should not be the same as in (C) and double the tension in (B).

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## marked as duplicate by Qmechanic♦, Manishearth♦Dec 11 '12 at 10:54

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

Have you drawn the free-body diagram? – tpg2114 Nov 27 '12 at 23:31
Possible duplicates: physics.stackexchange.com/q/41291/2451 and links therein. – Qmechanic Nov 27 '12 at 23:51

## 1 Answer

As you can see a) and c) are different because in a) the spring is part of the rope and in c) it is part of the support.

The tension in a) and b) is the same as the systems are in static equilibrium.

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But why is the spring being part of the rope significant? I just don't get it conceptually. – Chris Ballance Nov 28 '12 at 23:47
In c) glue together the pulley and two weights and you will see that the force on the spring is $2 m g$. If you agree that a) and b) are the same, then you agree that for b) the force is $m g$ only. – ja72 Nov 29 '12 at 4:33
I specifically don't understand why a) & b) are the same. In the free-body diagram, the forces are equal and opposing and the anchor point (of sorts) is the spring itself. – Chris Ballance Nov 29 '12 at 19:06
As I said, a) and b) are in static equilibrium, with equal and opposite forces acting on each side of the spring. Each one is $m g$ from the weights that are hanging. – ja72 Nov 29 '12 at 20:46
Right, but were this a game of "tug of war" would not two groups pulling in opposite directions create double the tension of a single group pulling against a fixed rope? – Chris Ballance Nov 30 '12 at 18:52