# Simple pulley/tension problem

1. Jul 26, 2006

### kristen151027

Here is the question:

Two blocks of different mass are connected by a massless rope which goes over a massless, frictionless pulley. The rope is free to move, and both of the blocks hang vertically. What is the magnitude of the tension in the rope?

(a) the weight of the heavier block
(b) the weight of the lighter block
(c) their combined weight
(d) a value between the two weights
(e) zero

It's probably pretty easy for most of you, but I'm new to this stuff :-)

2. Jul 26, 2006

### Hootenanny

Staff Emeritus
Can you use Newton's second law to formulate an equation to represent the motion of each mass? I would recommend starting with a free body diagram showing all the forces acting.

3. Jul 26, 2006

### kristen151027

I understand that, but I can't decide whether to add the two tension forces or average them.

4. Jul 26, 2006

### Hootenanny

Staff Emeritus
Can you show me your equations?

5. Jul 26, 2006

### kristen151027

Well, Newton's second law states that: F = ma
For each one, the sum of the forces is the T + (-mg), so T + (-mg) = ma
T = mg + ma
T = m (g + a)
So, the heavier block creates more tension on the rope.
Other than that, I'm clueless.

6. Jul 26, 2006

### HallsofIvy

Staff Emeritus
There are NOT two tensions - there is only one. To have "tension" in a wire or rope, you have to have forces in both directions, of equal strength and opposite directions. First calculate the weight (downward force) of each mass. The difference in weights will be net force on the system- the heavier block moves downward, the lighter block moves upward. That means that the heavier block is moving downward but not as fast as if the smaller block weren't there. That's because its weight is offset by the weight of the smaller block. Of course, the smaller weight is not going down but up- that's because the weight of the smaller block minus the weight of the larger is negative. That common value, ignoring the direction, is the tension.

7. Jul 26, 2006

### Hootenanny

Staff Emeritus
You should have two separate equations. The equation you have there represents the lighter block, where the tension is in the same direction as the motion (i.e. upwards towards the pulley) and the force of gravity opposes motion. Can you now write a separate equation representing the heavier block?

8. Jul 26, 2006

### kristen151027

T + mg = ma
T = m(a-g)

And in response to HallsOfIvy, the "common value" you're referring to is, in effect, a value between the weight of the two blocks?

9. Jul 26, 2006

### Hootenanny

Staff Emeritus
You may wish to reconsider this. In the case of the heavier block the sum of the forces is thus; $\sum F = Mg - T$ can you see why?
This is correct

10. Jul 26, 2006

### kristen151027

Yes, because, as HallsofIvy said, "the heavier block is moving downward but not as fast as if the smaller block weren't there. That's because its weight is offset by the weight of the smaller block." Is that the correct reasoning?

And the answer, I just want to make clear, is (d) a value between the two weights.

11. Jul 26, 2006

### Hootenanny

Staff Emeritus
In a round-a-bout way, yes. If you were to draw a free body diagram of the heavier block you would see two forces acting; gravity pulling the block downwards (in the same direction as the motion) and the tension acting upwards.
That is correct.

12. Jul 26, 2006

### kristen151027

Thank you very much for your help!!

13. Jul 26, 2006

### Hootenanny

Staff Emeritus
My pleasure