# Force applied to 2 connected mass.

1. Mar 29, 2014

### negation

A force is applied to m1 connected to m2 by a "massless" cord. where m2>m1.

the acceleration, a, is a = F/m1+m2

By decomposing the system into A and B where A refers to the tension on m1 upon having the force applied onto m2, the tension TA is to the positive x-direction.

However, in the case of the system B, the tension TB is to the negative x-direction as the force F is applied directly to m2.

Why does TB arises in the block m2 but yet not in m1?

2. Mar 29, 2014

### Staff: Mentor

I'm not sure I understand this statement. The cord connecting the two masses exerts a tension force on both m1 and m2. The cord exerts the same force on both blocks (different directions, of course).

3. Mar 29, 2014

### negation

I shall rephrase. I was lacking clarity in the OP.

When force F is applied on m2, m1 experiences a tension TA in the positive x-direction. (why shouldn't m1 experience another tension in the negative x-direction. By my understanding, there should be an equal and opposite reaction.

In the case of m2, upon the appliance of the force F, the m2 experience not only the force F in the positive x-direction but also a tension TB in the negative x-direction.

4. Mar 29, 2014

### Staff: Mentor

Right. Tension is created in the cord, which acts on both masses.

What would create that tension?

A reaction from what?

And if the mass m1 were connected to a third mass, then it too would experience a tension force acting to the left. But there is no third mass.

(Note that the force F and the tension in the cord are not equal.)

5. Mar 29, 2014

### negation

A reaction from m1 having experience a force F applied to m2?

Suppose, as you've suggested that a third mass m3 was attached to m1 to the left. If the same force F were applied, then aside from m2 having experienced the tension TB, m1 would now experience a tension in the negative x-direction, am I right?

6. Mar 29, 2014

### Staff: Mentor

I'm not sure what you mean by a 'reaction'. m2 pulls on m1 (via the massless cord) and m1 pulls back on m2.

Sure. The force F will end up creating various tensions in the cords connecting the masses.

7. Mar 29, 2014

### dauto

Yes there is a reaction (Newton's 3rd law). What made you believe there isn't one?

8. Mar 29, 2014

### dauto

Come on, Doc. he's talking about Newton's 3rd law. quoting from your post " m2 pulls on m1" - That's the action - "and m1 pulls back on m2" - That's the reaction.

9. Mar 29, 2014

### negation

I understand there to be one but it isnt outlined by my lecturer.

10. Mar 29, 2014

### Staff: Mentor

That's why I used it as an example. But I don't think that's what was meant. (I think he was looking for a second force acting on the trailing block.)

Last edited: Mar 29, 2014
11. Mar 29, 2014

### dauto

That doesn't mean it's not there.