Force applied to 2 connected mass.

AI Thread Summary
A force applied to mass m2, which is connected to mass m1 by a massless cord, results in different tension effects on both masses. When force F is applied to m2, it experiences tension TB in the negative x-direction, while m1 experiences tension TA in the positive x-direction. The discussion revolves around the confusion regarding why m1 does not experience a tension in the negative direction as well. It is clarified that m1 does indeed experience a reaction force due to Newton's third law, where m2 pulls on m1 through the cord. The overall understanding emphasizes that the tensions in the system are not equal and that the interactions are dictated by the applied force and the connection between the masses.
negation
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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?
 
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negation said:
Why does TB arises in the block m2 but yet not in m1?
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).
 
Doc Al said:
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).

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.
 
negation said:
When force F is applied on m2, m1 experiences a tension TA in the positive x-direction.
Right. Tension is created in the cord, which acts on both masses.

(why shouldn't m1 experience another tension in the negative x-direction.
What would create that tension?

By my understanding, there should be an equal and opposite reaction.
A reaction from what?

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.
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.)
 
Doc Al said:
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.)


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?
 
negation said:
A reaction from m1 having experience a force F applied to m2?
I'm not sure what you mean by a 'reaction'. m2 pulls on m1 (via the massless cord) and m1 pulls back on 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?
Sure. The force F will end up creating various tensions in the cords connecting the masses.
 
negation said:
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.

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

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


I understand there to be one but it isn't outlined by my lecturer.
 
  • #10
dauto said:
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.
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.)
 
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  • #11
negation said:
I understand there to be one but it isn't outlined by my lecturer.

That doesn't mean it's not there.
 
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