Explain Tension to me

[ravupadh]

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Hey all. So I was recently working on the TBR Physics Passage 2-8 and I find myself confused with tension. Let me get this straight (based on how they treat the elevator and counterweight in the passage).

If the rope is massless and frictionless, as well as accelerating, tension is uniform throughout the rope and the tension for an object equals the downward forces on that object? Does it matter if the system is accelerating uniformly?

Also, if the rope is massless and frictionless, and the system is NOT accelerating, then you have to treat each side of the rope differently UNLESS m1g = m2g?

Am I getting this right? One more nitpicking question, if two objects are bieng held by a rope, and that rope is accelerating, how do you account for the fact that one object is going up and another is going down? Do you set one tension negative or something in the T-mg equation?
Thanks.

 

[chemtopper]

If the rope is mass less and frictionless, as well as accelerating, the tension will remain uniform in every case. In the above mentioned case the tension for an object need not to be equal to the downward forces on that object. For the above scenario the system should be accelerating [it may be uniform or non-uniform].

If the rope is mass less and frictionless, and the system is not accelerating, then we'll treat each side of the rope in a same manner as the two part of the rope have same tension. If m1 is not equals to m2, so tension will be equals to {[m1+m2]*g}/2. It doesn't matter on the equaility or non equality of the masses.

If two objects are being held by a rope, and that rope is accelerating, then it is obvious that one mass is going up and the other mass is going down. But the tension can't be negative.
Let us assumed masses be m1[going up] and m2[going down], acceleration be 'a'.
T-m1g=m1a -I
m2g-T=m2a -II
If in the above equation 'I' a comes out to be negative that means the mass m1 is going down and m2 is going up.

I hope you get a clear answer from my explanation.