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I have the following problem (see picture). The mass of $A$ is 3kg and the mass of $B$ is 2kg. I am told the pulley is not smooth and the difference in tension on either side $\Delta T$ is given by $$\Delta T = 3 + 0.3a$$ where $a$ is the acceleration of the two particles.

Now I am unsure why the tension on the left has to be bigger. The reason given in the book is that $A$ is more massive. However, if I do the calculation supposing that the tension of the right is greater, I have $$A : 3g - T = 3a$$$$B : T+ 3 + 0.3a - 2g = 2a$$ and combining these I find $a=2.78$ m/s$^2$. This is different to the answer of 1.26 m/s$^2$ we get if we presume the tension on the left is greater. But why do I even get an answer?

This is part of a bigger issue I am having, as I am trying to apply a similar approach to considering the problem of two masses, one on a table, and one hanging freely, both connected by a string that passes over a non-smooth pulley. However, again, I am not sure on which side of the string the tension will be greatest.

Any help is thoroughly appreciated. enter image description here

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  • $\begingroup$ Look again at this question. It is not a homework question! It is in fact an intuitive question wondering why exactly the tension on the left "has to be bigger" and has shown work explaining the OP's concerns. $\endgroup$ – heather Mar 22 '18 at 0:31
  • $\begingroup$ With regards to this question being flagged: I think this is a thoughtful question, and provides value to other readers also. $\endgroup$ – Siva Mar 26 '18 at 0:42
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Assuming that you have provided all of the information in the problem, both solutions are possible. However, I suspect that you have omitted to say that the system is released from rest.

Case 1 : If the system was released from rest then A would go down, because it is heavier. Friction $f=\Delta T$ in the pulley opposes motion and acts down on the right. Because the pulley is massless the forces on it must balance at all times - otherwise, with a finite resultant torque it would have infinite acceleration. So the tension would be greater on the left : $T_L=f+T_R$. In this case you find that $a=1.26m/s^2$.

Case 2 : However, if the system was not released from rest then B could be moving down initially - eg propelled by an impulse. The friction force then acts towards the left, so tension on the right is greater : $T_L+f = T_R$. In this case you find that $a=2.78m/s^2$.

In both cases the acceleration is down on the left. In Case 2 block A is accelerating downwards (=decelerating upwards) faster than in Case 1 because friction is acting in the same direction as gravity on A, whereas in the Case 1 friction is opposing gravity on A. In Case 2 the two blocks will eventually change direction, friction will change direction also, and we're back to the situation in Case 1.

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The setup of the problem seems to be telling you that the pulley only resists motion via inertia and friction.

So the unequal gravitational forces on A and B will result in A going down, or the system being stuck. A lighter mass can’t lift a heavier mass without help!

Alternately, if you don’t trust a physical argument, you can assume B is going down. (Be careful with the sign of the “a” term for the pulley.). You’ll find an internally-inconsistent answer, which tells you your assumption about direction was wrong.

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  • $\begingroup$ Hi, thanks for that. I know that $A$ goes downwards, but why does that mean the tension is bigger on that side? $\endgroup$ – PhysicsMathsLove Mar 3 '18 at 17:37
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    $\begingroup$ The string is moving toward the A side, against the resistance of the pulley. That means the tension has to be higher on that side. Try pulling a string through your fingers, and then clamp your fingers down on it: Is the tension higher before the fingers exerting force on it, or after? $\endgroup$ – Bob Jacobsen Mar 4 '18 at 19:02

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