# Pulley - on which side is tension bigger? Two different answers

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.

• 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. – heather Mar 22 '18 at 0:31
• With regards to this question being flagged: I think this is a thoughtful question, and provides value to other readers also. – Siva Mar 26 '18 at 0:42

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.

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.

• Hi, thanks for that. I know that $A$ goes downwards, but why does that mean the tension is bigger on that side? – PhysicsMathsLove Mar 3 '18 at 17:37
• 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? – Bob Jacobsen Mar 4 '18 at 19:02