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frictionless pulley question

MEPS1996

Why does a frictionless and massless pulley mean that the tension in strings on either side of the pulley are the same? What effect would using a pulley with mass and friction have on the tensions?
thanks

Reply 1

Stonebridge

Original post by MEPS1996

Do this thought experiment and work it out for yourself.

Lets make the friction and mass of the pulley so great that it can't move.
eg.Put your finger on the top of the pulley and hold the string so it can't move. That's like very large friction and pulley with lots of mass.
The result is that now you effectively have two individual strings either side. The tension in the one will depend on the mass hanging there, and the tension in the other will depend on the mass there. You could have two widely different tensions.

Reply 2

natninja

Original post by MEPS1996

Basically what Stonebridge has said, I might add that the difference in tensions is because a force is required to accelerate the pulley so that it turns.

Reply 3

MEPS1996

Original post by Stonebridge

thanks for your reply stonebridge. I understand you are equating putting your finger on the string to a large friction in the axis of a pulley and large enough coefficient of friction for the string not to move over the pulley. However, how exactly does the friction in the axis of the pulley cause different tensions in each section of the string. Can this friction in the axis of the pulley be considered to act on the string? Can you offer an explanation of the different tensions in terms of F=ma and forces?
thanks a lot for your time.

Reply 4

MEPS1996

Original post by natninja

Basically what Stonebridge has said, I might add that the difference in tensions is because a force is required to accelerate the pulley so that it turns.

Thanks for your reply. Could you elaborate on that please?

Reply 5

natninja

Original post by MEPS1996

Thanks for your reply. Could you elaborate on that please?

Firstly, for up to M3 on most A-level boards and most A-level physics boards this is outside the spec so probably best not to go in-depth with it on papers....

Ok so if you have a stationary wheel, it can't just suddenly start spinning. You have to apply a torque to it (T=Fxr [vector cross product if you've done it]) and there is an equation for rotational motion of torque=moment of inertia x angular acceleration equivalent to F=ma for linear motion. The moment of inertia of a body is to do with how it's mass is distributed in space. So if the pulley has any mass at all then it requires a force to make it turn (mass 0 moment of inertia=0 so T=0 sor any angular acceleration) and this force is the difference between the tensions in the two pulleys (or maybe half that but that seems silly?). So that is the massless bit dealt with. The frictionless bit means that no force is required to overcome friction from the pulley rotating around it's axis and and also means that the pulley turning or not has no influence over the problem. There is one other reason that requires a massless pulley - take for example a double atwood machine (http://farside.ph.utexas.edu/teaching/336k/Newtonhtml/node79.html scroll down for an example picture) if you have masses of 2,3 and 5m then at first glance it appears that only one pulley rotates... but if you analyse the system the centre of mass is constrained to move in specific ways only and taking the masses of the pulleys into account makes the problem far harder to solve...

Reply 6

Stonebridge

Original post by MEPS1996

If you have a massless inextensible string and pull at the two opposite ends there is a tension set up in it. This is due to the molecules being slightly separated and trying to pull back together. There is only one value of the tension in this string and the tension is the same all along the string. Do you dispute this?
The only way you can set up a different tension in one part of a string from another part, is to have a third point at which a force acts, effectively dividing the string in two.
This is achieved on the pulley if there is friction, because at the point of contact there is now another force acting on the string in addition to the two forces (masses) at the ends. It actually doesn't matter in this case if the pulley has mass or not. The frictional force at the point of contact provides you effectively with "two" strings. It interrupts the transmission of tension through the whole string that would be there if the string were isolated. The mass of the pulley in this case will help to determine the actual value of the tensions from consideration of the acceleration produced in the pulley and the hanging masses. (F=ma on the masses, and torque on the pulley producing rotational acceleration.)