The system is in equilibria, I could have probably done it if there was acceleration going on and the masses were two with T and a in the equations like T -1g = 1 x a, but how can I solve this one.
------------- Note: actually from Math-Mechanics but no one responded there and besides Mechanics is better known in Physics.
Just to derail for one second sorry, but am I right in saying tension is only constant throughout a string if the pulleys and the surface is smooth (i.e. no friction involved)?
Just to derail for one second sorry, but am I right in saying tension is only constant throughout a string if the pulleys and the surface is smooth (i.e. no friction involved)?
Erm I think it should be the same when friction is involved, but there are two strings here hence the different tensions on the left and right sides
Erm I think it should be the same when friction is involved, but there are two strings here hence the different tensions on the left and right sides
But if friction was negligible in this model wouldnt the tension throughout the system be equal? (This would stop the object being in equilibrium but lets ignore that for now lol).
But if friction was negligible in this model wouldnt the tension throughout the system be equal? (This would stop the object being in equilibrium but lets ignore that for now lol).
Oh right, I dunno, we would expect the block to move to the left if there were no friction so I guess the difference in tensions must provide this force?
Oh right thanks for clearing it all up, and does that apply to the (even more unphysical...) second problem when there is no friction involved at all and the system is no longer in equilibrium?
If there were no friction it would be a standard problem. The 2kg would accelerate down, the 10kg to the left and the 1 kg up - all with the same acc assuming a light inextensible string. (borrowed from Maths dept)