Consider now the case where the string of length l has mass per length λ ̸= 0. Show that the instantaneous acceleration of m2 is given by a2 = (m1 −m2)+2x0λg/(M + m1 + m2)
where M = λ l is the total mass of the string and find the tension in both ends of the string. (please see picture!)
I have the solution and it starts off with three equations
T2 - m2g = m2a
T1 - m1g = -m1a
T1 + 2x0λg - T2 = Ma2
Although i completely understand where the first two come from I'm a little confused my the last one. Although i get that it is supposed to show the force on the string, I can't figure out how the left hand side is true. IF someone could please explain where this equation came from that would be much appreciated.
Also when i was attempting this problem before seeing the solution i came up with two equations:
T1- m1g - (l/2 + x0)λg = m1a and the equivalent for the second mass. Would these also be correct?
The Attempt at a Solution
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