1the horizontal component will be equal to Tcos θ
This is a pretty easy one....but I need some help.
Initially the system of masses is held
motionless. All surfaces, and pulley are frictionless.
At the instant after the system of objects is released,
find the accelerations of m1, m2 and m3.
The pulley exerts a force say R along its joint, so it will have 2 perpendicular components out of which one is responsible for the acceleration of m1.
I had doubt whether this horizontal component of R will be same as the tension in the rope?
Moreover, its not given that the pulley is massless, so do we have to take tensions on both sides of the rope different.?
some one please solve
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UP 0 DOWN 0 0 3
3 Answers
23force on m1 is T to the left horizontally
if u are confused then recall that accn of com in horizontal direction will be zero , so m1ax=m2ax
