Laws of MotionMedium
Question
In the system shown in figure mA = 4m, mB = 3m and mC = 8m. Friction is absent everywhere. String is light and inextensible. If the system is released from rest find the acceleration of block B


Options
A.
(leftward)
B.
(leftward)
C.
(rightward)
D.
(rightward)
Solution

a = b + c [string constrained]
T = mBb [Newtons II law for B in horizontal direction]
T = [mA + mc] c [Newton′s I law for A and C in horizontal direction]
mAg − T = mAa [Newton′s II law for A in vertical direction].
⇒ T =
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