Q: In the figure shown all the surface are smooth.All the blocks A, B and C are movable, x-axis is horizontal and
y-axis vertical as shown. Just after the system is released from the position as shown.
(A)Acceleration of ‘A’ relative to ground is in negative y-direction
(B) Acceleration of ‘A’ relative to B is in positive x-direction
(C) The horizontal acceleration of ‘B’ relative to ground is in negative x-direction.
(D) The acceleration of ‘B’ relative to ground along the inclined surface of ‘C’ is greater than g sinθ
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Ans: (A), (B) , (C) , (D)
Sol: There is no horizontal force on block A, therefore it does not move in x-direction, whereas there is net
downward force (mg ñ N) is acting on it, making its acceleration along negative y-direction. Block B moves downward as well as in negative x- direction. Downward acceleration of A and B will be equal due to constrain, thus w.r.t. B,A moves in positive x-direction.
Due to the component of normal exterted by C on B, it moves in negative x-direction.
The force acting vertically downward on block B are mg and NA (normal reaction due to block A). Hence the component of net force on block B along the inclined surface of Bis greater than mg sinθ . Therefore the acceleration of ‘B’ relative to ground directed along the inclined surface of ‘C’ is greater than g sinθ