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Home / Doubts and Solutions/JEE/Physics

In the given situation wedge is moving towards right with an acceleration 5m/s^2 with the help of external force F.(m=1kg).Find acceleration of sphere,normal between sphere and wedge, and value of F.

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Asked by m.nilu 3rd August 2018, 10:24 AM
Answered by Expert
Answer:
Free body diagrams for wedge and ball are shown in figure.
Let N1 be the normal reaction force between wedge and bottom support.
Let N2 be the normal reaction force between wedge and ball.
Let N3 be the normal reaction force between ball and side support.
 
For wedge,  in vertical direction we have,   2×m×g + N2 × cos53 = N1 ...........................(1)
For wedge, in horizontal direction we have, F - N2 × sin53 = 2×m×a ..........................(2)
where a is acceleration of wedge
 
For ball, in vertical direction we have,  N2 x cos53 = m×g ...........................(3)
due to wedge acceleration, reaction force N3 is given by, m×a = N3 ...................(4)
for ball, in horizontal direction, we have, N3 - N2 × sin53 = m×ab ...............(5)
 
from eqn.(3), we have, N2 = m×g×sec53 = 16.3 Newton ............(6)
we get force F, from eqn.(2) by using eqn.(6), F = 2×m×a + m×g×tan53 = 23 Newton  ( because m = 1kg and a = 5 m/s2)
 
we get N1 from eqn.(1) using eqn.(6)  N1 = 3×m×g = 29.4 Newton
acceleration of ball ab is obtained from eqn.(5), using eqns.(4) and (6), we get ab = -8 m/s2
 
Answered by Expert 12th August 2018, 5:12 PM
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