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Question: In the figure shown ‘\(P\)’ is a plate on which a wedge \(B\) is placed and on \(B\), a block \(A\) ...

In the figure shown ‘PP’ is a plate on which a wedge BB is placed and on BB, a block AA of mass mm is placed. The plate is suddenly removed and the system of BB and AA is allowed to fall under gravity. Neglecting any force due to air on AA and BB, the normal force on AA due to BB is

A. mgcosθ\dfrac{{mg}}{{\cos \theta }}
B. mgcosθmg\cos \theta
C. Zero
D. 2mgcosθ\dfrac{{2mg}}{{\cos \theta }}

Explanation

Solution

In physics, when two bodies are in contact with each other, then the perpendicular force between the surface of two bodies is called normal reaction force, in the given question we will first draw the free body diagram and observe various force acting on AA and BB under free fall of gravity and thus find the magnitude of normal reaction force which is denoted by NN.

Complete step by step answer:
Let us first draw the free body diagram of the given system consisting of wedges and blocks while falling under gravity. Let θ\theta be the angle between sliding part and bottom of wedge, now weight of block A of mass m acting downward is given by

F=mgF = mg
Now the reaction force in upward direction from the geometry of diagram can be seen as
FN=Ncosθ{F_N} = N\cos \theta Where NN is the normal reaction force between block AA and wedge BB. Since, weight F=mg=0F = mg = 0 under the free fall of gravity is always zero so,
By equilibrium of force we get,
F=FNF = {F_N}
Ncosθ=mg\Rightarrow N\cos \theta = mg
Since F=mg=0F = mg = 0 we get,
Ncosθ=0N\cos \theta = 0
cosθ0\therefore \cos \theta \ne 0
Therefore, the normal reaction force is N=0N = 0 between block A and wedge B.
Hence, the correct option is (C).

Note: It should be remembered that, the actual weight of block A of mass m remains mgmg in its own reference frame but when falling freely under gravity with the contact wedge B, the weight acting on wedge B of block A will be zero as, under free fall, force of gravity gives equal and opposite force in upward direction hence, wedge will read the weight of block as zero.