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ln the figure shown 'P' is a plate on which a wedge B is placed and on B a block A of mass m is placed. The plate is suddenly removed and system of B and A is allowed to fall under gravity. Neglecting any force due to air on A and B, the normal force on A due to B is

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$\frac{mg}{\cos θ}$

$mg \cos θ$

zero

$\frac{2 mg}{\cos θ}$

answer is .

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Detailed Solution

F.B.D. of block A

Applying Newton's second law for block A in vertical direction

$m_{A} g - N \cos θ = 0$

where $θ$ is the angle of the wedge

$\Rightarrow N \cos θ = 0$

as $θ < 90^{0} ∴ N = 0$

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