Question

what horizontal acceleration should be provided to the wedge so that the block of mass m placed on the wedge falls freely ?​

Answer 1

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for free fall, normal reaction should be zero.

$n = 0$

for equilibrium perpendicular to the wedge,

$0 + ma \sin \alpha = mgcos \alpha$

$= > a = \frac{g}{tan \alpha }$

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Answer 2

Answer:

Explanation:

for free fall, normal reaction should be zero.

n = 0

for equilibrium perpendicular to the wedge,

0 + ma \sin \alpha  = mgcos \alpha  

=  > a =  \frac{g}{tan \alpha }