# a particle slides down a smooth inclined plane of elevation Ө, fixed in an elevator going up with an acceleration a. The base of the inclined plane has length L. Find the time taken by the particle to reach the bottom.

the general figure of the problem

now, due to the movement of the elevator in the upward direction a pseudo force is generated in the inside frame, let it be f0 = ma0

So, we can now draw a free body diagram which shows the forces on the mass m as it moves with elevator and slides down. Thus, the following figure forms

now, for forces parallel to the inclined plane we will develop the following the equation of motion. Here we shall concern ourselves with the sine components. So,

mg.Sinθ + ma0Sinθ = ma

or acceleration of the particle wrt elevator will be

a = (g+a0) Sinθ

now,

cosθ = L/x

where x is the total distance moved by particle, so

x = L/cosθ

now, form

s = ut + (1/2)at2

and as u = 0

s = (1/2)at2

so, we get

L/cosθ = (1/2)[(g+a0) Sinθ]t2

or

t2 = [2L/cosθ] / [(g+a0) Sinθ]

or

time taken will be

t = [ 2L / [(g+a0) Sinθ.Cosθ] ]1/2

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