Question

A block of mass m is kept on a inclined plain of a lift moving down with an acceleration of 2 metre per second

Answer 1

mg cos θ − ma cos θ=N

⇒N=m(g−a) cos θ

∴Ffri=μN=μ{m(g−a) cos θ}

⇒mg sin θ − ma sin θ

=μ{m(g−a) cos θ}

⇒sin θ=μcos θ

⇒μ=sin θ/cos θ=tan θ

⇒μ=tan θ

Answer 2

Correct Question is like this :-

A block of mass m is kept on an inclined plane of a lift moving down with acceleration of 2 ms-².What should be the coefficient of Friction to let the block move down with constant velocity relative to lift.

As acceleration is acting downward it means lift is going downward and block is inclined on Ø.

RSinØ = m(g-a)SinØ

fCosØ = µNCosØ

As also,

RSinØ = fCosØ

m(g-a)SinØ = µm(g-a)CosØ

Since,

N = m(g-a)

µ = tanØ

µ = tan30°

µ = 1/√3

For simple, We also know that

Angle of Response $(\alpha)$

µmg Cos$\alpha$ = mg Sin$\alpha$

tan$\alpha$ = µ

$\alpha$ = tan-¹(µ)