A block of mass m rests on a rough inclined plane. The coefficient of friction between the surface and the block is μ. At what angle of inclination θ of the plane to the horizontal will the block just start to slide down the plane?

Question

A block of mass m rests on a rough inclined plane. The coefficient of friction between the surface and the block is μ. At what angle of inclination θ of the plane to the horizontal will the block just start to slide down the plane? (A) θ =tan-1 μ (B) θ =cos-1 μ (C) θ =sin-1 μ (D) θ =sec-1 μ

Tardigrade Admin · Accepted Answer

The various forces acting on the block are as shown in the figure.
From figure

m g

s in θ = f

\dots (i)

m g

cos θ = N

\dots (ii) < b r / >

Divide (i) by (ii), we get

t an θ

= \frac{f}{N}

= \frac{μ N}{N}

or

θ

= t a n^{- 1}

(μ)

Q. A block of mass $m$ rests on a rough inclined plane. The coefficient of friction between the surface and the block is $μ$ . At what angle of inclination $θ$ of the plane to the horizontal will the block just start to slide down the plane?

$θ$ $= t a n^{- 1}$ $μ$

$θ$ $= co s^{- 1}$ $μ$

$θ$ $= s i n^{- 1}$ $μ$

$θ$ $= se c^{- 1}$ $μ$

The various forces acting on the block are as shown in the figure.
From figure
$m g$ $s in θ = f$ $\dots (i)$
$m g$ $cos θ = N$ $\dots (ii) < b r / >$
Divide (i) by (ii), we get
$t an θ$ $= \frac{f}{N}$ $= \frac{μ N}{N}$ or $θ$ $= t a n^{- 1}$ $(μ)$

Q. A block of mass m rests on a rough inclined plane. The coefficient of friction between the surface and the block is μ. At what angle of inclination θ of the plane to the horizontal will the block just start to slide down the plane?

θ =tan−1 μ

θ =cos−1 μ

θ =sin−1 μ

θ =sec−1 μ