First slide

Friction on inclined plane of angle more than angle of repose

Question

A body of mass m is placed on a rough surface with coefficient of friction $μ$ inclined at $θ$ . If the mass is in equilibrium, then

Moderate

A

$θ = \tan^{- 1} μ$
B

$θ = \tan^{- 1} (\frac{1}{μ})$
C

$θ = \tan^{- 1} \frac{m}{μ}$
D

$θ = \tan^{- 1} \frac{μ}{m}$

Solution

Consider a body of mass m placed on a rough inclined surface and it is just on the point of sliding down, with coefficient of friction $μ$ inclined at angle $θ$ , as shown in figure.
At the equilibrium point O,
In case of limiting condition, F = mg sin $θ$ …….(i)
Normal force, R = mg cos $θ$ …….(ii)
On dividing Eq. (i) by Eq. (ii), we get
$\frac{F}{R} = \frac{m g sinθ}{m g cosθ}$
$μ = tanθ$ ( $∵$ Force of friction, F = $μ$ R)
$\Rightarrow θ = \tan^{- 1} (μ)$
This is maximum value of $θ$ for mass m to be at rest.
For smaller $θ$ , body will be at rest, i.e. in equilibrium.
So, angle of repose, i.e. $θ = \tan^{- 1} μ$

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