# Angle of Friction and Angle of Repose

**Angle of friction:**

The angle made by the resultant of limiting friction with the normal reaction is called angle of friction. In the figure below, θ is the angle of friction.

In triangle OAR,

$$Tan \theta = \frac{AR}{OR} = \frac{OF}{OR} = \frac{F}{R} $$

$$ \text{Here,} \frac{F}{R} = \mu \text{is the coefficient of friction.}$$

$$ \therefore Tan \theta = \mu$$

Hence, the coefficient of friction is equal to the tangent of angle of friction.

**Angle of Repose: **The angle made by the inclined plane with horizontal such that the body just begin to slide is called angle of repose. In figure below, α is the angle of repose.

Let us consider a body of mass ‘m’ placed over the surface of inclined plane OB. Here, mg is the weight of the body acting vertically downward. R is the normal reaction. mgsinθ and mgcosθ are the rectangular components of ‘mg’ as shown in figure. F is the limiting friction acting upward the plane. As the body just begin to slide, then from figure, we can write:

R = mgcosα……(i)

F = mgsinα……(ii)

Dividing equation (ii) by (i), we get:

$$\frac{F}{R} = \frac{mgsin \alpha }{mgcos \alpha }$$

$$or, \mu = tan \alpha [\because \mu = \frac{F}{R} ]$$

Here, α is the angle of repose. So, we can conclude that coefficient of friction is equal to the tangent of angle of repose.

Since tanθ = tanα

∴ θ = α

Hence it is proved that the angle of friction is equal to the angle of repose.

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## One Response to “Angle of Friction and Angle of Repose”

Why does the angle of inclination change when the body is sliding on a inclined plane?