Class 11 > Unit # 03: Dynamics > Relation between Angle of Friction & Repose

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Unit # 3: Dynamics — Angle of Friction & Angle of Repose

Lecture Overview

This lecture from Talha's Physics Academy explains the concepts of the angle of friction and the angle of repose, and proves mathematically that both angles are equal (Class 11 Physics, Unit 3: Dynamics).

Source Video: Relation between Angle of Friction and Angle of Repose - Unit 3 Dynamics - #TP11 32

Angle of Friction

Angle of friction is defined as the angle made by the resultant of frictional force and the normal reaction with the normal reaction as shown in figure.

In \(\Delta BCD\),

$$\tan \alpha = \frac{\text{Perpendicular}}{\text{Base}} = \frac{BC}{CD} = \frac{f}{R}$$

But the coefficient of friction is:

$$\mu = \frac{f}{R}$$

Therefore,

$$\tan \alpha = \mu$$

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

Angle of Repose

It is defined as the minimum angle made by an inclined plane with the horizontal such that an object placed on the inclined surface just begins to slide.

Relation Between Angle of Friction and Angle of Repose

Let us consider a body of mass 'm' resting on an inclined plane. Also, consider when the plane makes an angle '\(\theta\)' with the horizontal, the body just begins to move.

Let 'R' be the normal reaction of the body and 'f' be the frictional force.

Here, balancing the forces along and perpendicular to the inclined plane:

$$f = mg \sin\theta \quad \text{--- (i)}$$

$$R = mg \cos\theta \quad \text{--- (ii)}$$

Dividing equation (i) by equation (ii):

$$\frac{f}{R} = \frac{mg \sin\theta}{mg \cos\theta}$$

$$\frac{f}{R} = \tan\theta$$

Since \(\frac{f}{R} = \mu\):

$$\mu = \tan\theta \quad \text{--- (iv)}$$

From the angle of friction derivation, we also know that:

$$\mu = \tan\alpha \quad \text{--- (v)}$$

Comparing equation (iv) and equation (v):

$$\tan\theta = \tan\alpha$$

$$\theta = \alpha$$

Hence, the angle of repose is equal to the angle of friction.

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