Angle of Repose or Angle of Sliding

The angle of repose is defined as the angle of the inclined plane at which a body placed on it just begins to slide.

Consider an inclined plane, whose inclination with horizontal is gradually increasing, till the body placed on its surface just begins to slide down. If θ is the inclination, at which the body just begins to slide down, then θ is called angle of repose.

The body is under the action of the following forces :

(i) The weight Mg of the body acting vertically downwards.

(ii) The limiting friction F in upward direction along the inclined plane, which in magnitude is equal to the component of weight Mg acting along the inclined plane i.e.

$F = Mg \sin \theta$ … (2)

(iii) The normal reaction R acting at right angle to the inclined plane in upward direction, which is equal to the com-ponent of weight acting perpendicular to the inclined plane i.e.

$R = Mg \cos \theta$ … (3)

Dividing equation (2) by (3), we have

$\dfrac{F}{R} = \dfrac{Mg \sin \theta}{Mg \cos \theta} = \tan \theta$

Since, $\dfrac{F}{R} = \mu$

$\tan \theta = \mu$ … (4)

Therefore, coefficient of limiting friction is equal to the tangent of the angle of repose.

From equations (1) and (4), it follows that

θ = α

i.e. angle of repose is equal to angle of friction.

Example 2: A body placed on a rough inclined plane just begins to slide, when the slope of the plane is 1 in 4. Calculate the coefficient of friction.

Solution:

Here sin θ = 1/4

From figure, it follows that

$\tan \theta = \frac{1}{\sqrt{15}}$

From relation between coefficient of friction and angle of repose, we have

$\mu = \tan \theta = \frac{1}{\sqrt{15}} = 0.258$

Example 3: A body starts rolling down an inclined plane, the top half of which is perfectly smooth and the lower half is rough. Find the ratio of force of friction and the weight of the body, if the body is brought to rest just when it reaches the bottom. Given that inclination of plane with the horizontal is 30°.

Solution:

Consider that a body of weight Mg is placed at the top of the plane at point A. The length of the inclined plane is 2l so that upper half AB is perfectly smooth and the lower half is rough. As the body comes to rest at C, it follows that work done against friction along lower half (rough) of plane
= potential energy of body at A

If F is limiting friction, then work done against friction along the lower half BC = F l

Potential energy of body at A = M g (OA)

$ \displaystyle Mg\left( {2l\sin {{{30}}^{\circ }}} \right)$

$ \displaystyle \therefore Fl=Mgl$

or $\dfrac{F}{Mg} = 1$