Question

Derive relation for coefficient of friction in terms of tangent of angle of repose ?

Answer 1

Assume a body is placed on a horizontal surface and a force F is applied. The frictional force f eliminates the applied force and the body is on the brink of movement.

The angle made by frictional force as a consequence of general force and with normal is named as the angle of friction.

Let us consider, N = W  

f = µW = F (As the body is on the brink of movement)  

Hence,  

tanθ = f/N = µW/W = µ

= > θ = tan µ  

θ = angle of friction

Answer 2

Suppose angle of friction = alpha and angle of repose = theta

Let us suppose a body is placed on an inclined plane as in the above figure.

Various forces involved are:-

1. weight,mg of the body,acting vertically downwards.

2.normal reaction,R,acting perpendicular to inclined plane.

3.Force of friction,F, acting up the plane.

Now, mg can be resolved in two components:-

mgcos theta opposite to R

and mgsin theta opposite to F

In equilibrium,

F = mgsin theta -------- eq.1

R = mgcos theta ---------eq.2

On dividing eq.1 by eq.2 we get,

F/R = mgsin theta/mgcos theta

mu = tan theta