Arun
Last Activity: 6 Years ago
Dear Mahesh
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
Where mu = coefficient of limiting friction.
This is equal to the tangent of the angle of repose between two surfaces in contact.
Now, since mu = tan alpha i.e coefficient of limiting friction between any two surfaces in contact is equal to tangent of the angle of friction between them.
Thus, Theta = alpha i.e angle of friction is equal to angle of repose.