Question

what is the relation between coefficient of friction and angle of repose

Answer 1

if u is coefficient of static friction and @ is the angle of repose then relation between is ------
@ = tan^-1 (u)

angle of repose :- it is the minimum angle that an inclined plane makes with the horizontal when a body pleced on it just begins to slide down .
let any body which mass is m placed in inclined which inclination with horizontal is @ .
then ,
you resolve the mg along inclination and perpendicular to inclination.
you see .
in equilibrium condition.

mg.sin@ = friction.
but we know,
friction =uN
where N is normal reaction.
also N =mg.cos@
then,
mg.sin@ = umg.cos@
tan@ = u
hence,
@ =tan^-1 (u)

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