Question

1 derive the relation between angle of friction and angle of repose

Answer 1

Here is your answer ......

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.

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