Question

prove that the angle of friction=angle of repose.

Answer 1

Hi friend,

Good question !

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 are involved are :-
1. weight,mg of 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----------eg.1.
R = mgcos theta--------- e.g. 2.

now dividing e.g 1 by e.g 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

Thus,

Theta = alpha
i.e. the angle of friction = the angle of repose...i hope you understand ( proved )

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Mark me as a brainlist.

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