Question

define the angle of friction and angle of repose and show that both are numerically equal

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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Answer 2

Angle of friction:

physics the angle of a plane to the horizontal when a body placed on the plane will just start to slide. The tangent of the angle of friction is the coefficient of static friction.

Angle of response:

DescriptionThe angle of repose, or critical angle of repose, of a granular material is the steepest angle of descent or dip relative to the horizontal plane to which a material can be piled without slumping. At this angle, the material on the slope face is on the verge of sliding. The angle of repose can range from 0° to 90°.

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