Question

Show that angle of friction is numerically equal to angle of repose

Answer 1

prove that angle of friction is equal to angle of repose in the limiting friction. ... The component W cos f balances the normal reaction R while the component W sin f is equal to the limiting friction flimiting. We notice that the angle of repose or sliding and the angle of friction are both equal.

Answer 2

The angle of friction is numerically equal to the angle of repose is sown.

We are required to show that the angle of friction is numerically equal to the angle of repose.

• The angle between the resultant frictional force and the normal reaction made with the normal force(vertical force) is called the angle of friction. It is denoted by the symbol α. (tan α = μ).

• The angle between the surface of the inclined plane and the horizontal plane is called the angle of repose. It is denoted by the symbol (θr).

From the picture given below

By equating horizontal and vertical components we get

mgcosθr = N    ------(1)      

mgsinθr = f  

mgsinθr = µN   -------(2)         [∵ f = µN]

Divide equation(2) by equation(1)

sinθr/cosθr = µN/N

tanθr = µ --------(3)         [∵ sinθ/cosθ =  tanθ ]

We know that

tanα = µ, Substitute in equation(3)

tanθr = tanα

⇒  θr =α

Hence shown.

Therefore, The angle of friction is numerically equal to the angle of repose is sown.

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