Question

calculate the angle of repose of an inclined plane when coefficient of friction is 0.6​

Answer 1

I assume that you have an object of mass m on an inclined plane of angle θ.

The force straight down is F = mg

The force parallel to the plane is Fp = mgSinθ and the force normal to the plane Fn = mgCosθ

The coefficient of friction is defined as μ = Fp/Fn

So μ =mgSinθ/mgCosθ = Tanθ

Now the question states that μ = 0.6, so Tanθ = 0.6

Then θ = Tan^-1(0.6) = 30.96°

So the plane is inclined at 30.96°



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Answer 2

Answer:

30.96° .

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