Question

A body weighing 20 kg just slides down a rough inclined plane that rises 5 in 12, what is the coefficient of friction?

Answer 1

As the plane rises 5 in 12, therefore

sinθ = 5/12

∴ cosθ = √{1 - sin²θ}

= √{1 - (5/12)²} = {√119}/{12}

Now, coefficient of friction,

μ = tanθ = sinθ/cosθ = {5}/{12} x {12}/{√119}

= {5}/{√119} = 0.46