WAEC Physics 2012 Theory — Question 1

Model answer

On the inclined plane, the forces acting on the body are: the weight mg (acting vertically downward), the normal reaction R (perpendicular to the plane), and the frictional force Fp (acting along the plane, opposing motion). Resolving mg along and perpendicular to the plane: the component along the plane = mg sinθ, and the component perpendicular to the plane = mg cosθ. For the body on the verge of sliding (limiting friction), the frictional force Fp = mg sinθ, and the normal reaction R = mg cosθ. The coefficient of friction μ = Fp/R = (mg sinθ)/(mg cosθ) = sinθ/cosθ = tanθ. Hence, μ = tanθ.