A 400 N box is being pushed across a level floor at a constant speed by a force P of 100 N at an angle of 30.0° to the horizontal, as shown in the the diagram below. What is the coefficient of kinetic friction between the box and the floor?
W = 400 N; P = 100 N; θ = 30o; μ = ? Frictional force (Fr) = μR (where R is the normal reaction) The forces acting along the horizontal direction are Fr and Px ∴ Pcos 30° - Fr = ma (Pcos 30° is acting in the +ve x-axis while Fr in the -ve x-axis) ⇒ 100cos 30° - μR = ma Since the box is moving at constant speed, its acceleration is zero ⇒ 100cos 30° - μR = 0 ⇒ 100cos 30o = μR ----- (i) The forces acting in the vertical direction are W, Py and R ∴ R - Psin 30° - W = 0 (R is acting upward (+ve) while Py and W are acting downward (-ve) and they are at equilibrium) ⇒ R - 100sin 30° - 400 = 0 ⇒ R = 100sin 30° + 400
⇒ R = 50 + 400 = 450 N From equation (i) ⇒ 100cos 30° = 450μ ⇒μ=100cos30° N = 100cos30°450
= μ = 0.19
