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 diagr...

Answer Details

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 $\frac{\mathrm{<br>}}{}$

        = μ = 0.19