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?
Attached Figure / Diagram
A
0.19
B
0.24
C
0.40
D
0.22
Verified Correct Answer: A.
Physical Principles & Numerical Solution
W = 400 N; P = 100 N; \theta = 30o; \mu = ?
Frictional force (Fr) = \muR (where R is the normal reaction)
The forces acting along the horizontal direction are Fr and Px
∴ Pcos 30^\circ - Fr = ma (Pcos 30^\circ is acting in the +ve x-axis while Fr in the -ve x-axis)
⇒ 100cos 30^\circ - \muR = ma
Since the box is moving at constant speed, its acceleration is zero
⇒ 100cos 30^\circ - \muR = 0
⇒ 100cos 30o = \muR ----- (i)
The forces acting in the vertical direction are W, Py and R
∴ R - Psin 30^\circ - W = 0 (R is acting upward (+ve) while Py and W are acting downward (-ve) and they are at equilibrium)
⇒ R - 100sin 30^\circ - 400 = 0
⇒ R = 100sin 30^\circ + 400
⇒ R = 50 + 400 = 450 N
From equation (i)
⇒ 100cos 30^\circ = 450\mu
⇒\mu=100cos30^\circ
N = \(\frac{100cos30^\circ}{450}\)
= \mu = 0.19
Source: WAEC / JAMB / NECO Classroom Archive (Record #67893)