A particle of mass 0.8 kg is held at rest on a rough plane - Edexcel - A-Level Maths Mechanics - Question 5 - 2010 - Paper 1

To find the coefficient of friction (μ), we first calculate the forces acting on the particle:

1. The weight component down the slope: $W_{ ext{down}} = mg \sin(30°) = 0.8 \cdot 9.8 \cdot 0.5 = 3.92 \, \text{N}$
2. The net force downwards is given by:

$F_{ ext{net}} = mg \sin(30°) - F_f$ Where (F_f = \mu R)

3. The normal force: $R = mg \cos(30°) = 0.8 \cdot 9.8 \cdot \frac{\sqrt{3}}{2} \approx 6.79 \, \text{N}$

4. Newton's second law gives us:

$F_{ ext{net}} = ma\n3.92 - \mu \cdot 6.79 = 0.8 \cdot 0.6\n3.92 - \mu \cdot 6.79 = 0.48\n\mu \cdot 6.79 = 3.92 - 0.48 = 3.44\n\mu = \frac{3.44}{6.79} \approx 0.51$

In this scenario, the forces acting on the particle in equilibrium can be described as:

1. The normal force is equal to the weight component:

$R \cos(30°) = \mu R \cos(60°) + 0.8$

2. Solve for R, we have:

$R = X \sin(30°) + 0.8 \sin(60°)\nR = \frac{X}{2} + 0.8 \cdot \frac{\sqrt{3}}{2} \approx 12.8$

3. Substituting this back gives:

$X = 12 + \mu \cdot 0.8 \cdot \sin(60°)\nX = 12$