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 determine the coefficient of friction (µ), we first analyze the forces acting on the particle.
The force due to gravity along the incline is given by:
$F_g = mg \sin(30°) = 0.8 \times 9.8 \times 0.5 = 3.92 \text{ N}$
The normal reaction force (R) acting perpendicular to the surface is:
$R = mg \cos(30°) = 0.8 \times 9.8 \times \frac{\sqrt{3}}{2} \approx 6.79 \text{ N}$
Then using the formula for frictional force, we have:
$F_f = \mu R$
Applying Newton's second law along the incline:
$mg \sin(30°) - \mu mg \cos(30°) = ma$
Substituting in the known values:
$3.92 - \mu (0.8 \times 9.8 \times \frac{\sqrt{3}}{2}) = 0.8 \times 0.6$
Solving this for µ, we find:
$\mu = 0.5$

In order to find the value of X, we analyze the forces when the particle is held in equilibrium.
The forces acting parallel and perpendicular to the incline can be resolved as follows:
$R \cos(30°) = \mu R \cos(60°) + 0.8$
From the normal reaction force:
$R = X \sin(30°) + 0.8 \times g \sin(30°)$
Substituting the known values:
$R = X \cdot 0.5 + 0.8 \cdot 9.8 \cdot 0.5$
Solving gives:
$X = 12$