A particle P of mass 5kg is held at rest in equilibrium on a rough inclined plane by a horizontal force of magnitude 10N - Edexcel - A-Level Maths Mechanics - Question 4 - 2017 - Paper 1

Resolving forces perpendicular to the plane, we have: $R = 10 \sin \alpha + 5 g \cos \alpha$ Substituting $g = 9.81 m/s^2$, we get: $R = 10 \sin(\alpha) + 5 \times 9.81 \cos(\alpha).$

For forces acting parallel to the plane, we have: $F = 5 g \sin \alpha - R \cos \alpha$ Substituting $g = 9.81 m/s^2$, we can express the equation as: $10 = 5 \times 9.81 \sin(\alpha) - R \cos(\alpha).$

Using the above equations, we substitute for $R$ and $F$: $\mu = \frac{g \sin \alpha - 2 \cos \alpha}{2 \sin \alpha + g \cos \alpha}.$

Given $\tan \alpha = \frac{3}{4}$, find $\sin \alpha$ and $\cos \alpha$: $\mu = \frac{g \sin(\alpha) - 2 \cos(\alpha)}{2 \sin(\alpha) + g \cos(\alpha)}$ Substituting values gives: $\mu = 0.47 \text{ or } 0.473.$