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

From Equation (2), the normal force is resolved as follows:

$R = 10 \sin \alpha + 5g \cos \alpha \; (3)$ Using $\alpha$ given by $\tan \alpha = \frac{3}{4}$: Calculate $\sin \alpha$ and $\cos \alpha$:

• $\sin \alpha = \frac{3}{5}$
• $\cos \alpha = \frac{4}{5}$

Substituting back: $R = 10 \cdot \frac{3}{5} + 5 \cdot 9.81 \cdot \frac{4}{5} = 6 + 39.24 = 45.24 \text{N}$

Using the formula for friction: $F_{friction} = \mu R \; (4)$ Substituting for $F_{friction}$ from Equation (1): $5g \sin \alpha - 10\cos \alpha = \mu R \; (5)$

Using values obtained for $R$: Substitute from previous work:

• For $F_{friction} = 5g \sin \alpha - 10 \cos \alpha$ and knowing $g \approx 9.81$: $\mu = \frac{5g \sin \alpha - 10 \cos \alpha}{R}$ Further calculations lead to: $\mu = \frac{5 \cdot 9.81 \cdot \frac{3}{5} - 10 \cdot \frac{4}{5}}{2 \sin \alpha + g \cos \alpha}$ After simplifications, this results in $\mu \approx 0.473$.