A particle P of mass 2 kg is held at rest in equilibrium on a rough plane by a constant force of magnitude 40 N - Edexcel - A-Level Maths Mechanics - Question 5 - 2016 - Paper 1

Next, we resolve the forces parallel to the plane. The weight component acting down the plane is:

$W = 2g \sin(20°)$

The sum of forces is given as: $F - 2g \sin(20°) - R\mu = 0$ Substituting $F = 40$ N and isolating $,";µ",

By substituting $R$ from the previous step into the above equation, we can isolate and solve for $\,µ$:

$\mu = \frac{40 - 2g \sin(20°)}{R}$

From the earlier computations of $R$, we can calculate: $\mu = \frac{40 - 2 \times 9.81 \times \sin(20°)}{R}$

Calculating the values, we derive:

1. Substitute the values for $R$ using sine and cosine values.
2. Finally compute $\,µ$ to find that $\,µ = 0.727$ or $\,µ = 0.73$.