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 $\mu$, we start by determining the forces acting on the particle:

• Weight component down the slope: $W_{parallel} = mg \sin(30°) = 0.8 \times 9.8 \times \frac{1}{2} = 3.92 N$
• Normal force: $R = mg \cos(30°) = 0.8 \times 9.8 \times \frac{\sqrt{3}}{2} \approx 6.79 N$

Using Newton's second law, we have:

ightarrow W_{parallel} - F_{friction} = ma$$ dwhere $F_{friction} = \mu R = \mu \times 6.79$ This results in: $$3.92 - \mu \times 6.79 = 0.8 \times 0.6$$ Solving for $\mu$: $$3.92 - 0.48 = \mu \times 6.79$$ => $$3.44 = 6.79 \mu$$ => $$\mu = \frac{3.44}{6.79} \approx 0.507$$ Thus, the coefficient of friction $\mu \approx 0.51$.

For equilibrium, the forces acting on the particle gives us:

$R \cos(30°) - \mu R \cos(60°) - 0.8g = 0$

From the previous step, we can calculate $R$ as:

$R = mg \cos(30°) = 0.8 \times 9.8 \times \frac{\sqrt{3}}{2} \approx 12.8 N$

Substituting $R$ back into the equation, we can now solve for $X$:

Using the equilibrium condition:

$X = R \sin(30°) + 0.8g \sin(60°)$ => Plugging values of $R$, we find:

$X = 12.8 \times \frac{1}{2} + 0.8 \times 9.8 \times \frac{\sqrt{3}}{2}$ => Solving gives:

$X \approx 12$