A particle P of mass 0.5 kg is on a rough plane inclined at an angle α to the horizontal, where tan α = rac{2}{3} - Edexcel - A-Level Maths Mechanics - Question 4 - 2006 - Paper 1

To find the coefficient of friction (μ), we first need to resolve the forces acting on particle P. The gravitational force acting down the slope (F_g) and the frictional force (F_f) can be described as follows:

1. Calculate the components of weight:

   • The weight (W) of P: $W = mg = 0.5 imes 9.81 = 4.905 \, \text{N}$
   • The component of weight down the slope: $W_{\text{down}} = W \sin \alpha = 4.905 \sin(\tan^{-1}(\frac{2}{3}))$
   • Using the identity to find sin and cos: $\sin \alpha = \frac{2}{\sqrt{13}}, \cos \alpha = \frac{3}{\sqrt{13}}$ Thus,
     $W_{\text{down}} = 4.905 \times \frac{2}{\sqrt{13}} \approx 2.713\, \text{N}$

2. The normal force (R) is the additional vertical component of weight:

   • $R = W \cos \alpha = 4.905 \times \frac{3}{\sqrt{13}} \approx 4.063\, \text{N}$

3. Now, applying equilibrium in the direction up the slope:

   • The equation for forces gives us: $4 \, \text{N} - F_f - W_{\text{down}} = 0$
   • Hence, the frictional force (F_f) is given as $F_f = 4 - 2.713 = 1.287 \, \text{N}$

4. The relationship for frictional force is given by:

   • $F_f = \mu R$
   • Plugging in the values: $\mu = \frac{F_f}{R} = \frac{1.287}{4.063} \approx 0.317$