A parcel of mass 5 kg lies on a rough plane inclined at an angle $\alpha$ to the horizontal, where $\tan \alpha = \frac{1}{4}$ - Edexcel - A-Level Maths Mechanics - Question 4 - 2003 - Paper 1

Using the horizontal force equilibrium condition:

$F + 20 \cos \alpha = 5g \sin \alpha$

Substituting $\cos \alpha = \frac{4}{\sqrt{17}}$ and $\sin \alpha = \frac{1}{\sqrt{17}}$, we rewrite the equation as:

$F + 20 \left(\frac{4}{\sqrt{17}}\right) = 5(9.81) \left(\frac{1}{\sqrt{17}}\right)$

Rearranging and solving for $F$ gives us:

$$ F = 5(9.81) \left(\frac{1}{\sqrt{17}}\right) - 20 \left(\frac{4}{\sqrt{17}}\right) = 51.2 , \text{N} - 13.4 , \text{N} = 37.8 , \text{N} $

Using the equation of friction:

$F = \mu R$

We can solve for the coefficient of friction $\mu$:

$\mu = \frac{F}{R}$

After substituting the values of $F$ and $R$, we find:

$\mu = \frac{37.8 \, \text{N}}{R}$

Based on the previously calculated value of $R$, we find that $\mu = 0.262$, which can be accepted as an approximate value.