A lifeboat slides down a straight ramp inclined at an angle of 15° to the horizontal - Edexcel - A-Level Maths Mechanics - Question 4 - 2013 - Paper 1

The forces acting on the lifeboat include:

• The gravitational force down the incline: $F_g = mg \sin(15°)$
• The normal force acting perpendicular to the ramp: $R = mg \cos(15°)$
• The frictional force opposing the motion: $f = \mu R$

Where:

• $m = 800 \, ext{kg}$
• $g = 9.81 \, ext{m/s}^2$

According to Newton's second law, we set up the equation:

$800g \sin(15°) - f = 800a$

Substituting the expression for the frictional force gives:

$800g \sin(15°) - \mu(800g \cos(15°)) = 800(1.5876)$

This can be rearranged to solve for the coefficient of friction, $\, \mu$.

Continuing with the previous equation:

$800g \sin(15°) - 800 \mu g \cos(15°) = 800 \times 1.5876$

Factoring out $800g$:

$g \sin(15°) - \mu g \cos(15°) = 1.5876$

Rearranging gives:

$\mu = \frac{g \sin(15°) - 1.5876}{g \cos(15°)}$

Substituting $g = 9.81 \, ext{m/s}^2$ will yield the value of $\mu$.