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:

• Gravitational force down the ramp: $F_g = mg \sin(\theta)$
• Normal force perpendicular to the ramp: $R = mg \cos(\theta)$
• Frictional force: $F_f = \mu R$

Where:

• $m = 800 \, \text{kg}$ (mass of the lifeboat)
• $g = 9.81 \, \text{m/s}^2$ (acceleration due to gravity)
• $\theta = 15°$ (angle of incline)

1. Calculate the normal force:

   $R = 800 \times 9.81 \times \cos(15°) = 800 \times 9.81 \times 0.9659 = 7580.77 \, \text{N}$

2. Now, express the net force equation:

   $800 \times 9.81 \sin(15°) - F_f = 800 \times a$

   This becomes:

   $800 \times 9.81 \sin(15°) - \mu \times 7580.77 = 800 \times 1.5876$

By substituting the known values:

$800 \times 9.81 \sin(15°) - \mu \times 7580.77 = 800 \times 1.5876$

To find $\mu$, rearranging terms will give us:

$\mu \times 7580.77 = 800 \times 9.81 \sin(15°) - 800 \times 1.5876$

Calculate:

$\mu = \frac{800 \times (9.81 \sin(15°) - 1.5876)}{7580.77}$

Substituting the values of $heta$:

$\mu \approx 0.10$.