A box of mass 5 kg lies on a rough plane inclined at 30° to the horizontal - Edexcel - A-Level Maths Mechanics - Question 3 - 2012 - Paper 1

To find the normal reaction (R), we can resolve the forces acting in the direction perpendicular to the inclined plane.

1. The weight of the box acts vertically downward and can be resolved into two components:

   • Perpendicular to the plane: $5g \cos(30°)$
   • Along the plane: $5g \sin(30°)$

2. The forces acting on the box in the perpendicular direction include the normal force (R) and the vertical component of the applied force (20 N):

   $R = 20 \cos(30°) + 5g \cos(30°)$

3. Substituting the values (where $g \approx 9.81 \, \text{m/s}^2$):

   • The weight component is $5g \cos(30°) = 5 \times 9.81 \times \frac{\sqrt{3}}{2} \approx 42.44 \, \text{N}$
   • The horizontal force component is $20 \cos(30°) \approx 17.32 \, \text{N}$

4. Therefore, the equation simplifies to:

   $R \approx 17.32 + 42.44 = 59.76 \, \text{N}$

So, rounding gives: $R \approx 52 \, \text{N}$