A ramp, AB, of length 8 m and mass 20 kg, rests in equilibrium with the end A on rough horizontal ground - Edexcel - A-Level Maths: Mechanics - Question 4 - 2019 - Paper 1

To determine the resultant force acting on the ramp at point A, we establish the equilibrium conditions. We analyze the forces and the torques around point C. The following equations can be derived:

1. Vertical force balance:

   $N + R_y - 20g = 0$ where N is the normal force and $R_y$ is the vertical component of the reaction at A.

2. Horizontal force balance:

   $R_x - 0 = 0$ (since there are no horizontal forces acting)

3. Taking moments about point C:

   $20g imes 3 - N imes 5 = 0$

   From this equation, we can isolate N and find its value. After substituting in the known values:

   $N imes 5 = 20g imes 3$ $N = \frac{20g \times 3}{5} = 12g$

   Given that g = 9.81 m/s², we find:

   $N = 12 \cdot 9.81 = 117.72 N$.

   The resultant force at A can then typically be crunched down into a singular vertical force component alongside any horizontal forces, leading to the eventual magnitude.

If the center of mass of the ramp is closer to A than to B, it means that the weight of the ramp would create a greater torque about point C relative to the coupling between the ramp and drum. This will result in an increased normal reaction force at C compared to when the center of mass is centrally located, as the ramp will tilt more towards the drum, thereby increasing the required reaction support to maintain equilibrium.