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 A, we begin by resolving the forces acting on the ramp. Using free-body diagram analysis, we can derive the following equations:

1. Vertical force balance:

   $N - 20g = 0$ where $N$ is the normal force at A and $g$ is the acceleration due to gravity (approximately 9.81 m/s²). Thus, we find that: $N = 20g$ = 20 × 9.81 = 196.2 ext{ N}.

2. Applying moments about point C, we establish:

   $R imes 5 - 20g imes 4 = 0$ (assuming $R$ is the reaction from the drum at C). Rearranging, we find: R = rac{20g imes 4}{5} = rac{80g}{5} = 16g = 16 × 9.81 = 156.96 ext{ N}.

3. The resultant force, $F$, acting on the ramp at A can be determined using Pythagorean Theorem between the normal force and the reaction from the drum:

   F = rac{ ext{Normal force}^{2} + R^{2}}{A} = rac{(20g)^{2} + (16g)^{2}}{A}. On solving, the resultant force at point A would be approximately: $F ext{ (final value)} ext{ N}$.

If the center of mass of the ramp is closer to A than to B, this means that the ramp will tend to tip towards B due to the weighted balance. Consequently, the normal reaction at point C will decrease because the distribution of weight will shift more towards point A, leading to a smaller load on the drum at C. Thus, the magnitude of the normal reaction between the ramp and the drum at C will be less than when the ramp is uniformly distributed.