Hello. I'm a bit confused by part a. The answer says that it's because of the smoothness of the drum (no friction). I don't get what does friction have to do with normal reaction when they are in different plane? I googled it to understand and it said (If there is friction opposing motion, it counteracts a portion of the weight along the plane. As a result, the normal force must increase to counterbalance the remaining component of the weight acting perpendicular to the plane.). Could you please explain it to me. Thanks!

sin 8 = _E 13 casn iz AL '3 ronlh = \\ 2% Figure 2 A ramp, 4B, of length 8 m and mass 20 kg, rests in equilibrium with the end 4 on rough horizontal ground. The ramp rests on a smooth solid cylindrical dram which is partly under the ground. The drum is fixed with its axis at the same horizontal level as 4. The point of contact between the ramp and the drum is C, where AC = 5m, as shown in Figure 2. The distance from the point of contact to the end of the ramp is CB =3 m. The ramp is resting in a vertical plane which is perpendicular to the axis of the drum, at an angle & to the horizontal, where tan 6= 152- The ramp is modelled as a uniform rod. (a) Explain why the reaction from the drum on the ramp at point C acts in a direction which is perpendicular to the ramp. 1) (b) Find the magnitude of the resultant force acting on the ramp at 4. ) The ramp is still in equilibrium in the position shown in Figure 2 but the ramp is not now modelled as being uniform. o Given that the centre of mass of the ramp is assumed to be closer to 4 than to B, (c) state how this would affect the magnitude of the normal reaction between the ramp and the drum at C. 1) (Total for Question 4 is 11 marks)