AL MECH

(a) Let R be the normal reaction and Ff be the frictional force.
Resolve forces in the vertical direction:
R.cos(10) = Ff.sin(10) + 80g ---------- (1)
where g is the acceleration due to gravity

Take moment about the centre of gravity of the cyclist
R.L.sin(20) = Ff.L.cos(20)
where L is the distance from the centre of gravity to the road surface
hence, Ff = R.tan(20) ---------------- (2)
substitute into (1):
R.cos(10) = R.tan(20).sin(10) + 80g ------------ (3)
solve for R
then solve for Ff using (2)

(b) Resolve forces in the horizontal direction:
R.sin(10) + Ff.cos(10) = 80v^2/10 ------------------- (4)
where v is the speed of the cyclist
solve for v using results of R and Ff from part (a)

(c) Let B be the angle of [beta]
thus, equation (2) becomes, 0.2R = R.tan(B)
i.e. tan(B) = 0.2
B = arc-tan(0.2)

Using equation (3),
R.cos(10) = R.tan(B).sin(10) + 80g
solve for R
then using (4),
R.sin(10) + 0.2R.cos(10) = 80v^2/10
solve for v