A track consists of two circular parts ABC and CDE of equal radius...

Since the cycle moves with a constant speed, its tangential acceleration is zero ($a_t = 0$).

The net force along the tangential direction must be zero at all points on the track. The forces acting along the tangent are the tangential component of gravity and the force of friction. To maintain a constant speed, the friction force must exactly balance the tangential component of gravity.

The magnitude of the friction force is given by $f = mg \sin\theta$, where $\theta$ is the angle made by the tangent to the track with the horizontal.

At point B (the highest point of the track) and point D (the lowest point of the track), the tangent is horizontal. Thus, \theta_B = 0^{}^{{}^{\circ}} and \theta_D = 0^{}^{{}^{\circ}}.

f_B = mg \sin 0^{}^{{}^{\circ}} = 0

f_D = mg \sin 0^{}^{{}^{\circ}} = 0

At point C, the track joins smoothly. Since the circular part ABC subtends a right angle (90^{}^{{}^{\circ}}) at its centre and B is the highest point (midpoint of the arc), the angle subtended by arc BC at the centre is 45^{}^{{}^{\circ}}. Therefore, the tangent at C makes an angle of 45^{}^{{}^{\circ}} with the horizontal.

\theta_C = 45^{}^{{}^{\circ}}

f_C = mg \sin 45^{}^{{}^{\circ}} = 100 \times 10 \times \dfrac{1}{\sqrt{2}} = 500\sqrt{2} \approx 707 \text{ N}