Answer:
(a) [tex]F_f=173.05N[/tex]
(b) [tex]F=793.30N[/tex]
Explanation:
According to the free body diagram of the bicyclist, we have:
[tex]\sum F_x:F_f=F_c\\\sum F_y:N=mg[/tex]
Here [tex]F_c[/tex] is the centripetal force, which is defined as:
[tex]F_c=ma_c=m\frac{v^2}{r}[/tex]
(a) Replacing this in the sum of the force in the x-axis, we calculate the force of friction on the bicycle from the road:
[tex]F_f=m\frac{v^2}{r}\\F_f=79kg\frac{(7.19\frac{m}{s})^2}{23.6m}\\F_f=173.05N[/tex]
(b) The forces on the bycicle from the road are the friction force and the normal force. This forces are perpendicular. So, the net force is:
[tex]F=\sqrt{F_f^2+N^2}\\F=\sqrt{F_f^2+(mg)^2}\\F=\sqrt{(173.05N)^2+(79kg*9.8\frac{m}{s^2})^2}\\F=793.30N[/tex]
