91Ó°ÊÓ

• Radius of the circle: R = 25.0 m.
• Velocity of the bicycle: v = 9.00 m/s.
• Mass of bicycle-rider: m = 85.0 kg.

It is the force of friction that provides the centripetal force required for circular motion. The free-body diagram is shown below. The magnitude of the acceleration of the cyclist as it moves along the horizontal circular path is given by$\frac{{\mathbf{v}}^{\mathbf{2}}}{\mathbf{R}}$, where v is the speed of the cyclist and R is the radius of the curve.

Formula:

$f_s=\frac{m{v}^2}{R}$

The horizontal component of Newton’s second law is $f_s=\frac{m{v}^2}{R}$, where $f_s$is the static friction exerted horizontally by the ground on the tires.

Similarly, $F_N$if is the vertical force of the ground on the bicycle and m is the mass of the bicycle and rider, the vertical component of Newton’s second law leads to $F_N=mg=85×9.8=833\mathrm{N}$.

Force of friction on the bicycle from the road:

$f_s=\frac{m{v}^2}{R}$

Substitute the values in the above expression, and we get,

role="math" localid="1660976789922" $$\begin{gathered} f_s=\frac{85.0\mathrm{kg}×\left(9.00\mathrm{m}/{\mathrm{s}}^2\right)}{25.0\mathrm{m}} \\ {f}_s=275.4\mathrm{N} \\ ≈275 \end{gathered}$$

Thus, the force of friction on the bicycle from the road is 275 N.

Since the frictional force $f_s$and the normal force exerted by the road $F_N$are perpendicular to each other, the magnitude of the force exerted by the ground on the bicycle is,

$F=\sqrt{f_s^2+F_N^2}$

Substitute the values in the above expression, and we get,

$$\begin{gathered} F=\sqrt{{275}^2+{833}^2} \\ F=877\mathrm{N} \end{gathered}$$

The force exerted by the ground on the bicycle F is at an angle:

$$\begin{gathered} \mathrm{θ}={\tan }^{-1}\left(\frac{275\mathrm{N}}{833\mathrm{N}}\right) \\ =18.3° \end{gathered}$$

It will make an 18.3-degree angle with respect to the vertical axis.

Thus, the net force is 877 N.