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the magnitudes of (a) the force of friction on the bicycle from the

road and (b) the net force on the bicycle from the road.

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Solution

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The horizontal component of Newton’s second law is $$f-s=\dfrac{mv^2}{R}$$where fs is the static

friction exerted horizontally by the ground on the tires. Similarly, if $$F_N$$ is the vertical the 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 $$F_N=mg=833N$$.

(a) The frictional force is

$$f_s=\dfrac{mv^2}{R}=\dfrac{(85.0 kg) (9.00 m/s)^2}{25.0 m}=275 N$$

(b) Since the frictional force $$\vec f_s$$ and $$\vec F_N$$, the normal force exerted by the road, is

perpendicular to each other, the magnitude of the force exerted by the ground on the

bicycle is therefore

$$F=\sqrt{{f_s}^2+{F_N}^2}=\sqrt{(275 N)^2 +(833 N)^2}=877 N$$

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