42. The magnitude of the acceleration of the cyclist as it moves along the horizontal circular path is given
by v2/R,wherevis the speed of the cyclist and Ris the radius of the curve.
(a) The horizontal component of Newton’s second law is f=mv2/R,wherefis the static friction
exerted horizontally by the ground on the tires. Thus,
f=(85.0)(9.00)2
25.0= 275 N .
(b) If Nis the vertical force of the ground on the bicycle and mis the mass of the bicycle and rider,
the vertical component of Newton’s second law leads to N=mg = 833 N. The magnitude of the
force exerted by the ground on the bicycle is therefore
f2+N2=2752+ 8332= 877 N .