Question

Wings on race cars push them into the track. The increased normal force makes large friction forces possible. At one Formula One racetrack, cars turn around a half-circle with diameter 190m at 68m/s .
Part A
For a 610 kg vehicle, the approximate minimum static friction force to complete this turn is. Select best answer
a.15000N
b.6000N
c.30000N
d.18000N
e.24000N

Answer 1

Answer:

a) F = 15000 N

Explanation:

At flat race track the friction force between the tyre and the road will provide the required frictional force in the form of centripetal force

So we will have

$F_f = \frac{mv^2}{R}$

so here we know that

m = 610 kg

v = 68 m/s

R = 190 m

now we have

$F = \frac{610 (68^2)}{190}$

$F = 14845.5 N$

this is nearly

$F = 15000 N$