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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

User Cesans
by
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1 Answer

3 votes

Answer:

a) F = 15000 N

Step-by-step 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 = (mv^2)/(R)

so here we know that

m = 610 kg

v = 68 m/s

R = 190 m

now we have


F = (610 (68^2))/(190)


F = 14845.5 N

this is nearly


F = 15000 N

User Thomas Wagenaar
by
6.2k points