A highway curves to the left with radius of curvature of 42 m and is banked at 20 ◦ so that cars can take this curve at higher speeds. Consider a car of mass 1869 kg whose tires have a static friction coefficient 0.51 against the pavement.

Question

Physics

Grady Wilkinson

31 May, 02:34

A highway curves to the left with radius of curvature of 42 m and is banked at 20 ◦ so that cars can take this curve at higher speeds. Consider a car of mass 1869 kg whose tires have a static friction coefficient 0.51 against the pavement.

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Answers (1)

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Chico · Answer 1

The maximum speed the car can attain is 14 m/s

Explanation:

Assuming that the velocity is to be calculated.

Here, the centripetal force of car must equal the frictional force, in order to take the turn without skidding.

Therefore,

Centripetal Force = Frictional force

mv²/r = f

where,

f = frictional force

m = mass of car = 1896 kg

r = radius of curvature = 42 m

v = speed of car

Now,

f = μR

where,

μ = coefficient of friction = 0.51

R = Normal Reaction = Normal component of the weight of the car

R = W Cos θ = mg Cos θ

Therefore,

f = μmg Cos θ

Therefore,

mv²/r = μmg Cos θ

v² = rμg Cos θ

v = √rμg Cos θ

v = √ (42 m) (0.51) (9.8 m/s²) Cos (20°)

v = 14 m/s