Acar of mass 1200 kg drives around a curve with a radius of 25.0 m. if the driver maintains a speed of 20.0 km/hr, what is the force of friction between the tires and the road? what is the minimum coefficient of static friction required to keep the car in this turn?

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explanation:

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F=1487.52 N

μ = 0.12

Explanation:

Given that

m= 1200 kg

r= 25 m

v= 20 km/hr = 20 x 5/18 m/s

v= 5.55 m/s

To maintain the speed 5.55 m/s ,friction force should be equal to the radial force

Friction force = Radial force(F)

[tex]F=\dfrac{mv^2}{r}[/tex]

By putting the values

[tex]F=\dfrac{mv^2}{r}[/tex]

[tex]F=\dfrac{1200\times 5.55^2}{25}[/tex]

F=1487.52 N

So friction force = 1478.52 N

We also know that

Friction force = μ m g

1478.52 N = μ x 1200 x 10 ( take g =10 m/s²)

μ = 0.12

This is minimum coefficient of static friction required to keep the car in this turn.