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3 years ago

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A 600-kg car traveling at 30.0 m/s is going around a curve having a radius of 120 m that is banked at an angle of 25.0°. The coe

fficient of static friction between the car's tires and the road is 0.300. What is the magnitude of the force exerted by friction on the car?

1 answer:

ikadub [295]3 years ago

5 0

Answer:

$f_{fr}=1590.85 N$

Explanation:

Here the total force at the horizontal components will be equal to the centripetal force on the car. So we will have:

$f_{fr}cos(25)+Nsin(25)=m\frac{v^{2}}{r}$ (1)

f(fr) is the friction force
N is the normal force

Now, the sum of forces at the vertical direction is equal to 0.

$Ncos(25)-mg-f_{fr}sin(25)=0$ (2)

Let's combine (1) and (2) to find f(fr)

$f_{fr}=\frac{(mv^{2}/r)-mgtan(25)}{cos(25)+tan(25)sin(25)}$

$f_{fr}=\frac{(600*30^{2}/120)-600*9.81*tan(25)}{cos(25)+tan(25)sin(25)}$

$f_{fr}=1590.85 N$

I hope it helps you!

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