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V125BC [204]
3 years ago
14

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?
Physics
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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