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 coefficient of

Question

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 coefficient 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?

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Thu Hương 6 months 2021-08-04T19:41:19+00:00 1 Answers 74 views 0

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    2021-08-04T19:43:06+00:00

    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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