Physics

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A 600-kg car is going around a curve with a radius of 120 m that is banked at an angle of 25.0° with a speed of 30.0 m/s.The coefficient of static friction between the car and the road is 0.300. What is the force exerted by friction on the car?

  • Physics -

    Inward force needed =m a = m v^2/R
    = 600 (900)/120 = 4500 Newtons inward
    down slope component = 4500 cos 25

    Force up slope = Ff friction
    weight down slope = m g sin 25
    = 2488

    Net force down slope = 2488 - Ff

    so
    2488-Ff = 4500 cos 25
    Ff = 2488 - 4078
    Ff = -1590 N so 1590 DOWN the slope

    Can that .3 coef of friction provide the 1590 N ?
    normal component of weight = 600 * 9.81 * cos 25 = 5334, yes, good, we do not slip up or down

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