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A car (m = 1540 kg) is parked on a road that rises 18.9 ° above the horizontal. What are the magnitudes of (a) the normal force and (b) the static frictional force that the ground exerts on the tires?

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    Wc = mg = 1540kg * 9.8N/kg = 15,092 N. = Wt of car.

    Fc = 15092N. @ 18.9 Deg.
    Fp = 15092*sin18.9 Deg. = 4889 N. = Force parallel to road.

    a. Fv = 15092*cos18.9 = 14,278 N. = Force perpendicular to road = The normal.

    b. Fn = Fp - Fs = 0,
    4889 - Fs = 0,
    Fs = 4889 N. = Force of static friction.

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