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March 25, 2017

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A 2000 pound car is parked on a street that makes an angle of 12* with the horizontal (think it's a street on a hill). a) find the magnitude of the force required to keep the car from rolling down the hill. b) Find the force perpendicular to the street.

  • trig - ,

    mass = 2000lbs * 0.454kg/lb = 908kg.

    Fc = mg = 908kg * 9.8N/kg = 8898.4N @
    12deg.

    a. Fp = 8898.4*sin(12) = 1850.4N. = Force parallel to the street = Force required to keep the car from roling down hill.

    b. Fv = 8898.4 * cos12 = 8704N = Force
    perpendicular to the street.

  • Pre-Cal: Vectors - ,

    a. Force to prevent car from rolling = v
    v=<cos12°,sin12°>
    F=projvF=(F•v/|v|^2)v= (F•v)v
    <0,2000>•<cos12,sin12>= 0+2000sin12 = 415.82lbs

    b. Find the angle of the perpendicular in relation to the force of 2000lbs. To do this, I drew a triangle, one line horizontal, one line being the 12° angle between it and the horizontal, and one being perpendicular to the 12° line(90°). Connect the lines and the third angle should be 78°.
    Then you do the same as the last.
    Force to prevent car from rolling(perpendicular) = c
    c=<cos78°,sin78°>
    F=projcF=(F•v/|v|^2)v= (F•v)v
    <0,2000>•<cos78,sin78>= 0+2000sin78 = 1956.3lbs

  • Yo MUM - ,

    Ask ur mom

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