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Homework Help: Equilibirum Forces

  1. Oct 10, 2004 #1
    1350kg car resting on a plane surface with its brakes locked.
    Unit vector [tex] e_n = <.231,.923,.308> [/tex] is perpendicular to the surface. The y axis points upwards. The direction cosines of a cable supporting the car are [tex]<-.816,.408,-.408>[/tex] and the tension in the cable is 1.2KN. Determine the magnitude of the normal and friction forces the car's wheels exert on the surface.

    This is what I have so far:

    [tex]F_f = <.231F_x, .923F_y, .308F_z>[/tex]
    [tex]T_{AB} = 1200<-.816, .408, -.408>[/tex]
    [tex]W = -(1350)*(9.8)\hat{j}[/tex]
    [tex]N = |1200|<.231,.923,.308>[/tex]

    [tex]0 = F_f + T_{AB} + N - W[/tex]


    But when I solve the equations, I come up with N being 9719N, it is supposed to be about 2500 more.

    Thanks!
     
  2. jcsd
  3. Oct 10, 2004 #2
    Picture

    Here is a picture of the diagram and the math.
     

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    Last edited: Oct 10, 2004
  4. Oct 11, 2004 #3
    .231N + X -.816*1200=0
    .923N+Y-1350*9.8+1200*.408=0
    .308N + Z -.408*1200=0
    The fourth is correct.
     
  5. Oct 11, 2004 #4
    Got it

    Thanks, I figured this out late last night right before i went to bed.

    I forgot that <fx,fy,fz> were already in the form of the frictional vector and that the equation .231Fx + .923Fy + .308Fz was just meant to relate the components, they are not the actual components.

    Thanks!
     
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