Angle of a Vector

  1. 1. The problem statement, all variables and given/known data
    two forces, 406N at 17 degrees and 256N at -26 degrees are applied to a a 3400kg car. Find resultant of these two forces and the direction of the resultant force between -180 and 180 degrees.


    2. Relevant equations



    3. The attempt at a solution
    406cos(17) = 388.26, 256cos(26) = 230.09, 230.09 = 388.26 = 618.35N
    arctan(230.09/388.26) = 30.65 degrees
     
  2. jcsd
  3. cronxeh

    cronxeh 1,232
    Gold Member

    What about the y component?
     
  4. because it's a car and it doesn't move on the y-axis. I'm still not sure how to get the angle though.
     
  5. cronxeh

    cronxeh 1,232
    Gold Member

    The car has nothing to do with anything. And changing -26 degrees to 26 was a nice trick, but you need to follow the rules because the next step requires the actual angle.
     
  6. cronxeh

    cronxeh 1,232
    Gold Member

    Does your answer make sense that the highest forces pulling at 17 degrees somehow generates a force vector that is 30 degress? Your answer is obviously somewhere between (17+26)/2=21.5 degree spread between the two forces. You have 1.58 times the force at 17 degrees than at -26 degrees, so you should be slightly higher off a midpoint mark (17-21.5 = -4.5 deg). So your answer should be above 0 degree mark at least, but not at 30 degrees! Your feasible region is thus between (-4.5, 17) degrees

    Look at it this way.. if 17 degree force has (406/256)= 1.5859375 times more weight than the negative force, then your resultant should be around 1.5859375*17 + 1*(-26) ~ 0.96 degrees. Or more closely to the answer now, 1*17 + 0.60591138*(-26) ~ 0.60 degrees

    You need to calculate all resultant forces and follow rules of trigonometry.
     
    Last edited: Oct 14, 2010
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