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I need some help on a force equilibrium problem!

  1. Mar 11, 2009 #1
    1. The problem statement, all variables and given/known data

    A particular object is moving on a horizontal (x,y) plane with a velocity of 10 m/s in the (+x) direction. It is simulaneously acted upon by two horizontal forces. F1= 225 N at an angle of 30 degrees with respect to (+x) and F2= 135N at an angle of -75 degrees with respect to (+x). Find the magnitude and direction of a third applied force that will leave the system in a state of equilibrium. Is the equilibrium static or dynamic? Explain your answer.


    2. Relevant equations

    I only know how to get started on this... I need help o:)

    3. The attempt at a solution

    I understand that the y components are ignored since there is a constant velocity in this system and acceleration would therefore equal 0. I drew a free-body diagram with force of gravity pointing downard, normal force pointing up, and the two other force vectors. Now what?

    Thank you for your help!
     
  2. jcsd
  3. Mar 12, 2009 #2
    In my opinion, we should not consider gravity force (at least, we don't receive its mass to compute mg). Just calculate 2 force in the horizontal plane to find the 3rd force to make object is in equilibrium.

    x-axis
    [tex]\sum[/tex]Fx = 0
    F1(cos30)+F2(cos75)+F3x = 0
    225(0.866)+135(0.259)+F3x = 0
    F3x = -229.796 N

    y-axis
    [tex]\sum[/tex]Fy = 0
    F1(sin30)+F2(sin75)=F3y = 0
    225(0.5)+(135(0.966)+F3y = 0
    F3y = 17.9N

    magnitude
    F3 = sqrt(F3x^2+F3y^2)
    F3 = 230.492N

    angle
    F3 angle = arctan(F3y/F3x)
    F3 angle = arctan(17.9/229.796) = 4.45 degrees with respect to (-x).
    (or 175.55 degrees from +x axis)

    System will be dynamic equilibrium with a constant velocity of 10 m/s in the (+x) direction
     
  4. Apr 29, 2009 #3
    Sorry for the very late reply, pafala... thank you so much for your help on this problem. I truly appreciate it!
     
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