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Homework Help: The angle between two horizontal forces?

  1. Oct 17, 2006 #1
    I need help with this problem (my work is at the bottom):

    http://img96.imageshack.us/img96/9823/physicsquestiondd9.jpg [Broken]

    Here is what I tried:

    [tex]a_x[/tex] from graph = 3 m/s^2

    [tex]Fnetx[/tex] = m* [tex]a_x[/tex]

    [tex]F_n_e_t_x[/tex] = 3.9 N

    [tex]F_n_e_t_x[/tex] = -3 + [tex]F_2*\cos(\theta)[/tex]

    6.9 = 9.0*[tex]\cos(\theta)[/tex]

    [tex]\cos(\theta)[/tex] = 0.7667
    [tex]\theta[/tex] = 39.9 degrees from x-axis
    degrees between forces = 180+39.9 = 219.9 degrees

    This is wrong and I've been trying to figure out what to do for a long time so I'm hoping someone can help me out please? :confused: Thanks
     
    Last edited by a moderator: May 2, 2017
  2. jcsd
  3. Oct 17, 2006 #2
    Typically you use the smaller angle between two things. Have you tried that?
     
  4. Oct 17, 2006 #3

    radou

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    Homework Helper

    If the force F1 points in the negative x direction, and F2 has an angle of 39.9 with the positive x direction, then the angle between them is 90 + (90 - 39.9) = 180 - 39.9 = 140.1 degree, unless I missed something.
     
  5. Oct 17, 2006 #4
    Yes, I believe that is right. However, I tried inputting 140 and it is incorrect. I am unsure if [tex]F_1[/tex] is actually going at 180 degrees. It only says it is going in the negative x direction? Also, I am unsure if I am setting up the equation correctly, and also if I am supposed to use [tex]F_2\cos(\theta)[/tex].

    Can anyone confirm or deny this?
     
    Last edited: Oct 17, 2006
  6. Oct 17, 2006 #5

    radou

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    Using [tex]F_{2} \cos(\theta)[/tex] is correct, since [tex]\vec{F}_{2}[/tex] makes some angle [tex]\theta[/tex] with the x axis. Further on, [tex]\vec{F}_{1}[/tex] is not 'going' at 180 degrees. It is a simple calculation of the angle between the two force vectors, and 180 arises in it.
     
  7. Oct 17, 2006 #6
    Well I am not sure what I did wrong because neither 140 or 220 degrees is correct. Is the equation supposed to be [tex]F_n_e_t_x[/tex] = [tex]-F_1[/tex] + [tex]F_2\cos(\theta)[/tex]? Hopefully someone can show me where I went wrong
     
    Last edited: Oct 17, 2006
  8. Oct 17, 2006 #7
    I also get that the angle between them is 140.06 degrees.
     
  9. Oct 17, 2006 #8
    Thanks guys, the answer WAS 140 degrees :rofl: . I thought I had tried that answer before I tried 220, but I only tried 220 :grumpy: Thanks for making my physics experience a little better :rolleyes:

    At least I understand why I got that answer now
     
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