Statics - Coplanar Force Systems

  1. The problem statement:

    My relevant equation:

    [itex]\phi[/itex] will be the angle between the X axis and [itex]F_{CO}[/itex]

    [tex]\theta = \phi + \arcsin\left(\frac{3}{5}\right)[/tex]

    My attempt at a solution:

    [tex]\Sigma F_{x} = 0:[/tex]

    [tex]F_{CO}\cos\phi - F_{BO}\frac{4}{5} = 0[/tex]

    [tex]F_{CO} = \frac{F_{BO}\frac{4}{5}}{\cos\phi}[/tex]

    [tex]\Sigma F_{y} = 0:[/tex]

    [tex]F_{AO} - F_{BO}\frac{3}{5} - F_{CO}\sin\phi = 0[/tex]

    Combining terms and substituting the equation found for [itex]\Sigma F_{x} = 0[/itex] into [itex]\Sigma F_{x} = 0:[/itex]

    [tex]F_{AO} - \frac{3}{5}F_{BO} - \frac{4}{5}F_{BO}\tan\phi = 0[/tex]

    [tex]9kN - \frac{3}{5}8kN - \frac{4}{5}8kN\tan\phi = 0[/tex]

    [tex]\phi = \arctan\left(\left(9+\frac{24}{5}\right)*\frac{5}{32}\right)[/tex]

    [tex]\phi = 65.12^{\circ}[/tex]

    [tex]\theta = 102^{\circ}[/tex]

    The published value of [itex]\theta[/itex]:

    [tex]\theta = 70.1^{\circ}[/tex]

    I don't know what I did wrong.

    TIA for any response.
    Last edited: Sep 17, 2009
  2. jcsd
  3. PhanthomJay

    PhanthomJay 6,334
    Science Advisor
    Homework Helper
    Gold Member

    that plus 24/5 should be a minus 24/5.
    you did well, just missed the sign.
  4. Jay, thank you sir!
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