- #1
EtherealMonkey
- 41
- 0
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.
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: