Statics - Coplanar Force Systems


by EtherealMonkey
Tags: coplanar, force, statics, systems
EtherealMonkey
EtherealMonkey is offline
#1
Sep17-09, 08:20 PM
P: 41
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.
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PhanthomJay
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#2
Sep17-09, 09:25 PM
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Quote Quote by EtherealMonke View Post
[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]
that plus 24/5 should be a minus 24/5.
I don't know what I did wrong.

TIA for any response.
you did well, just missed the sign.
EtherealMonkey
EtherealMonkey is offline
#3
Sep17-09, 09:54 PM
P: 41
Jay, thank you sir!


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