Statics - Coplanar Force Systems

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SUMMARY

The discussion focuses on solving a coplanar force system problem involving angles and forces. The user derived equations for the forces in the x and y directions, leading to the calculation of angles φ and θ. The calculated angle θ was found to be 102° while the published value is 70.1°. The error was identified as a sign mistake in the equation, specifically in the term involving 24/5. The correct approach was confirmed by another user, emphasizing the importance of sign accuracy in force equations.

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EtherealMonkey
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The problem statement:
Statics_0311_PF.png


My relevant equation:

\phi will be the angle between the X axis and F_{CO}

\theta = \phi + \arcsin\left(\frac{3}{5}\right)

My attempt at a solution:

\Sigma F_{x} = 0:

F_{CO}\cos\phi - F_{BO}\frac{4}{5} = 0

F_{CO} = \frac{F_{BO}\frac{4}{5}}{\cos\phi}

\Sigma F_{y} = 0:

F_{AO} - F_{BO}\frac{3}{5} - F_{CO}\sin\phi = 0

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

F_{AO} - \frac{3}{5}F_{BO} - \frac{4}{5}F_{BO}\tan\phi = 0

9kN - \frac{3}{5}8kN - \frac{4}{5}8kN\tan\phi = 0

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

\phi = 65.12^{\circ}

\theta = 102^{\circ}

The published value of \theta:

\theta = 70.1^{\circ}

I don't know what I did wrong.

TIA for any response.
 
Last edited:
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EtherealMonke said:
9kN - \frac{3}{5}8kN - \frac{4}{5}8kN\tan\phi = 0

\phi = \arctan\left(\left(9+\frac{24}{5}\right)*\frac{5}{32}\right)
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.
 
Jay, thank you sir!
 

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