MecMovies 4.0 : M1.4: Load Capacity of Two-Bar Assembly

$\sum_{}^{} F_{x} = F_{1} \cos (49 . 3^{\circ}) + F_{2} \cos (24 . 9^{\circ}) = 0 \sum_{}^{} F_{y} = F_{1} \sin (49 . 3^{\circ}) + F_{2} \sin (24 . 9^{\circ}) - P = 0$

Solve $\sum_{}^{} F_{x}$ for $F_{1}$ in terms of $F_{2}$ :

$F_{1} = - F_{2} \frac{\cos (24 . 9^{\circ})}{\cos (49 . 3^{\circ})} = - 1.39096 F_{2}$

Substitute into $\sum_{}^{} F_{y}$ to determine $F_{2}$ in terms of the unknown $P$ :

$F_{2} = - 0.67770 P$

and backsubstitute to determine $F_{1}$ in terms of the $P$ :

$F_{1} = 0.94266 P$

The unknown $P$ can now be stated in terms of the two member forces:

$P = 1.06083 F_{1} and P = - 1.47557 F_{2}$

The figure shows a diagonal member labeled 1 and a horizontal member labeled fixed to a column on the right at the points A on top and C below. The diagonal member runs between
points A and B and the horizontal member runs between the points B and C. Joint B is at a distance of 4.3 meters to the left
of A and 5 meters below A. Joint C is at a distance of 7 meters directly below A. The load at joint B is labeled P and it is
directed downwards.