Problem 016 The magnitude of vertical force F shown in Fig. P-016 is 8000 N. Resolve F into components parallel to the bars AB and AC.

Solution 016

$F_{AB} = 4256.71 \, \text{ N}$ answer

$\dfrac{F_{AC}}{\sin 120^\circ} = \dfrac{8000}{\sin 40^\circ}$

$F_{AC} = 10\,778.37 \, \text{ N}$ answer

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