# 016 Components of a force parallel to supporting bars

**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**

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By Sine Law:

$\dfrac{F_{AB}}{\sin 20^\circ} = \dfrac{8000}{\sin 40^\circ}$

$\dfrac{F_{AB}}{\sin 20^\circ} = \dfrac{8000}{\sin 40^\circ}$

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

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

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