# 017 Computation of force with given component parallel to a frame member

**Problem 017**

If the force F shown in Fig. P-017 is resolved into components parallel to the bars AB and BC, the magnitude of the component parallel to bar BC is 4 kN. What are the magnitudes of F and its component parallel to AB?

**Solution 017**

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$\alpha = 56.31^\circ$

$\tan \beta = \dfrac{1.5}{4.0}$

$\beta = 20.56^\circ$

$\theta = 90^\circ - \alpha = 90^\circ - 56.31^\circ$

$\theta = 33.69^\circ$

$\phi = 90^\circ - \beta = 90^\circ - 20.56^\circ$

$\phi = 69.44^\circ$

$\varphi = 180^\circ - \theta - \phi = 180^\circ - 33.69^\circ - 69.44^\circ$

$\varphi = 76.87^\circ$

By Sine law

$\dfrac{F}{\sin \varphi} = \dfrac{F_{BC}}{\sin \theta}$

$F = \dfrac{F_{BC} \, \sin \varphi}{\sin \theta}$

$F = \dfrac{4 \sin 76.87^\circ}{\sin 33.69^\circ}$

$F = 7.02 \, \text{ kN}$ *answer*

$\dfrac{F_{AB}}{\sin \phi} = \dfrac{F_{BC}}{\sin \theta}$

$F_{AB} = \dfrac{F_{BC} \, \sin \phi}{\sin \theta}$

$F_{AB} = \dfrac{4 \sin 69.44^\circ}{\sin 33.69^\circ}$

$F_{AB} = 6.75 \, \text{ kN}$ *answer*

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