#### Answer

$\theta=53.5^{\circ}$
$F_{AB}=621lb$

#### Work Step by Step

Given that $\phi=40^{\circ}$ and the component of the 400lb force in member AC should be 500lb, we are asked to find the component along member AB.
Using this information, we can apply the law of sines to find $ \theta $
$\sin40^{\circ} /400 = \sin \theta / 500$
Solving for $\theta$, we obtain:
$\theta=53.5^{\circ}$
Therefore: $\psi = 180^{\circ} -40^{\circ}-53.5^{\circ}=86.5^{\circ}$
Now we can use the law of sines again to find the component along member AB, $F_{AB}$
$\sin40^{\circ} /400 = \sin 86.5^{\circ} / F_{AB}$
Solving for $F_{AB}$, we obtain:
$F_{AB}=621lb$