## Engineering Mechanics: Statics & Dynamics (14th Edition)

$F_a=30.6lb$ $F_b=26.9lb$
We are given that the force acting on the gear tooth is $F=20lb$. Using the law of sines with the given angles, we can resolve the F into it's components. $\sin40^{\circ} / 20lb = \sin 80^{\circ} / F_a$ Solving for $F_a$, we obtain: $F_a=30.6lb$ $\sin40^{\circ} / 20lb = \sin 60^{\circ} / F_b$ Solving for $F_b$, we obtain: $F_b=26.9lb$