# The force acting on the gear tooth is

The force acting on the gear tooth is F = 20 lb. Resolve this force into two components acting along the lines aa and bb.

Image from: Hibbeler, R. C., S. C. Fan, Kai Beng. Yap, and Peter Schiavone. Statics: Mechanics for Engineers. Singapore: Pearson, 2013.

#### Solution:

Show me the final answer↓

We will first draw vector component diagram showing our forces along the aa and bb lines.

We can now draw these vectors tail to tail and apply the sine law to figure out the values.

Using the sine law, we can write the following:

$\dfrac{20}{\sin40^0}=\dfrac{F_a}{\sin80^0}$

(Solve for $F_a$)

$F_a\,=\,30.6$ lb

$\dfrac{20}{\sin40^0}=\dfrac{F_b}{\sin60^0}$

(Solve for $F_b$)

$F_b\,=\,26.9$ lb

#### Final Answers:

$F_a\,=\,30.6$ lb

$F_b\,=\,26.9$ lb