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.

The force acting on the gear tooth is

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.

The force acting on the gear tooth is

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

The force acting on the gear tooth is

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

 

This question can be found in Engineering Mechanics: Statics, 13th edition, chapter 2, question 2-11.

Leave a comment

Your email address will not be published.