Accounting for the average car weight, each arm must support a weight of $1,500lb,$ that is$,W=1,500lb.$ The value of reaction R is $10,000lb$ and that of force

F is $6,666\mathrm{.}66lb.$ The diameter of the pin is $2$ inches. The material of the pin is AISI $1020$ steel with a yield strength of $47\mathrm{.}9$ ksi $,$ an ultimate tensile strength of

$65\mathrm{.}3$ ksi and a compressive strength of $60$ ksi $.$

For the pin, focusing on the small section between forces R and F$,$ find the safety factor for the critical point at the surface of the pin. Since the two forces are

only $1\mathrm{.}5$ inches apart, consider direct shear over the entire cross$-$section instead of shear due to a transverse load.