free body diagram of cruches

from the first diagram (\(\displaystyle{n}_{{{g}{r}{o}{u}{n}{d}}}={\frac{{{170}}}{{{2}}}}={85}\ {l}{b}\))

\(\displaystyle\sum{F}_{{x}}={F}_{{1}}{\sin{{22}}}-{F}_{{2}}{\sin{{22}}}={0}\)

\(\displaystyle\sum{F}_{{y}}={2}{F}{\cos{{22}}}+{85}{l}{b}{s}-{170}{l}{b}{s}\)

so F=45.8lbs

From the second diagram

\(\displaystyle\sum{F}_{{x}}={f}-{45.8}{l}{b}{s}{\sin{{22}}}={0}\) f=17.2lbs

\(\displaystyle\sum{F}_{{y}}={n}_{{{t}{i}{p}}}-{45.8}{\cos{{22}}}={0}\) \(\displaystyle{n}_{{{t}{i}{p}}}={42.5}{l}{b}{s}\)

for minimum coefficient of friction

\(\displaystyle\mu={\frac{{{f}}}{{{n}_{{{t}{i}{p}}}}}}={0.404}\)

b) compressing force F =45.8lbs.