F \(\displaystyle{\cos{}}\) x = u.m.g then

\(\displaystyle{F}={u}{m}\frac{{g}}{{\cos{{x}}}}={\left({0.35}\right)}{\left({1.3}\right)}\frac{{{9.8}}}{{\cos{{x}}}}\)

\(\displaystyle{F}=\frac{{4.46}}{{\cos{{x}}}}\)

a. taking differential of this we have :

\(\displaystyle{F}'={0}{\left({\cos{{x}}}\right)}-{\left({4.46}\right)}\frac{{-{\sin{{x}}}}}{{{\cos}^{{{2}}}{x}}}={4.46}\frac{{\sin{{x}}}}{{{\cos}^{{{2}}}{x}}}\)

taken its stationary we'll have

\(\displaystyle{F}'={4.46}\frac{{\sin{{x}}}}{{{\cos}^{{{2}}}{x}}}={0}\rightarrow{\sin{{x}}}={0}\rightarrow{x}={0}^{{{o}}}\)

b. \(\displaystyle{F}=\frac{{4.46}}{{\cos{{0}}}}=\frac{{4.46}}{{1}}={4.46}{N}\)

\(\displaystyle{F}={u}{m}\frac{{g}}{{\cos{{x}}}}={\left({0.35}\right)}{\left({1.3}\right)}\frac{{{9.8}}}{{\cos{{x}}}}\)

\(\displaystyle{F}=\frac{{4.46}}{{\cos{{x}}}}\)

a. taking differential of this we have :

\(\displaystyle{F}'={0}{\left({\cos{{x}}}\right)}-{\left({4.46}\right)}\frac{{-{\sin{{x}}}}}{{{\cos}^{{{2}}}{x}}}={4.46}\frac{{\sin{{x}}}}{{{\cos}^{{{2}}}{x}}}\)

taken its stationary we'll have

\(\displaystyle{F}'={4.46}\frac{{\sin{{x}}}}{{{\cos}^{{{2}}}{x}}}={0}\rightarrow{\sin{{x}}}={0}\rightarrow{x}={0}^{{{o}}}\)

b. \(\displaystyle{F}=\frac{{4.46}}{{\cos{{0}}}}=\frac{{4.46}}{{1}}={4.46}{N}\)