We will use the next diagram
Where Fr, is the frictional force, F is the applied force, W is the weight and N is the normal
For a)
We analyze the horizontal forces
[tex]\begin{gathered} \sum ^{}_{}F_x=F_{}-Fr=0 \\ F=Fr \end{gathered}[/tex]Then we analyze the vertical forces
[tex]\begin{gathered} \sum ^{}_{}F_y=N-W=0 \\ N=W \end{gathered}[/tex]then for the frictional force
[tex]Fr=\mu_sN[/tex]Then for the force we have
[tex]F=\mu_sN=0.21(80)(9.8)=164.64N[/tex]Then for b)
[tex]F=\mu_kN=0.18(80)(9.8)=141.12N[/tex]We
