The area of the cylinder can be given as,
[tex]A=\pi(\frac{d}{2})^2[/tex]Substitute the known value,
[tex]\begin{gathered} A=(3.14)(\frac{5\text{ in}}{2})^2 \\ =19.625in^2 \end{gathered}[/tex]The extension force acting on the cylinder is,
[tex]F=PA[/tex]Substitute the known values,
[tex]\begin{gathered} F=(200psi)(19.625in^2)(\frac{1\text{ lbf}}{1psiin^2})_{} \\ =3925\text{ lbf} \end{gathered}[/tex]Thus, the force exerted on the cylinder is 3925 lbf.