Answer
1.5 cm
Work Step by Step
Use equation 25–6b.
$$M=\frac{N \mathcal{l}}{f_of_e}$$
$$f_e=\frac{N \mathcal{l}}{f_oM}$$
$$=\frac{(25cm)(17.5cm)}{(0.40cm)(720)}=1.5cm$$
This answer satisfies the underlying assumption for the equation, that $f_e \ll \mathcal{l}$.
