Given,
The length of the side of the square, L=80 cm=0.8 m
The force applied, F=60 N
The distance between the centre and the point where the force is applied is given by,
[tex]\begin{gathered} d=\sqrt[]{0.4^2+0.4^2} \\ =0.16\text{ m} \end{gathered}[/tex]
The torque at the centre is given by,
[tex]\tau=F\times d[/tex]
On substituting the know values,
[tex]\begin{gathered} \tau=60\times0.16 \\ =9.6\text{ Nm} \end{gathered}[/tex]
Thus the torque about the center of the square is 9.6 Nm
