Solution:
1) Using the net torque by taking into account that the counterclockwise rotation is positive and clockwise rotation is negative:
[tex]\sf \tau_{net}=\Sigma \tau\\\sf{}\hspace{15}=(F_A\cdot 0\;m)+(F_B\cdot 0.2\;m)+[-F_C\cdot(0.2+0.6)m]\\\sf{}\hspace{40}+[-F_D\cdot(0.2+0.6+0.2)m]\\\sf{}\hspace{15}=0+(50\;N\cdot 0.2\;m)+[-70\;N\cdot(0.2+0.6)m]\\\sf{}\hspace{40}+[80\;sin30^\circ\;N\cdot(0.2+0.6+0.2)m]\\\sf{}\hspace{15}=0+10\;\textsf{N-m}-56\;\textsf{N-m}+40\;\textsf{N-m}\\\sf{}\hspace{15}=-6\;\textsf{N-m}\\\tau_{net}=6\;\textsf{N-m},\;down\quad(ANSWER)[/tex]
2) ANSWER: From the solution, taking the forces individually for each torque, thus, force at point C has the greatest torque of 56 N-m in magnitude.
