Chapter 10 - Dynamics of Rotational Motion - Problems - Exercises - Page 329: 10.3

$\sum \tau = 2.50~N~m$

We can use the lever arm to find each torque. $\sum \tau = \tau_1+\tau_2+\tau_3$ $\sum \tau = -F_1~d_1+F_2~d_2+F_3~d_3$ $\sum \tau = -(18.0~N)(0.090~m)+(26.0~N)(0.090~m)+(14.0~N)(\sqrt{2}\times 0.090~m)$ $\sum \tau = 2.50~N~m$