Chapter 12 - Rotation of a Rigid Body - Exercises and Problems - Page 331: 30

$F_1 = 120~N$ $F_2 = 80~N$

Let's consider the torque about an axis located at the position of the force $F_2$. $\sum \tau = 0$ $-(3.0~m)(40~N)+(1.0~m)~F_1 = 0$ $F_1 = \frac{(3.0~m)(40~N)}{1.0~m}$ $F_1 = 120~N$ We can use the vertical forces to find $F_2$. $\sum F_y = 0$ $40~N-F_1+F_2 = 0$ $F_2 = F_1-40~N$ $F_2 = 120~N-40~N$ $F_2 = 80~N$