#### Answer

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

#### Work Step by Step

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 $F_2$.
$\sum F_y = 0$
$40~N-F_1+F_2 = 0$
$F_2 = F_1-40~N$
$F_2 = 120~N-40~N$
$F_2 = 80~N$