Answer
The tension in the long wire is 50.0 N.
The tension in the short wire is 25.0 N.
Work Step by Step
Let $T_L$ be the tension in the long wire. Let's consider an axis of rotation at the left end of the shelf. Note that the tension in the short wire exerts zero torque about this axis.
$\sum \tau = 0$
$(T_L)(0.40~m) - (25~N)(0.20~m)- (50~N)(0.30~m) = 0$
$(T_L)(0.40~m) = (25~N)(0.20~m)+ (50~N)(0.30~m)$
$T_L = \frac{(25.0~N)(0.20~m)+ (50.0~N)(0.30~m)}{0.40~m}$
$T_L = 50.0~N$
The sum of the tension in the wires must be equal to the sum of the weights (tool and shelf). Let $T_S$ be the tension in the short wire.
$T_S+T_L = 25.0~N+50.0~N$
$T_S = 25.0~N+50.0~N - T_L$
$T_S = 25.0~N+50.0~N - 50.0~N$
$T_S = 25.0~N$
The tension in the long wire is 50.0 N.
The tension in the short wire is 25.0 N.
