#### Answer

The sum of these two forces on the barge is 1080 N
The sum of the two forces found above is probably not equal to the net force.

#### Work Step by Step

By symmetry, the component of each force directed perpendicular to the direction of motion cancel out. The sum of the two forces is the sum of the component of each force in the direction of motion.
$F_{sum} = 2\times F~cos(\theta)$
$F_{sum} = 2\times (560~N)~cos(15^{\circ})$
$F_{sum} = 1080~N$
The sum of these two forces on the barge is 1080 N
This sum would be equal to the net force if the water exerted a force of zero on the barge in the direction of motion or against the direction of motion. However, since we can assume that the water exerts some kind of force (example: drag force) in the direction of motion or against the direction of motion, the sum of the two forces found above is probably not equal to the net force.