#### Answer

(a) The component of the net force that is parallel to the floor is 0.40 N
(b) The component of the net force that is perpendicular to the floor is 1.7 N
(c) The magnitude of $F_{net}$ is 1.7 N and it is directed at an angle of $76.8^{\circ}$ above the x-axis.

#### Work Step by Step

(a) $F_{net,x} = F_{1x}+F_{2x}+F_{3x}$
$F_{net,x} = (-3.0~N)~cos(30^{\circ})+(6.0~N)~cos(60^{\circ})+0$
$F_{net,x} = 0.40~N$
The component of the net force that is parallel to the floor is 0.40 N.
(b) $F_{net,y} = F_{1y}+F_{2y}+F_{3y}$
$F_{net,y} = (3.0~N)~sin(30^{\circ})+(6.0~N)~sin(60^{\circ})-5.0~N$
$F_{net,y} = 1.7~N$
The component of the net force that is perpendicular to the floor is 1.7 N.
(c) We can find the magnitude of $F_{net}$.
$F_{net} = \sqrt{(0.40~N)^2+(1.7~N)^2}$
$F_{net} = 1.7~N$
We can find the angle $\theta$ above the positive x-axis.
$tan(\theta) = \frac{1.7}{0.4}$
$\theta = arctan(\frac{1.7}{0.4})$
$\theta = 76.8^{\circ}$
The magnitude of $F_{net}$ is 1.7 N and it is directed at an angle of $76.8^{\circ}$ above the x-axis.