## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

(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.
(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.