Answer
(a) $F_x = 49.1~N$
$F_y = 2.95~N$
(b) $F = 49.2~N$
(c) The sum of these force is directed at an angle of $3.4^{\circ}$ above the +x-axis.
Work Step by Step
(a) We can find the horizontal component of the sum of the forces:
$F_x = (22.0~N)~cos~30.0^{\circ}+ (22.0~N)~cos~30.0^{\circ}+ (22.0~N)~cos~60.0^{\circ}$
$F_x = 49.1~N$
We can find the vertical component of the sum of the forces:
$F_y = (22.0~N)~sin~30.0^{\circ}+ (22.0~N)~sin~30.0^{\circ}+ (-22.0~N)~sin~60.0^{\circ}$
$F_y = 2.95~N$
(b) We can find the magnitude of the sum of the forces:
$F = \sqrt{F_x^2+F_y^2} = \sqrt{(49.1~N)^2+(2.95~N)^2} = 49.2~N$
(c) We can find the angle $\theta$ above the positive x-axis:
$tan~\theta = \frac{2.95~N}{49.1~N}$
$\theta = tan^{-1}(\frac{2.95~N}{49.1~N})$
$\theta = 3.4^{\circ}$
The sum of these force is directed at an angle of $3.4^{\circ}$ above the +x-axis.
