## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

The magnitude of the net force exerted on A is 7.3 N and it is directed at an angle of $79.2^{\circ}$ below the negative x-axis.
We can find the angle $\theta$ below the positive x-axis from A to D. $tan(\theta) = \frac{100}{141}$ $\theta = arctan(\frac{100}{141})$ $\theta = 35.3^{\circ}$ We then find the horizontal component of the force exerted on A. $F_x = F_B+F_{Dx}$ $F_x = -3.0~N+(2.0~N)~cos(35.3^{\circ})$ $F_x = -1.37~N$ We then find the vertical component of the force exerted on A. $F_y = F_C+F_{Dx}$ $F_y = -6.0~N-(2.0~N)~sin(35.3^{\circ})$ $F_y = -7.16~N$ We then find the magnitude of the force F. $F = \sqrt{(F_x)^2+(F_y)^2}$ $F = \sqrt{(-1.37~N)^2+(-7.16~N)^2}$ $F = 7.3~N$ We find the angle below the negative x-axis. $tan(\theta) = \frac{7.16}{1.37}$ $\theta = arctan(\frac{7.16}{1.37})$ $\theta = 79.2^{\circ}$ The magnitude of the net force exerted on A is 7.3 N and it is directed at an angle of $79.2^{\circ}$ below the negative x-axis.