A 5.0 Kg mass is subject to three forces: an 8.0N force pulling due north, a 5.0N force pulling due South, and a 4.0N force pulling due West. What is the magnitude and direction of the resulting acceleration?

Question

anonymous · Answer

Fx = 8.0n*cos(90)+5.0n*cos(270)+4.0n*cos(180) = -4
Fy = 8.0n*sin(90)+5.0n*sin(270)+4.0n*sin(180) = 3

sqrt((-4)^2 + (3)^2) = 5

arctan(3/-4) = 138.59 degrees

direction = 5/5 = 1

Can you tell me if I am on the right path for this question.