Question

Compute the direction and magnitude of the resultant of the following system of coplanar and concurrent forces; 20 N, 40 N, 25N, 42N, and 12 N making an angle of 30

Answer 1

In unit vector form

$F_x = 20 cos 30 - 40 cos 60 -25 +12 cos 45 = - 19.194 N\\ F_y = 20 sin 30 - 40 sin 60 -42 +12 sin 45 = - 5.844 N\\ F= \sqrt{F_x^2+F_y^2}\\ F= \sqrt{(-19.194)^2+(-5.844)^2}\\ F= 20.064 N\\ \alpha = \tan^{-1}(\frac{-5.844}{-19.194})\\ \alpha = 16.934^0\\$

Therefore, the angle is 180⁰+16.934⁰ = 196.934⁰