Answer:
Magnitude= 129.55N
Direction= 60°
Step-by-step explanation:
In order to find the magnitude, we must find the magnitude in both vertical and horizontal direction.
In horizontal direction: \(f_{x}\)=150cos30°+ 75cos150°
=\(150\)×\(\frac{\sqrt{3} }{2}\) +\(75\)×\(-\frac{\sqrt{3} }{2}\)
=\(129.90-64.95\)
=\(64.95N\)
In vertical direction: \(f_{y}\)= 150sin30°+75sin150°
=\(150\)×\(\frac{1}{2}\)+\(75\)×\(\frac{1}{2}\)
=\(112.50N\)
Thus, magnitude is given by: \(\sqrt{f_{x} ^{2}+f_{y} ^{2} }\)
=\(\sqrt{64.25^{2}+112.50^{2} }\)
=\(\sqrt{16784.31}\)
=\(129.55N\)
Direction of the force is calculated as: tanα=\(\frac{f_{y} }{f_{x} }\)
=\(\frac{112.50}{64.95}\)
=\(1.7321\)
α=60°
