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: [tex]f_{x}[/tex]=150cos30°+ 75cos150°
=[tex]150[/tex]×[tex]\frac{\sqrt{3} }{2}[/tex] +[tex]75[/tex]×[tex]-\frac{\sqrt{3} }{2}[/tex]
=[tex]129.90-64.95[/tex]
=[tex]64.95N[/tex]
In vertical direction: [tex]f_{y}[/tex]= 150sin30°+75sin150°
=[tex]150[/tex]×[tex]\frac{1}{2}[/tex]+[tex]75[/tex]×[tex]\frac{1}{2}[/tex]
=[tex]112.50N[/tex]
Thus, magnitude is given by: [tex]\sqrt{f_{x} ^{2}+f_{y} ^{2} }[/tex]
=[tex]\sqrt{64.25^{2}+112.50^{2} }[/tex]
=[tex]\sqrt{16784.31}[/tex]
=[tex]129.55N[/tex]
Direction of the force is calculated as: tanα=[tex]\frac{f_{y} }{f_{x} }[/tex]
=[tex]\frac{112.50}{64.95}[/tex]
=[tex]1.7321[/tex]
α=60°
