What would the graphs look like for part 2, 3, and 4. I was able to do it for Part 1 (second pic) but now I’m not sure if it’s the same for all the others or if they require more components based on the degree of the line. ComponentMethodGraphicalMethod2x9.8Fi = 1,96 @ 60°FR = 3.5 wnitsm1 =FR =(200g)90°90°Part 1.2X9.8F2 = \. 96@@ 120°m2 =(200g)25 x9.8F1 = 2.45 @ 120°FR =m1 =FR =(250g)·2×9.8Part 2F2 =1.96 @ 200°m2 =(200g). 15x9.8F1 =].47 @0°FR =m1 =FR =(150g)2X 9.8Part 3F2 = \, 96 @ 90°m2 =(200g)·15 X9.8·2x 9.8Fi = 1.47m1 =@ 50°(150g)FR =FR =F2 = 1.96@ 90°Part 4m2 =@(200g)25x9.8F3= 2.45 @ 140°m3 =(250g)EP =3 = in equilibriumEfx=0Forces acting onthe ring balan ce Graphical method3.5 units1.96 units120°1.96 units60°Component method0=tanFx = F,x +F;X = F,COS60 - F, COS 60° =Fy = FY + F;Y =F,Sin60° – F, Sin 60°=Fy

Solution: The graphical representation of the vectors given in part 2, is drawn below: The given…

Answered: Component Method Graphical Method 2×9.8…

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