25 46 20 1. Calculate the Torque ("Moment") around the pivot shown in each example. a) So call of Ø cose= 0 = tax²¹ (²/2) =22.020 d) .1 m .3 m 20 150 mm 40 N 25" 100 mm -4 m- 4 N 6N 8 N 60% 45° 100 lb 3 52 N 12 996 = cor" (²) = 36 1/3m B =7 20 N Y = 7² x F (1 = 144 sings -.5 m- 36.20 = 25.100. sin (30.2°) 1500 lb-inches 90-36.9-53.10 beak into Components TESKE 5 = (dxd) ₁0) F = ( Fx, ty, 0) indersection of lines of action Fy=52.3ing = 26 Fx= 52.co3g = 78 T= fd 20N-3m= 60Nm 140N-4m = 160NM mag (Eds) k = (b 31-(0)²) - (2-1-2) ^ -52 S-2 mN clockwise taking - as لاع dicection T₁=60NM clockwise que) Y ₂ = 160NM cow(+torque) ( = 100wm coul

Any functioning component, such as machine parts, structural members, pressure vessels, and so on, are all under the influence of more than one force. When anybody is under the influence of more than one force such as shear forces, bending moments, axial…