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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

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

Elements Of Electromagnetics
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
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25 20 1. Calculate the Torque ("Moment") around the pivot shown in each example. a) So call of Ø cose= b) x = tan²¹ (12) =22.62° c) d) m .3 m 20" 150 mm 40 N 25" 100 mm 4 N 6N 8 N 60% 45° 100 lb 644 3 B 8=cos" (²) = 36.2° - 52 N 12 20 N Y = 7x F |Y₁| = 1/sing .5 m- 0=25.100. sin (30.2°) 1500 lb-inches 90-36-9-53.10 break into components T=SKE F = (axy ₁0) F = ( Fx, ty ₁0) indersection of lines of action Fy= 52.sing = Fx= 52. =c058 = 98 26 T=Fd mag 20N-3m = 60wm 1140N-4m = 160NM 0 0 *} (134 = 01-01 (dr) k dy fx (k) = (6731-(0)²) = (2-7.2) 1^ ---5.20 3.N 5.2 MN clockwise taking - as cw & dicection T₁=60NM clockwise (que) Y2 = 160NM cew = 100wm cou (+ torque)
Transcribed Image Text:25 20 1. Calculate the Torque ("Moment") around the pivot shown in each example. a) So call of Ø cose= b) x = tan²¹ (12) =22.62° c) d) m .3 m 20" 150 mm 40 N 25" 100 mm 4 N 6N 8 N 60% 45° 100 lb 644 3 B 8=cos" (²) = 36.2° - 52 N 12 20 N Y = 7x F |Y₁| = 1/sing .5 m- 0=25.100. sin (30.2°) 1500 lb-inches 90-36-9-53.10 break into components T=SKE F = (axy ₁0) F = ( Fx, ty ₁0) indersection of lines of action Fy= 52.sing = Fx= 52. =c058 = 98 26 T=Fd mag 20N-3m = 60wm 1140N-4m = 160NM 0 0 *} (134 = 01-01 (dr) k dy fx (k) = (6731-(0)²) = (2-7.2) 1^ ---5.20 3.N 5.2 MN clockwise taking - as cw & dicection T₁=60NM clockwise (que) Y2 = 160NM cew = 100wm cou (+ torque)
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