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
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
Related questions
Question
100%
Part c only
![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)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7f432baf-00d8-43fb-936b-3da4a70ae1a5%2Fc905f05e-b380-4d25-a4a7-78b9e8a7c471%2Fd5pg76_processed.jpeg&w=3840&q=75)
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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