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
icon
Related questions
icon
Concept explainers
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)
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)
Expert Solution
steps

Step by step

Solved in 4 steps with 3 images

Blurred answer
Knowledge Booster
Combined Loading
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, mechanical-engineering and related others by exploring similar questions and additional content below.
Recommended textbooks for you
Elements Of Electromagnetics
Elements Of Electromagnetics
Mechanical Engineering
ISBN:
9780190698614
Author:
Sadiku, Matthew N. O.
Publisher:
Oxford University Press
Mechanics of Materials (10th Edition)
Mechanics of Materials (10th Edition)
Mechanical Engineering
ISBN:
9780134319650
Author:
Russell C. Hibbeler
Publisher:
PEARSON
Thermodynamics: An Engineering Approach
Thermodynamics: An Engineering Approach
Mechanical Engineering
ISBN:
9781259822674
Author:
Yunus A. Cengel Dr., Michael A. Boles
Publisher:
McGraw-Hill Education
Control Systems Engineering
Control Systems Engineering
Mechanical Engineering
ISBN:
9781118170519
Author:
Norman S. Nise
Publisher:
WILEY
Mechanics of Materials (MindTap Course List)
Mechanics of Materials (MindTap Course List)
Mechanical Engineering
ISBN:
9781337093347
Author:
Barry J. Goodno, James M. Gere
Publisher:
Cengage Learning
Engineering Mechanics: Statics
Engineering Mechanics: Statics
Mechanical Engineering
ISBN:
9781118807330
Author:
James L. Meriam, L. G. Kraige, J. N. Bolton
Publisher:
WILEY