Please explain how the moment of FAB and FAD were calculated in this picture. For example, I dont know where (0.24i and 720j) came from in calculating for FAB. I also dont know how -172.8 and 74.051 were obtained. Please explain it because I want to understand how to calculate for these two moments. The original item and diagram are shown on the right. ●•MomentOF FAB @ line AD(MFAB) = (0.24 ₁² x=720 j). UADAD172.8 k==74.0571Nmm=) AD = (TAC XTCE). UAD= [(0.48+² +0.16j) X TCE]M TCE) AD==0.48 ₁²ª + 0·16 j— 0-24k²0.56.UAD-0.16 TCE-0.0609 TCE1 +0.1829 TCEj+0.3+33 TCE K[0.481"0.481 +0.160.24 k0.56... (1)... (2).ADThe pipe ABCDE is supported by ball-and-socket jointsat A and D and by cable ECF that passes through aring C with negligible friction and is attached to hooksat E and F. Knowing that the frame supports auniformly distributed load of 1500 N/m along segmentAB, answer the following questions.490 mm480 mm1500 N/mEB180 mmD160 mm240 mmRing C

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Answered: Please explain how the moment of FAB…

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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Please explain how the moment of F_AB and F_AD were calculated in this picture. For example, I dont know where (0.24i and 720j) came from in calculating for F_AB. I also dont know how -172.8 and 74.051 were obtained. Please explain it because I want to understand how to calculate for these two moments. The original item and diagram are shown on the right.

●
•Moment
OF FAB @ line AD
(MFAB) = (0.24 ₁² x=720 j). UAD
AD
172.8 k
=
=
74.0571
Nmm
=) AD = (TAC XTCE). UAD
= [(0.48+² +0.16j) X TCE]
M TCE) AD
=
=
0.48 ₁²ª + 0·16 j— 0-24k²
0.56
.
UAD
-0.16 TCE
-0.0609 TCE1 +0.1829 TCEj
+0.3+33 TCE K
[0.481"
0.481 +0.160.24 k
0.56
... (1)
... (2)
.
AD
The pipe ABCDE is supported by ball-and-socket joints
at A and D and by cable ECF that passes through a
ring C with negligible friction and is attached to hooks
at E and F. Knowing that the frame supports a
uniformly distributed load of 1500 N/m along segment
AB, answer the following questions.
490 mm
480 mm
1500 N/m
E
B
180 mm
D
160 mm
240 mm
Ring C

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