Find the magnitude and direction of the resultant force and moment at A
Find the magnitude and direction of the resultant force and moment at A
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
Please fill the data in the table with the correct signs.
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4) Find the magnitude and direction of the resultant force and moment at A
FORCE
lbs
R
20
=
2.53'
Rx
2.53'
e F = F.Cos (0)
1414.21lb
1000lb
45
---
45
F= F.Sin(Ⓒ)
RY
-1000lb
+y
MR
Forces
=
+MA
A
MA= Fxd
Couples
-M
A"
Transcribed Image Text:1-4
4) Find the magnitude and direction of the resultant force and moment at A
FORCE
lbs
R
20
=
2.53'
Rx
2.53'
e F = F.Cos (0)
1414.21lb
1000lb
45
---
45
F= F.Sin(Ⓒ)
RY
-1000lb
+y
MR
Forces
=
+MA
A
MA= Fxd
Couples
-M
A
Expert Solution
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Follow-up Questions
Read through expert solutions to related follow-up questions below.
Follow-up Question
What about the other force. I only see the cos and sin of force 1414.21 and 45 degrees. I don't see the -1000 force with the degrees. This data might be wrong but I want to see the other data to full the table that it's in the beginning. Thank tou

Transcribed Image Text:1-4
4) Find the magnitude and direction of the resultant force and moment at A
(999.997) (1991.9) R =
Tan (1999.977
999.997)
2.53'
2.53'
FORCE Ⓒ F= F.Cos(e)
lbs
X
-10.00 -90 + 7070197
1414.21
R
= 63.435°
MR = -9999.985 lb
1414.21lb
1000lb
-X
———-
2'
4'
45°
3.-225
4'-
45°
2
2.53
A
Fy= F.Sin(e)
Forces
19
=1000°
- 1000 × 4 = -4000
-70+ 5.7492953-1783.980
-999.99776= -5999.985
45° 9 99.997
-999.997
Rx 999,997 16 Ry - 1999.997 16 MR = -9999.985.16
X
+y
2/236.064 lb
+MA
-1000lb
MR=
-9999.985 lb
63.4°
999.997.lb.
+X
4'
-1999.997 16
MA= Fxd
Couples
-MA
Solution
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