• Real =0: • Imag=0: a cos 02 + b cos 03 - CCS 0₁ = 0 1.0296 cos 60.95° + 2.4 cos 160° - c cos 0₁ = 0 c cos 0₁ = -1.755 (Eq 2) a sin 02 + b sin 03 - c sin 0₁ = 0 1.0296 sin 60.95° + 2.4 sin 160° csin₁ = 0 c sin 0₁ = 1.7209 (Eq 3) then, solve the two equations 2 and 3 with the two unknowns, you get c sin 01 CCOS 01 =tan 0₁ = 1.7209 -1.755 => 0₁ = 135.56° and using equation 3, you find (c) => c sin 135.56° = 1.7209 => c = 2.458 m Derivative of Eq 1: bwзjejo -ċejo - c w₁jejo₁ = 0 b w3j (cos 03+j sin 03) - ċ(cos 0₁ + j sin 0₁) - c w₁j (cos 0₁ + j sin 0₁) = 0 b w3 (sin 03 +j cos 03) - ċ(cos 0₁+j sin 0₁) - cw₁(- sin 0₁ + j cos 0₁) = 0 We need to determine w₁ and ċ. • Real =0: -b w3 sin 03-ċ cos 0₁ + c w₁ sin 0₁ = 0 -2.4 (-0.1) sin 160 ċ cos 135.56° + 2.458 w₁ sin 135.56° = 0 1.7209 w₁ +0.7139 ċ = -0.082 (Eq. 4) Imag=0: 2.4 (-0.1) cos 160 ċ sin 135.56° 2.458 w₁ cos 135.56° = 0 1.755 w₁ 0.7 ċ = -0.2255 (Eq. 5) 7 Solution a) Vector loop of DAB 5 m 2.4 m b B 0.9 m x b) Firstly, determine the input angle and link length Angle and angular velocity: 02 = tan¹ (0) = 60.95° 03 = 1800 = 180-20 = 160° -0.1 rad/s • W3 Length a=AD = √0.52 + 0.92 = 1.0296 m b AB 2.4 m DA = a eje₂ Vector loop of DAB 0.5 m AB = beje, DB = c eje₁ DAAB-DB=0 a ejzbejsce₁ = 0 (Eq 1) a (cos B₂+j sin 8₂) + b (cos 0, +j sin 0₁) c (cos 0₁+j sin 0₁)=0 6
• Real =0: • Imag=0: a cos 02 + b cos 03 - CCS 0₁ = 0 1.0296 cos 60.95° + 2.4 cos 160° - c cos 0₁ = 0 c cos 0₁ = -1.755 (Eq 2) a sin 02 + b sin 03 - c sin 0₁ = 0 1.0296 sin 60.95° + 2.4 sin 160° csin₁ = 0 c sin 0₁ = 1.7209 (Eq 3) then, solve the two equations 2 and 3 with the two unknowns, you get c sin 01 CCOS 01 =tan 0₁ = 1.7209 -1.755 => 0₁ = 135.56° and using equation 3, you find (c) => c sin 135.56° = 1.7209 => c = 2.458 m Derivative of Eq 1: bwзjejo -ċejo - c w₁jejo₁ = 0 b w3j (cos 03+j sin 03) - ċ(cos 0₁ + j sin 0₁) - c w₁j (cos 0₁ + j sin 0₁) = 0 b w3 (sin 03 +j cos 03) - ċ(cos 0₁+j sin 0₁) - cw₁(- sin 0₁ + j cos 0₁) = 0 We need to determine w₁ and ċ. • Real =0: -b w3 sin 03-ċ cos 0₁ + c w₁ sin 0₁ = 0 -2.4 (-0.1) sin 160 ċ cos 135.56° + 2.458 w₁ sin 135.56° = 0 1.7209 w₁ +0.7139 ċ = -0.082 (Eq. 4) Imag=0: 2.4 (-0.1) cos 160 ċ sin 135.56° 2.458 w₁ cos 135.56° = 0 1.755 w₁ 0.7 ċ = -0.2255 (Eq. 5) 7 Solution a) Vector loop of DAB 5 m 2.4 m b B 0.9 m x b) Firstly, determine the input angle and link length Angle and angular velocity: 02 = tan¹ (0) = 60.95° 03 = 1800 = 180-20 = 160° -0.1 rad/s • W3 Length a=AD = √0.52 + 0.92 = 1.0296 m b AB 2.4 m DA = a eje₂ Vector loop of DAB 0.5 m AB = beje, DB = c eje₁ DAAB-DB=0 a ejzbejsce₁ = 0 (Eq 1) a (cos B₂+j sin 8₂) + b (cos 0, +j sin 0₁) c (cos 0₁+j sin 0₁)=0 6
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
I want a detailed explanation of the two areas marked in red. Firstly, why + and - became in derivation all - Secondly, why did we put it above C dot and why did we delete a? I want an explanation about this derivation with a detailed explanation of its drawing, please.
Transcribed Image Text:• Real =0:
•
Imag=0:
a cos 02 + b cos 03
-
CCS 0₁ = 0
1.0296 cos 60.95° + 2.4 cos 160° - c cos 0₁ = 0
c cos 0₁ = -1.755 (Eq 2)
a sin 02 + b sin 03 - c sin 0₁ = 0
1.0296 sin 60.95° + 2.4 sin 160° csin₁ = 0
c sin 0₁ = 1.7209 (Eq 3)
then, solve the two equations 2 and 3 with the two unknowns, you get
c sin 01
CCOS 01
=tan 0₁ =
1.7209
-1.755
=> 0₁ = 135.56°
and using equation 3, you find (c) => c sin 135.56° = 1.7209 => c = 2.458 m
Derivative of Eq 1:
bwзjejo -ċejo - c w₁jejo₁ = 0
b w3j (cos 03+j sin 03) - ċ(cos 0₁ + j sin 0₁) - c w₁j (cos 0₁ + j sin 0₁) = 0
b w3 (sin 03 +j cos 03) - ċ(cos 0₁+j sin 0₁) - cw₁(- sin 0₁ + j cos 0₁) = 0
We need to determine w₁ and ċ.
• Real =0:
-b w3 sin 03-ċ cos 0₁ + c w₁ sin 0₁ = 0
-2.4 (-0.1) sin 160 ċ cos 135.56° + 2.458 w₁ sin 135.56° = 0
1.7209 w₁ +0.7139 ċ = -0.082 (Eq. 4)
Imag=0:
2.4 (-0.1) cos 160 ċ sin 135.56° 2.458 w₁ cos 135.56° = 0
1.755 w₁ 0.7 ċ = -0.2255 (Eq. 5)
7
Transcribed Image Text:Solution
a) Vector loop of DAB
5 m
2.4 m
b
B
0.9 m
x
b) Firstly, determine the input angle and link length
Angle and angular velocity:
02 = tan¹ (0) = 60.95°
03 = 1800 = 180-20 = 160°
-0.1 rad/s
•
W3
Length
a=AD = √0.52 + 0.92 = 1.0296 m
b AB 2.4 m
DA = a eje₂
Vector loop of DAB
0.5 m
AB = beje,
DB = c eje₁
DAAB-DB=0
a ejzbejsce₁ = 0 (Eq 1)
a (cos B₂+j sin 8₂) + b (cos 0, +j sin 0₁) c (cos 0₁+j sin 0₁)=0
6
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