• Real =0: • Imag=0: a cos 02 + b cos 03 - c cos 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°c sin 0₁ = 0 c sin ₁ = 1.7209 (Eq 3) then, solve the two equations 2 and 3 with the two unknowns, you get c sin 01 C COS 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: bw3j ejes -ċeje- c w₁jeje₁ = 0 bw3j (cos 03+j sin 03)- c(cos 0₁ +j sin 0₁) cw₁j (cos 0₁+j sin 0₁) = 0 b w3 (-sin 03 +j cos 03) - è(cos 0₁+j sin 0₁) c w₁(-sin 0₁ +j cos 0₁) = 0 We need to determine w₁ and c. • Real=0: -b w3 sin 03 - è cos 0₁ + c w₁ sin ₁ = 0 -2.4 (-0.1) sin 160 - è cos 135.56° + 2.458 w₁ sin 135.56° = 0 1.7209 w₁+0.7139 -0.082 (Eq. 4) • Imag=0: 2.4 (-0.1) cos 160-è sin 135.56° 2.458 w₁ cos 135.56° 0 1.755 w₁-0.7 è = -0.2255 (Eq. 5) Solution a) Vector loop of DAB 5 m b) Firstly, determine the input angle and link length Angle and angular velocity: 0.5 m 0₂tan 1 = 60.95° 031800 180-20 = 160° • W3 = -0.1 rad/s 0.9 m Length • a=AD = √0.52 +0.92 = 1.0296 m b= AB 2.4 m DA = a el Vector loop of DAB AB=bele DB= cele DA+ AB- DB = 0 ael+bel-cel = 0 (Eq 1) a (cos 02+jsin 02) + b (cos 03 +j sin 03) c (cos 0₁ + j sin 0₁) = 0
• Real =0: • Imag=0: a cos 02 + b cos 03 - c cos 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°c sin 0₁ = 0 c sin ₁ = 1.7209 (Eq 3) then, solve the two equations 2 and 3 with the two unknowns, you get c sin 01 C COS 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: bw3j ejes -ċeje- c w₁jeje₁ = 0 bw3j (cos 03+j sin 03)- c(cos 0₁ +j sin 0₁) cw₁j (cos 0₁+j sin 0₁) = 0 b w3 (-sin 03 +j cos 03) - è(cos 0₁+j sin 0₁) c w₁(-sin 0₁ +j cos 0₁) = 0 We need to determine w₁ and c. • Real=0: -b w3 sin 03 - è cos 0₁ + c w₁ sin ₁ = 0 -2.4 (-0.1) sin 160 - è cos 135.56° + 2.458 w₁ sin 135.56° = 0 1.7209 w₁+0.7139 -0.082 (Eq. 4) • Imag=0: 2.4 (-0.1) cos 160-è sin 135.56° 2.458 w₁ cos 135.56° 0 1.755 w₁-0.7 è = -0.2255 (Eq. 5) Solution a) Vector loop of DAB 5 m b) Firstly, determine the input angle and link length Angle and angular velocity: 0.5 m 0₂tan 1 = 60.95° 031800 180-20 = 160° • W3 = -0.1 rad/s 0.9 m Length • a=AD = √0.52 +0.92 = 1.0296 m b= AB 2.4 m DA = a el Vector loop of DAB AB=bele DB= cele DA+ AB- DB = 0 ael+bel-cel = 0 (Eq 1) a (cos 02+jsin 02) + b (cos 03 +j sin 03) c (cos 0₁ + j sin 0₁) = 0
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 to explain the parameter in green where it came from and how to determine its derivative, with an explanation on this graph and what mean clockwise and anticlockwise, and what the - and + are on this drawing? x and y. I want a detailed explanation please..
Transcribed Image Text:• Real =0:
•
Imag=0:
a cos 02 + b cos 03 - c cos 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°c sin 0₁ = 0
c sin ₁ = 1.7209 (Eq 3)
then, solve the two equations 2 and 3 with the two unknowns, you get
c sin 01
C COS 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:
bw3j ejes -ċeje- c w₁jeje₁ = 0
bw3j (cos 03+j sin 03)- c(cos 0₁ +j sin 0₁) cw₁j (cos 0₁+j sin 0₁) = 0
b w3 (-sin 03 +j cos 03) - è(cos 0₁+j sin 0₁) c w₁(-sin 0₁ +j cos 0₁) = 0
We need to determine w₁ and c.
•
Real=0:
-b w3 sin 03 - è cos 0₁ + c w₁ sin ₁ = 0
-2.4 (-0.1) sin 160 - è cos 135.56° + 2.458 w₁ sin 135.56° = 0
1.7209 w₁+0.7139 -0.082 (Eq. 4)
•
Imag=0:
2.4 (-0.1) cos 160-è sin 135.56° 2.458 w₁ cos 135.56° 0
1.755 w₁-0.7 è = -0.2255 (Eq. 5)
Transcribed Image Text:Solution
a) Vector loop of DAB
5 m
b) Firstly, determine the input angle and link length
Angle and angular velocity:
0.5 m
0₂tan 1
= 60.95°
031800
180-20 = 160°
•
W3 = -0.1 rad/s
0.9 m
Length
•
a=AD = √0.52 +0.92 = 1.0296 m
b= AB 2.4 m
DA = a el
Vector loop of DAB
AB=bele
DB= cele
DA+ AB- DB = 0
ael+bel-cel = 0 (Eq 1)
a (cos 02+jsin 02) + b (cos 03 +j sin 03) c (cos 0₁ + j sin 0₁) = 0
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