Please include a kinamatic diagram (one for velocity and one for acceleration)Please DO NOT solve this using velocity analysis (cartesian vector analysis j k i). I would like it to be solved using scalar method. We dont use 3D in this course. We only use scalar analysis for the relative velocity equation as requested in the question (writing the x and y components of the equation and solving the equations for the unknowns). Below I have attached a sample question with solution in order to get an idea on how to use the scalar method to solve my question. I would like question this to be solved in a similar way. Thank you for your understanding.If you can solve it as soon as possible that would be great and I will give you a thumps up and positive feedback :) Question [5]: If link AB has the angular velocity and angular acceleration shown, determine:a) the angular velocity of link BC and CD at the instant shownradIf WBC = 6 cw and wcD = 12ccw at this istant, determine:radSSb) the angular acceleration of link CD (acD).c) and the acceleration of point C and its angle to horizontal at the instant shown.0.5 mABABсD= 6 rad/s²= 3 rad/s1 m-0.5 m-B1m 2. If the angular velocity of link AB is 3 rad/s. determine the velocity of the block at C and the angular velocity andangular acceleration of the connecting link CB at the instant shown.BKimmetic Diagram(Velocity)First solve relativeVelocity8=45⁰an,B201Link BC60A=3 radis"=30"VBVB = WAB· TAB = 3 (2) = 6 Ft/√48Vc#743°Vc = V8+ VCI3Y-Direction0 = -√₂ Sin 60° + V Sin 45°VC18:VB Sin 60°Sings"Link ABADic=> WBC = TBCVes7-353с-7.35 Ft/= 2.45 radCCVKinematic diogram(Acceleration))30°School of Energy / Mechanical Engineering(0²12 = WAB TAB = 18 F²7/3²Ac=AB+ A430nDA₁ = a + ² + ₁ + 0²/0²=² Foc=18 F√/5²BY-DinationA₂ = ₂² +² +Link BCLink ABat=d-√√8=0ABdet45"48OFENB048An0-0 Sin 30- Sin 45+ Sinus45°460 = 18 Sin30-18 Sin 45° + De Sinus⇒a=-5-27 ²X-Directions- Vc = V₂ Grobu - V2 C0-45°WAB2AB=dge TBCBCsection 16.5Q 16-6323Vc = -6 Cos6⁰ + 7-35 C-45Vc = 2.2 14/₂,Chers_ 5.27=1076 rad/2XBC = TBC3Page 2CCW

wCD=12 rad/secwBC=6 rad/secwAB=3 rad/sec

Answered: Question [5]: If link AB has the…

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 include a kinamatic diagram (one for velocity and one for acceleration)
Please DO NOT solve this using velocity analysis (cartesian vector analysis j k i). I would like it to be solved using scalar method. We dont use 3D in this course. We only use scalar analysis for the relative velocity equation as requested in the question (writing the x and y components of the equation and solving the equations for the unknowns). Below I have attached a sample question with solution in order to get an idea on how to use the scalar method to solve my question. I would like question this to be solved in a similar way. Thank you for your understanding.

If you can solve it as soon as possible that would be great and I will give you a thumps up and positive feedback :)

Question [5]: If link AB has the angular velocity and angular acceleration shown, determine:
a) the angular velocity of link BC and CD at the instant shown
rad
If WBC = 6 cw and wcD = 12ccw at this istant, determine:
rad
S
S
b) the angular acceleration of link CD (acD).
c) and the acceleration of point C and its angle to horizontal at the instant shown.
0.5 m
AB
AB
с
D
= 6 rad/s²
= 3 rad/s
1 m-
0.5 m-
B
1m

2. If the angular velocity of link AB is 3 rad/s. determine the velocity of the block at C and the angular velocity and
angular acceleration of the connecting link CB at the instant shown.
B
Kimmetic Diagram
(Velocity)
First solve relative
Velocity
8=45⁰
an,
B
201
Link BC
60
A=3 radis
"=30"
VB
VB = WAB· TAB = 3 (2) = 6 Ft/
√48
Vc
#7
43°
Vc = V8+ VCI3
Y-Direction
0 = -√₂ Sin 60° + V Sin 45°
VC18:
VB Sin 60°
Sings"
Link AB
A
Dic
=> WBC = TBC
Ves7-35
3
с
-7.35 Ft/
= 2.45 rad
CCV
Kinematic diogram
(Acceleration)
)30°
School of Energy / Mechanical Engineering
(0²12 = WAB TAB = 18 F²7/3²
Ac=AB+ A430
n
D
A₁ = a + ² + ₁ + 0²/0²=² Foc=18 F√/5²
B
Y-Dination
A₂ = ₂² +² +
Link BC
Link AB
at=d-√√8=0
AB
de
t
45
"48
OFENB
048
A
n
0-0 Sin 30- Sin 45+ Sinus
45°
46
0 = 18 Sin30-18 Sin 45° + De Sinus
⇒a=-5-27 ²
X-Directions
- Vc = V₂ Grobu - V</2 C0-45°
WAB
2
AB=dge TBC
BC
section 16.5
Q 16-63
23
Vc = -6 Cos6⁰ + 7-35 C-45
Vc = 2.2 14/₂,
Chers_ 5.27=1076 rad/2
XBC = TBC
3
Page 2
CCW

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