Question [4]: At the instant shown roller C has a velocity of 250 mm/s up the channel. Using relative motion analysis-velocity (scalar method) method, determine: a) the angular velocities of both bars (WBC & WAB), b) the velocity of pin B at this instant (VB). 60° 120 mm 30° A B 100 mm

Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
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  • Please include a kinamatic diagram (for velocity)
  • Please DO NOT solve this using velocity analysis (cartesian vector analysis). 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 [4]: At the instant shown roller C has a velocity of 250 mm/s up the channel. Using relative
motion analysis-velocity (scalar method) method, determine:
a) the angular velocities of both bars (WBC & WAB),
b) the velocity of pin B at this instant (VB).
60°
120 mm
30°
A
B
100 mm
Transcribed Image Text:Question [4]: At the instant shown roller C has a velocity of 250 mm/s up the channel. Using relative motion analysis-velocity (scalar method) method, determine: a) the angular velocities of both bars (WBC & WAB), b) the velocity of pin B at this instant (VB). 60° 120 mm 30° A B 100 mm
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
Transcribed Image Text: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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