ion and Force An X b Answered: Statics of Particle and x 401Asg01: Statics of Particles X + bartleby.com/questions-and-answers/statics-of-particle-and-forces-force-resolution-and-representation-equilibrium-free-body-diagram-etc/6009b08d-1085-402a-82cc-75e0f5ac5ecd Homework help starts here! Q Search Statics of Particle and Forces: Force resolution and r... Step 5: Part B) iii) Тве B Dividing both equation: tan 8 = 0.6 1.8 0 = 71.6⁰ Step 6: Part C) TBD cos 40 = 0.6mg TBD sin 40 = 1.8mg TBC TBD cos 0 0.6Tpc = 0.6mg TBD sin = 0.8TBC + W = 1.8mg Magnitude: √(0.6mg)² + (1.8mg)² = 1.9mg N DO Solution All solutions are provided in above steps BD & ASK AN EXPERT IS THIS ANSWER HELPFUL? √ MATH SOLVER
ion and Force An X b Answered: Statics of Particle and x 401Asg01: Statics of Particles X + bartleby.com/questions-and-answers/statics-of-particle-and-forces-force-resolution-and-representation-equilibrium-free-body-diagram-etc/6009b08d-1085-402a-82cc-75e0f5ac5ecd Homework help starts here! Q Search Statics of Particle and Forces: Force resolution and r... Step 5: Part B) iii) Тве B Dividing both equation: tan 8 = 0.6 1.8 0 = 71.6⁰ Step 6: Part C) TBD cos 40 = 0.6mg TBD sin 40 = 1.8mg TBC TBD cos 0 0.6Tpc = 0.6mg TBD sin = 0.8TBC + W = 1.8mg Magnitude: √(0.6mg)² + (1.8mg)² = 1.9mg N DO Solution All solutions are provided in above steps BD & ASK AN EXPERT IS THIS ANSWER HELPFUL? √ MATH SOLVER
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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![Ch01: Introduction and Force An X b Answered: Statics of Particle and X
← → C
27°C
Rain
401Asg01: Statics of Particles X +
bartleby.com/questions-and-answers/statics-of-particle-and-forces-force-resolution-and-representation-equilibrium-free-body-diagram-etc/6009b08d-1085-402a-82cc-75e0f5ac5ecd
Homework help starts here!
Q Search
Statics of Particle and Forces: Force resolution and r...
Step 5: Part B) iii)
Тас
Dividing both equation:
tan = 1.8
0.6
0 = 71.6º
TBC
TBD COS 0 = 0.6TBC = 0.6mg
TBD sin 0 = 0.8TBC + W = 1.8mg
Magnitude:
√(0.6mg)² + (1.8mg)² = 1.9mg N
Step 6: Part C)
TBD cos 40 = 0.6mg
TBD sin 40 = 1.8mg
B
Solution
BD
All solutions are provided in above steps
ASK AN EXPERT
IS THIS ANSWER HELPFUL?
√ MATH SOLVER
4
ENG
US
☐
10:54 PM
22/10/2023
⠀](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F5a264682-ba74-4b87-b848-c61521147a28%2F9842af03-4d21-4b95-a984-80762a74d928%2F1ljvu3q_processed.png&w=3840&q=75)
Transcribed Image Text:Ch01: Introduction and Force An X b Answered: Statics of Particle and X
← → C
27°C
Rain
401Asg01: Statics of Particles X +
bartleby.com/questions-and-answers/statics-of-particle-and-forces-force-resolution-and-representation-equilibrium-free-body-diagram-etc/6009b08d-1085-402a-82cc-75e0f5ac5ecd
Homework help starts here!
Q Search
Statics of Particle and Forces: Force resolution and r...
Step 5: Part B) iii)
Тас
Dividing both equation:
tan = 1.8
0.6
0 = 71.6º
TBC
TBD COS 0 = 0.6TBC = 0.6mg
TBD sin 0 = 0.8TBC + W = 1.8mg
Magnitude:
√(0.6mg)² + (1.8mg)² = 1.9mg N
Step 6: Part C)
TBD cos 40 = 0.6mg
TBD sin 40 = 1.8mg
B
Solution
BD
All solutions are provided in above steps
ASK AN EXPERT
IS THIS ANSWER HELPFUL?
√ MATH SOLVER
4
ENG
US
☐
10:54 PM
22/10/2023
⠀
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