The given system of knotted cords (label the left knot as J and the right knot as K) supports the indicated weights. a) What is the tension TA in cord A?b) What is the tension TB in cord B?c) What is the tension TC in cord C? please explain the solution in full details NowWeDiagramConsider Free BodyFree Body BodyFor XFree Body125°Тв155 80'SoodyneTA =TA =Con ConsiderJATA500=SmgoSm125°Now from équation500TA=SmrsSm80500 xTA =Sm/25°TA =5000.98480-8191492.4TA =0.81916011476 dyne601.15 dy neTBSmissxSm 80*AnsDSm80= 0.9848Sm 125° = 0·819) Again from equation (1)Тв500Sm/25Smi55TB =2 x Sm/155)Тв =1500Sm125°500 X 0.42260-8191211-3=0.8191257. 97 dyneTB 2Free Body DiagramJa175°100⁰ $145.1000 dyneJaw257.9660.dyneFor - YТсApplying SmeTeSmlooSm1451000 хSmlooTc =Sm115Tc= 1086 61 dyne1000 (2)Sm1151000 X 0.98480.9063

Lame's theorem It state that When three forces acting at a point are in equilibrium, then…

The given system of knotted cords (label the left knot as J and the right knot as K) supports the indicated weights. a) What is the tension TA in cord A? b) What is the tension TB in cord B? c) What is the tension TC in cord C?

The given system of knotted cords (label the left knot as J and the right knot as K) supports the indicated weights. a) What is the tension TA in cord A? b) What is the tension TB in cord B? c) What is the tension TC in cord C?

Principles of Physics: A Calculus-Based Text

5th Edition

ISBN:9781133104261

Author:Raymond A. Serway, John W. Jewett

Publisher:Raymond A. Serway, John W. Jewett

Chapter8: Momentum And Collisions

Section: Chapter Questions

Problem 24P

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Newton’s Third Law of Motion

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Question

The given system of knotted cords (label the left knot as J and the right knot as K) supports the indicated weights.

a) What is the tension T_A in cord A?

b) What is the tension T_B in cord B?

c) What is the tension T_C in cord C?

please explain the solution in full details

Now
We
Diagram
Consider Free Body
Free Body Body
For X
Free Body
125°
Тв
155 80'
Soodyne
TA =
TA =
Con Consider
JA
TA
500
=
Smgo
Sm125°
Now from équation
500
TA
=
Smrs
Sm80
500 x
TA =
Sm/25°
TA =
500
0.9848
0-8191
492.4
TA =
0.8191
6011476 dyne
601.15 dy ne
TB
Smiss
x
Sm 80*
Ans
D
Sm80= 0.9848
Sm 125° = 0·819)

Again from equation (1)
Тв
500
Sm/25
Smi55
TB =
2 x Sm/155)
Тв =
1
500
Sm125°
500 X 0.4226
0-8191
211-3
=
0.8191
257. 97 dyne
TB 2
Free Body Diagram
Ja
175°
100⁰ $145.
1000 dyne
Jaw
257.9660.dyne
For - Y
Тс
Applying Sme
Te
Smloo
Sm145
1000 х
Smloo
Tc =
Sm115
Tc= 1086 61 dyne
1000 (2)
Sm115
1000 X 0.9848
0.9063

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