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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?

College Physics
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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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 

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)
Transcribed Image Text: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
Transcribed Image Text: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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