Question to answer: Can you explain to me the process for this problem and its solution? I was confused and I need to understand how it end up and result like that. 1st pic is the problem then next is the solution. Please, respond immediately. Thank you so much! O665°A120500Nfore the knot Ki-Box,95°loooNIn the Horisontal Dix. Net Ford =0So,esin 15°B sin 80 = 0on,c sin 15° = B sin 80°c = 3.8 x8In the Vertical Dik. Net fasa = 050,e cos 15² + 8eos 80° <= 1000on, (3.8xB) x eos 15°OP, Bx [3.67+ 0.174] = 1000010,B= 260. 14 NewtonC = 3.8xB = 988.55 Newton.Horizontal Dise. Net foscee =0A sin 25⁰ =B sin 80A =2.33xB= 606.12 Newton.for the knot J:In the50,on,75°TA = 606-12 N10015°80°By simple geometrical Calculations.4 Beas 80°¿cos IsºIB→ sin 15°B sin 80°+ B cos 80 = 10009 TB =Asin 20.= 260.14N, Tc = 988-55NAm.80°1000 NFree Body DiagramA K2510080> Bsingo500NBPriee Body Diagram 1B. 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?65AC1000 dyn125°500 dyn8

The system is in equilibrium, so net force is zero or forces are balanced at each point.

O O 65° 120 500 № fore the knot K- ox, 07/ OP, 010, B K loooN In the Horisontal Dix. Net force =0 So, esin 15° - B sin 80 = 0 on, c sin 15° = 8 sin 80⁰° c = 3.8 x 8 In the Vertical Dik. Net fasa = 0 50, e cos 15² + 8 eos 80 = 1000 (3.8xB) x ws 15° + B cos 80 = 1000 Bx [3.67+ 0.174] <= 1000 260. 14 Newton C = 3.8x8 = 988.55 Newton. Horizontal Dis. Net forcee =0 A sin 250 8 sin 80 = 2.33xB = 606.12 Newton. for the knot J: In the 50, on, A= 95⁰ 75° с TA= 606-12 N 15° 100⁰ 80° By simple geometrical Calculations. 4 Beas 80 8 -csin 15° B sin 80° 80° Asin 25⁰ , TB = 260.14N, T₂ = 988.55N Am. A K هير بن - 1000 N Free Body Diagram 80 > Bsingo 500N B Pree Body Diagram

O O 65° 120 500 № fore the knot K- ox, 07/ OP, 010, B K loooN In the Horisontal Dix. Net force =0 So, esin 15° - B sin 80 = 0 on, c sin 15° = 8 sin 80⁰° c = 3.8 x 8 In the Vertical Dik. Net fasa = 0 50, e cos 15² + 8 eos 80 = 1000 (3.8xB) x ws 15° + B cos 80 = 1000 Bx [3.67+ 0.174] <= 1000 260. 14 Newton C = 3.8x8 = 988.55 Newton. Horizontal Dis. Net forcee =0 A sin 250 8 sin 80 = 2.33xB = 606.12 Newton. for the knot J: In the 50, on, A= 95⁰ 75° с TA= 606-12 N 15° 100⁰ 80° By simple geometrical Calculations. 4 Beas 80 8 -csin 15° B sin 80° 80° Asin 25⁰ , TB = 260.14N, T₂ = 988.55N Am. A K هير بن - 1000 N Free Body Diagram 80 > Bsingo 500N B Pree Body Diagram

Physics for Scientists and Engineers: Foundations and Connections

1st Edition

ISBN:9781133939146

Author:Katz, Debora M.

Publisher:Katz, Debora M.

Chapter6: Applications Of Newton’s Laws Of Motion

Section: Chapter Questions

Problem 11PQ: In Problem 10, the mass of the sign is 25.4 kg, and the mass of the potted plant is 66.7 kg. a....

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