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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1st pic is the problem then next is the solution. Please, respond immediately. Thank you so much!
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