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Where did the 4/5 and 3/5 came from on the solution? Can someone explain

Where did the 4/5 and 3/5 came from on the solution? Can someone explain

Elements Of Electromagnetics
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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Where did the 4/5 and 3/5 came from on the solution? Can someone explain, please. Thank you!
Two (2) huge circular mass are resting inside a larger container with a length equal to 9 meters. The radius of the ball weighing 10 kg is 2 meters, whereas the ball weighing 12 kg is 3 meters. Through your solutions, prove that the reactions at points A, B, C, and D are similar to the values shown below. Tip: first determine the angle of Reaction B with respect to x-axis (or the length components if you are familiar with using triangles) by using the radii and length of the container. Rb = 163.5 N Ra = 130.8 N Rd = 215.82 N Rc = 130.8 N Note: if you believe that the correct answer/s is/are not included above, justify your answer. 10 kg А 12 kg B 9.
Transcribed Image Text:Two (2) huge circular mass are resting inside a larger container with a length equal to 9 meters. The radius of the ball weighing 10 kg is 2 meters, whereas the ball weighing 12 kg is 3 meters. Through your solutions, prove that the reactions at points A, B, C, and D are similar to the values shown below. Tip: first determine the angle of Reaction B with respect to x-axis (or the length components if you are familiar with using triangles) by using the radii and length of the container. Rb = 163.5 N Ra = 130.8 N Rd = 215.82 N Rc = 130.8 N Note: if you believe that the correct answer/s is/are not included above, justify your answer. 10 kg А 12 kg B 9.
153.7 RAx ↑ RAY 152,09N T 划g N 12036N Now usirg Fore and Momert equilibruum equations At suffert A end B t EFX =0 RAX +4ax3 -120.36_132.ng TRey B EFyzo RAy + Rey -159.72+ 111.09 fis EMA =° (momenf at A=o) RBY X 30 + 141•09X7 – 4aX 3X(1s+1-5)– 159.79X4 RB4X {o =71.63 + 378 F 630.88 RBY X 1° = 239.25 RBy = 23.925N %3D how using equation Li) → RA 152.09 + 120-36 – 126 %3D = 147.26 using equntion lüv - Ray + Roy = 159.72 – 311. 69 %3D Ray = 4P.63- RB Ray = 48.63-23, §25 KBY= 23-9 5N) %3D (Ray = 24r705 N
Transcribed Image Text:153.7 RAx ↑ RAY 152,09N T 划g N 12036N Now usirg Fore and Momert equilibruum equations At suffert A end B t EFX =0 RAX +4ax3 -120.36_132.ng TRey B EFyzo RAy + Rey -159.72+ 111.09 fis EMA =° (momenf at A=o) RBY X 30 + 141•09X7 – 4aX 3X(1s+1-5)– 159.79X4 RB4X {o =71.63 + 378 F 630.88 RBY X 1° = 239.25 RBy = 23.925N %3D how using equation Li) → RA 152.09 + 120-36 – 126 %3D = 147.26 using equntion lüv - Ray + Roy = 159.72 – 311. 69 %3D Ray = 4P.63- RB Ray = 48.63-23, §25 KBY= 23-9 5N) %3D (Ray = 24r705 N
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