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BA? Pese ve que1.

BA? Pese ve que1.

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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The first photo was the screenshot of my question here in Bartleby. But in other site (chegg), I saw the same problem but the answers are different. On another site also, the answers are the same as of chegg—however, "member AB" was stated, and not BA.

I am hoping for clarification on what process to use, and why. Thank you

 

Sol:- Cosine Rle; FSA =I 60021 So02-2(600) (sco) cose) * 214.911b 500 Sine Rale,'- BA sinzo° ·Sin O SO0 21491 500 Sin / sifr0) 20°C 500 2149リ * 52.7
Transcribed Image Text:Sol:- Cosine Rle; FSA =I 60021 So02-2(600) (sco) cose) * 214.911b 500 Sine Rale,'- BA sinzo° ·Sin O SO0 21491 500 Sin / sifr0) 20°C 500 2149リ * 52.7
TNT 4G SUN ll ll O O N78% 1 6:51 = bartleby Q&A Engineering / Mechanical Engineeri... / Q&A Library / 1. Determine the design angle e (e<... Get live help whenever you Try bartleby tutor today need from online tutors! BAP Phase ven fque . need from online tutors! tutor today View transcribed image text ) Expand Expert Answer Step 1 Solution: Step 2 The free body diagram is shown below: TAB need from online tutors! tutor today 600 lb 20° 500 lb The equilibrium equation in the horizontal direction: ++£F, = 0 (6001bxcos 20°)+ 500 1b + [T,g × cos (20° +e)] =0 x cos (20°+0) =-1063.821b. -(1) The equilibrium equation in the vertical direction: ↑ +EF, =( (6001bx sin 20°) +[Tgx sin ( 20°+0)]=0 [Tg x sin (20° +0)]= +0)=-205.211b. -(2) AB Step 3 Divide equation (2) by equation (1) to determine the angle: sin (20°+0) cos( 20°+0) -1063.821b -205.211b (20°+0)= tan (0.193) e = -9.076° Thus, the force in member AB is calculated as: Tg X sin (20°+0)=-205.211b Тдв х(20°-9.076°) %3 -205.211b Negative sign indicates the direction of the force is pointing from B to A. Тів —1082.86 1b Was this solution helpful?
Transcribed Image Text:TNT 4G SUN ll ll O O N78% 1 6:51 = bartleby Q&A Engineering / Mechanical Engineeri... / Q&A Library / 1. Determine the design angle e (e<... Get live help whenever you Try bartleby tutor today need from online tutors! BAP Phase ven fque . need from online tutors! tutor today View transcribed image text ) Expand Expert Answer Step 1 Solution: Step 2 The free body diagram is shown below: TAB need from online tutors! tutor today 600 lb 20° 500 lb The equilibrium equation in the horizontal direction: ++£F, = 0 (6001bxcos 20°)+ 500 1b + [T,g × cos (20° +e)] =0 x cos (20°+0) =-1063.821b. -(1) The equilibrium equation in the vertical direction: ↑ +EF, =( (6001bx sin 20°) +[Tgx sin ( 20°+0)]=0 [Tg x sin (20° +0)]= +0)=-205.211b. -(2) AB Step 3 Divide equation (2) by equation (1) to determine the angle: sin (20°+0) cos( 20°+0) -1063.821b -205.211b (20°+0)= tan (0.193) e = -9.076° Thus, the force in member AB is calculated as: Tg X sin (20°+0)=-205.211b Тдв х(20°-9.076°) %3 -205.211b Negative sign indicates the direction of the force is pointing from B to A. Тів —1082.86 1b Was this solution helpful?
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