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)* 214.911b500Sine Rale,'-BAsinzo°·Sin OSO021491500Sin / sifr0)20°C5002149リ* 52.7 TNT 4GSUN ll llO O N78%1 6:51= bartlebyQ&AEngineering / Mechanical Engineeri... / Q&A Library /1. Determine the design angle e (e<...Get live help whenever you Try bartlebytutor todayneed from online tutors!BAP Phase ven fque .need from online tutors!tutor todayView transcribed image text ) ExpandExpert AnswerStep 1Solution:Step 2The free body diagram is shown below:TABneed from online tutors!tutor today600 lb20°500 lbThe equilibrium equation in the horizontal direction:++£F, = 0(6001bxcos 20°)+ 500 1b + [T,g × cos (20° +e)] =0x 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)ABStep 3Divide 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)=-205.211bТдв х(20°-9.076°) %3 -205.211bNegative sign indicates the directionof the force is pointing from B to A.Тів —1082.86 1bWas this solution helpful?

Answered: Image /qna-images/answer/f024baaa-b20c-466c-80c7-e29825b4d7e0.jpg

Answered: BA? Pese ve que1.

Engineering

Mechanical Engineering

BA? Pese ve que1.

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

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
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