5. For the structure shown, if h=200 mm, determine the reactions at A and B. Answer: RA= 519 N 12.6° RB-586 N 30° 300 mm 250 mm 250 mm 180 N

Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
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I got stuck with this problem. Can you help? I fear I didn't identify all the forces required

5. For the structure shown, if h=200 mm, determine
the reactions at A and B.
(Answer: RA=519 N
12.6° RB 586 N 30°
300 mm
T
250
mm
с
250
mm
180 N
oy.
160⁰
Transcribed Image Text:5. For the structure shown, if h=200 mm, determine the reactions at A and B. (Answer: RA=519 N 12.6° RB 586 N 30° 300 mm T 250 mm с 250 mm 180 N oy. 160⁰
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I was getting closer to the answer but for some reason, I get a different result despite I found all the right forces. Could you please tell me what I am doing wrong? As there are no all the steps how you got to the final result. Also, why Rbcos60 has a negative sign? Isn't it making a positive moment so it should have a positive sign instead?

Q3m
Freebody diagram
0.25m 0.25m
Ax
A4
A
G
Take
+5 EM₂₁₂=0
Mg
B
Mathematical Representation.
mament at
up 5 the
น
A
X
RB5n60
Mg-0.25
9
(sin60.0.2 + c0560.0.5.
160 RBC0560
+ (RBSINGO.0.2) + (RBC0660 0.5)-(Mg⋅ 0.25) = 0
RB (Sin60.0.2 + Cos60.0.5). Me .0.25
€0560.0.5) -
RB-
Transcribed Image Text:Q3m Freebody diagram 0.25m 0.25m Ax A4 A G Take +5 EM₂₁₂=0 Mg B Mathematical Representation. mament at up 5 the น A X RB5n60 Mg-0.25 9 (sin60.0.2 + c0560.0.5. 160 RBC0560 + (RBSINGO.0.2) + (RBC0660 0.5)-(Mg⋅ 0.25) = 0 RB (Sin60.0.2 + Cos60.0.5). Me .0.25 €0560.0.5) - RB-
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