How can I differentiate between internal and external forces in fbds. In problem 5.49 what is the difference between the two cases

Elements Of Electromagnetics
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How can I differentiate between internal and external forces in fbds. In problem 5.49 what is the difference between the two cases 

Problem 5.48 The tension in cable BC is 100 lb.
Determine the reactions at the built-in support.
FBD:
Ay
Ax
Ма
200
← Note tension
is internal
to this FBD.
2=
300
A
B
200 lb
► 3 ft →3 ft →◄
6 ft
6 ft
300 ft-lb
Ax
EFx = A₂ = 0 → Ax=0
εF₁ = Ay - 200=0 → Ay = 20016.
EMA² = M₁ - 300 - 200 (3) = 0 → MA = 900 lb·ft
Transcribed Image Text:Problem 5.48 The tension in cable BC is 100 lb. Determine the reactions at the built-in support. FBD: Ay Ax Ма 200 ← Note tension is internal to this FBD. 2= 300 A B 200 lb ► 3 ft →3 ft →◄ 6 ft 6 ft 300 ft-lb Ax EFx = A₂ = 0 → Ax=0 εF₁ = Ay - 200=0 → Ay = 20016. EMA² = M₁ - 300 - 200 (3) = 0 → MA = 900 lb·ft
Problem 5.49 The tension in cable AB is 2 kN. What
are the reactions at C in the two cases?
* Assume all components
are weight less.
a) FBD:
T
{√₁₂₁²= = =²/²₂T+C₁₂₁₂=0
Cx ==T=IKN
≤E₁₂x = C²₂₂=0
EFy = C₁ = 0
Cy
A
'Cy
60° B C
Mc
Jocx
-2 m
Me
(a)
1 m
2
EM₂²² = M₁ = 0
Cx
A
60° BC
- 2 m
✦1m
EM ₂²/² = M₁₂ - T (3) - VT (1) =0 → Mc = 7₁, 73 kN.m
b) FBD:
tension is
internal to this FBD
(b)
EFy=C₂₂+T+₂ ²/² T = 0
C₁ = -3.73 KN
the
be
4
Popular
ww
Transcribed Image Text:Problem 5.49 The tension in cable AB is 2 kN. What are the reactions at C in the two cases? * Assume all components are weight less. a) FBD: T {√₁₂₁²= = =²/²₂T+C₁₂₁₂=0 Cx ==T=IKN ≤E₁₂x = C²₂₂=0 EFy = C₁ = 0 Cy A 'Cy 60° B C Mc Jocx -2 m Me (a) 1 m 2 EM₂²² = M₁ = 0 Cx A 60° BC - 2 m ✦1m EM ₂²/² = M₁₂ - T (3) - VT (1) =0 → Mc = 7₁, 73 kN.m b) FBD: tension is internal to this FBD (b) EFy=C₂₂+T+₂ ²/² T = 0 C₁ = -3.73 KN the be 4 Popular ww
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