PROBLEM 4.23
Determine the reactions at A and B when (a) h = 0,
(b) h = 200 mm.
SOLUTION
Free-Body Diagram:
ΣM A = 0: ( B cos 60°)(0.5 m) − ( B sin 60°)h − (150 N)(0.25 m) = 0
37.5
B=
0.25 − 0.866h
(a)
(1)
When h = 0,
B=
From Eq. (1):
37.5
= 150 N
0.25
B = 150.0 N
30.0°
ΣFy = 0: Ax − B sin 60° = 0
Ax = (150)sin 60° = 129.9 N
A x = 129.9 N
ΣFy = 0: Ay − 150 + B cos 60° = 0
Ay = 150 − (150) cos 60° = 75 N
A y = 75 N
α = 30°
A = 150.0 N
(b)
A = 150.0 N
30.0°
When h = 200 mm = 0.2 m,
From Eq. (1):
B=
37.5
= 488.3 N
0.25 − 0.866(0.2)
B = 488 N
30.0°
ΣFx = 0: Ax − B sin 60° = 0
Ax = (488.3) sin 60° = 422.88 N
A x = 422.88 N
ΣFy = 0: Ay − 150 + B cos 60° = 0
Ay = 150 − (488.3) cos 60° = −94.15 N
A y = 94.15 N
α = 12.55°
A = 433.2 N
A = 433 N
12.55°
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