PROBLEM 4.4
For the bracket and loading of Problem 4.3, determine the smallest
distance a if the bracket is not to move.
PROBLEM 4.3 A T-shaped bracket supports the four loads shown.
Determine the reactions at A and B (a) if a = 10 in., (b) if a = 7 in.
SOLUTION
Free-Body Diagram:
For no motion, reaction at A must be downward or zero; smallest distance a for no motion
corresponds to A = 0.
ΣM B = 0: (40 lb)(6 in.) − (30 lb)a − (10 lb)(a + 8 in.) + (12 in.) A = 0
A=
(40a − 160)
12
A = 0: (40a − 160) = 0
a = 4.00 in. 
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348
PROBLEM 4.21
Determine the reactions at A and C when (a) α = 0, (b) α = 30°.
SOLUTION
(a)
α =0
From F.B.D. of member ABC:
ΣM C = 0: (300 N)(0.2 m) + (300 N)(0.4 m) − A(0.8 m) = 0
A = 225 N
or
A = 225 N 
ΣFy = 0: C y + 225 N = 0
C y = −225 N or C y = 225 N
ΣFx = 0: 300 N + 300 N + C x = 0
C x = −600 N or C x = 600 N
Then
C = C x2 + C y2 = (600) 2 + (225) 2 = 640.80 N
and
θ = tan −1 
 Cy 
−1  −225 
 = tan 
 = 20.556°
 −600 
 Cx 
or
(b)
C = 641 N
20.6° 
α = 30°
From F.B.D. of member ABC:
ΣM C = 0: (300 N)(0.2 m) + (300 N)(0.4 m) − ( A cos 30°)(0.8 m)
+ ( A sin 30°)(20 in.) = 0
A = 365.24 N
or
A = 365 N
60.0° 
ΣFx = 0: 300 N + 300 N + (365.24 N) sin 30° + C x = 0
C x = −782.62
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365
PROBLEM 4.21 (Continued)
ΣFy = 0: C y + (365.24 N) cos 30° = 0
C y = −316.31 N or C y = 316 N
Then
C = C x2 + C y2 = (782.62) 2 + (316.31) 2 = 884.12 N
and
θ = tan −1 
 Cy 
−1  −316.31 
 = tan 
 = 22.007°
 −782.62 
 Cx 
C = 884 N
or
22.0° 
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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366
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° 
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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369