Deportes

Problem 5.1 In Active Example 5.1, suppose that the
beam is subjected to a 6kN-m counterclockwise couple
at the right end in addition to the 4-kN downward force.
Draw a sketch of the beam showing its new loading.
Draw the free-body diagram of the beam and apply the
equilibrium equations to determine the reactions at A.

4 kN
A

2m

Solution: The equilibrium equations are
Fx : Ax D 0Fy : Ay

4 kN D 0

MA : MA

4 kN 2 m

C 6 kN-m D 0
Solving yields
Ax D 0
Ay D 4 kN
MA D 2 kN-m

Problem 5.2 The beam has a fixed support at A and is
loaded by two forces and a couple. Draw the free-body
diagram of the beam and apply equilibrium to determine
the reactions at A.

4 kN
A

2 kN

6 kN-m
60

1m

1.5 m

1.5 m

Solution: The free-body diagram is drawn.The equilibrium equations are
Fx : Ax C 2 kN cos 60° D 0
Fy : Ay C 4 kN C 2 kN sin 60° D 0
MA : MA C 6 kN-m C 4 kN 2.5 m C 2 kN sin 60° 4 m D 0
We obtain: Ax D

268

1 kN, Ay D

5.73 kN, MA D

22.9 kN-m

c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they
currently exist. No portion of thismaterial may be reproduced, in any form or by any means, without permission in writing from the publisher.

Problem 5.3 The beam is subjected to a load F D
400 N and is supported by the rope and the smooth
surfaces at A and B.
(a)
(b)

Draw the free-body diagram of the beam.
What are the magnitudes of the reactions at A
and B?

F
A

B
30°

45°

1.2 m
y

Solution:
FX D 0:

Acos 45°

FY D 0:

A sin 45° C B cos 30°

1.5 m

A

F
45°

x

B sin 30° D 0

1m

B

1.5 m

1.2 m

C

MA D 0:

1.2T

T

400 N D 0

1m
30°

T

2.7 400 C 3.7B cos 30° D 0

Solving, we get
A D 271 N
B D 383 N
T D 124 N

Problem 5.4 (a) Draw the free-body diagram of the
beam. (b) Determine the tension in the rope and the
reactions at B.

30

30 600 lbB

A

5 ft

9 ft

Solution: Let T be the tension in the rope.
The equilibrium equations are:
Fx :

T sin 30°

Fy : T cos 30°

600 lb sin 30° C Bx D 0
600 lb cos 30° C By D 0

MB : 600 lb cos 30° 9 ft

T cos 30° 14 ft D 0

Solving yields T D 368 lb, Bx D 493 lb, By D 186 lb

c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material isprotected under all copyright laws as they
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

269

Problem 5.5 (a) Draw the free-body diagram of the
60-lb drill press, assuming that the surfaces at A and B
are smooth.

Solution: The system is in equilibrium.
(a)
(b)

The free body diagram isshown.
The sum of the forces:

(b) Determine the reactions at A and B.
FX D 0,

FY D FA C FB

60 D 0

The sum of the moments about point A:
MA D

from which FB D

60 lb

10 60 C 24 FB D 0,
600
D 25 lb
24

Substitute into the force balance equation:
FA D 60

A

FB D 35 lb

B
10 in

14 in
60 lb

A

B
10 in

FA

270

14 in
FB

c 2008 Pearson Education,Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Problem 5.6 The masses of the person and the diving
board are 54 kg and 36 kg, respectively. Assume that
they are in equilibrium.
(a)
(b)Draw the free-body diagram of the diving board.
Determine the reactions at the supports A and B.

Solution:
(a)

AX D 0

FY D 0:

AY C BY

MA D 0:

A

FX D 0:

(b)

1.2BY

B

54 9.81

36 9.81 D 0

2.4 36 9.81
4.6 54 9.81 D 0

WP

WD

AX D 0 N

Solving:

1.2 m

AY D

1.85 kN

2.4 m
BY D 2.74 kN
4.6 m

4.6 m

2.4 m
1.2 m

AX

WP

y...