Unlock document.
This document is partially blurred.
Unlock all pages and 1 million more documents.
Get Access
Problem 5.109 The rocket launcher is supported by
the hydraulic jack DE and the bearings Aand B. The
bearings lie on the xaxis and support shafts parallel to
the xaxis. The hydraulic cylinder DE exerts a force on
the launcher that points along the line from Dto E. The
coordinates of Dare (7, 0, 7) ft, and the coordinates of
Eare (9, 6, 4) ft. The weight WD30 kip acts at (4.5,
5, 2) ft. What is the magnitude of the reaction on the
launcher at E?3 ft
W
AB D
E
3 ft
y
x
Solution: The position vectors of the points D,Eand Ware
The vector parallel to DE is
The unit vector parallel to DE is
Since the bearings cannot exert a moment about the xaxis, the sum of
the moments due to the weight and the jack force must be zero about
the xaxis. The sum of the moments about the xaxis is:
From which
348
Problem 5.110 Consider the rocket launcher described
in Problem 5.109. The bearings at Aand Bdo not exert
couples, and the bearing Bdoes not exert a force in the
xdirection. Determine the reactions at Aand B.
Solution: See the solution of Problem 5.109. The force FDE can
be written
The equilibrium equations are
M⊲origin⊳D
ijk
ij k
The components of the moment eq. are
5.9997FDE C60 D0,
Problem 5.111 The crane’s cable CD is attached to a
stationary object at D. The crane is supported by the
bearings Eand Fand the horizontal cable AB. The
tension in cable AB is 8 kN. Determine the tension in
the cable CD.
Strategy: Since the reactions exerted on the crane by
the bearings do not exert moments about the zaxis, the
sum of the moments about the zaxis due to the forces
exerted on the crane by the cables AB and CD equals
zero. (See the discussion at the end of Example 5.9.)
Solution: The position vector from Cto Dis
The moment about the origin due to the forces exerted by the two
cables is
ijk
ij k
2 m 2 m
3 m
x
3 m
D
C
y
D
3 m
z
350
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 5.112 In Example 5.9, suppose that the cable
CE is shortened and its attachment point Eis moved to
the point (0, 80, 0) mm. The plate remains in the same
position. Draw a sketch of the plate and its supports
showing the new position of cable CE. Draw the free-
body diagram of the plate and apply equilibrium to deter-
mine the reactions at the hinges and the tension in the
cable.
y
B
C
D
A
⫺400j (N)
200 mm
200 mm
80 mm
100
mm
z
x
E
Solution: The sketch and free-body diagram are shown.
The vector from Cto Eis
The force exerted by cable CE can be expresses as
The equilibrium equations for the plate are
Problem 5.113 The plate is supported by hinges at A
and Band the cable CE, and it is loaded by the force at
D. The edge of the plate to which the hinges are attached
lies in the yzplane, and the axes of the hinges are
parallel to the line through points Aand B. The hinges
do not exert couples on the plate. What is the tension in
cable CE?
2i – 6j (kN)
Dx
C
B
A
E
z
y
3 m
1 m
2 m
Solution:
However, we just want tension in CE. This quantity is the only
unknown in the moment equation about the line AB. To get this, we
need the unit vector along CE.
y
FD = 2i – 6j
352
Problem 5.114 In Problem 5.113, the hinge at Bdoes
not exert a force on the plate in the direction of the hinge
axis. What are the magnitudes of the forces exerted on
the plate by the hinges at Aand B?
Solution: From the solution to Problem 5.113, TCE D8.15 kN
Also, from that solution,
y
F = + 2i – 6j (kN)
Problem 5.115 The bar ABC is supported by ball and
socket supports at Aand Cand the cable BD. The
suspended mass is 1800 kg. Determine the tension in
the cable.
y
x
z
A
B
C
(⫺2, 2, ⫺1) m
2 m 4 m
4 m
D
Solution: We take moments about the line AC to eliminate the
reactions at Aand C.
We have
TBD
Az
354
Problem 5.116* In Problem 5.115, assume that the
ball and socket support at Ais designed so that it
exerts no force parallel to the straight line from Ato
C. Determine the reactions at Aand C.
Solution: We have
TBD
Ay
Ax
Az
Cz
Cx
17.66 kN
Problem 5.117 The bearings at A,B, and Cdo not
exert couples on the bar and do not exert forces in the
direction of the axis of the bar. Determine the reactions
at the bearings due to the two forces on the bar.
x
y
C
B
200 i (N)
300 mm
180 mm
Solution: The strategy is to take the moments about Aand solve
the resulting simultaneous equations. The position vectors of the bear-
ings relative to Aare:
The sum of the moments about Ais:
MADrA1ðF1CrAB ðBCrA2ðF2CrAC ðCD0
x
z
CY
BZ
BX
CX
200 N
356
Problem 5.118 The support that attaches the sailboat’s
mast to the deck behaves like a ball and socket support.
The line that attaches the spinnaker (the sail) to the top
of the mast exerts a 200-lb force on the mast. The force
is in the horizontal plane at 15°from the centerline of the
boat. (See the top view.) The spinnaker pole exerts a 50-
lb force on the mast at P. The force is in the horizontal
plane at 45°from the centerline. (See the top view.)
The mast is supported by two cables, the back stay AB
and the port shroud ACD. (The forestay AE and the
starboard shroud AFG are slack, and their tensions can
be neglected.) Determine the tensions in the cables AB
and CD and the reactions at the bottom of the mast.
A
x
z
A
P
F
G
B
Top View (Spinnaker not shown)
15°
45°C
D
200 lb
50 lb
6 ft
Aft View
D
D
EB
P
y
x
Spinnaker
50 ft
Side View 21 ft 15 ft
A
C
G
P
CF
A
E
Solution: Although the dimensions are not given in the sketch,
rAC DrCrAD25j7.5k.
The forces acting on the mast are: (1) The force due to the spinnaker
(4) The force acting on the cross spar CE:
(5) The force due to the spinnaker pole:
FTX FTY FTZ
C
000.2873jTACj
C
ijk
066
7.1829 D389.81 lb,
5.118 (Continued )
The tension in cable CD is the vertical component of the tension
in AC,
jTCDjDjTACj⊲jÐeAC⊳DjTACj⊲0.9578⊳D373.37 lb.
QD31.14iC823.26j87.12k(lb)
z
z
FA
TAB
TCE
TCD
TCD
FA
FPFP
QYQB
358
Problem 5.119* The bar AC is supported by the cable
BD and a bearing at Athat can rotate about the axis AE.
The person exerts a force FD50j(N) at C. Determine
the tension in the cable.
Strategy: Use the fact that the sum of the moments
about the axis AE due to the forces acting on the free-
body diagram of the bar must equal zero.
(0.3, 0.4, 0.3) m
(0.3, 0.5, 0) m
(0.82, 0.60, 0.40) m
(0.46, 0.46, 0.33) m
B
C
E
A
x
y
Problem 5.120* In Problem 5.119, determine the
reactions at the bearing A.
component of MAparallel to the axis AE to equal zero.
Solution: See the previous problem for setup. We add the reactions
(force, moment)
We have the following 6 equilibrium equations:
Fx:AxC0.440TBD D0
MADMACrAB ðTBD CrAC ðFD0
MAx :MAx 5 N-m C⊲0.0440 m⊳TBD D0
MAy :MAy ⊲0.0366 m⊳TBD D0
Problem 5.121 In Active Example 5.10, suppose that
the support at Ais moved so that the angle between the
bar AB and the vertical decreases from 45°to 30°. The
position of the rectangular plate does not change. Draw
the free-body diagram of the plate showing the point
Pwhere the lines of action of the three forces acting
on the plate intersect. Determine the magnitudes of the
reactions on the plate at Band C.
4 ft
B
45
AC
Solution: The equilibrium equations are
360
Problem 5.122 The magnitude of the reaction exerted
on the L-shaped bar at Bis 60 lb. (See Example 5.11.)
(a) What is the magnitude of the reaction exerted on the
bar by the support at A?
(b) What are the xand ycomponents of the reaction
exerted on the bar by the support at A?
14 in
17 in
x
y
B
A
Solution: The angle between the line AB and the xaxis is
Problem 5.123 The suspended load weighs 1000 lb.
The structure is a three-force member if its weight is
neglected. Use this fact to determine the magnitudes of
the reactions at Aand B.
5 ft
B
A
10 ft
Solution: The pin support at Ais a two-force reaction, and the
roller support at Bis a one force reaction. The moment about Ais
MAD5B10⊲1000⊳D0, from which the magnitude at Bis BD
2000 lb. The sums of the forces:
FXDAXCBDAXC2000 D0,from which AXD2000 lb.
5 ft
A
10 ft
Problem 5.124 The weight WD50 lb acts at the
center of the disk. Use the fact that the disk is a three-
force member to determine the tension in the cable and
the magnitude of the reaction at the pin support.
60°
Solution: Denote the magnitude of the reaction at the pinned joint
362
Problem 5.125 The weight WD40 N acts at the
center of the disk. The surfaces are rough. What force
Fis necessary to lift the disk off the floor?
50 mm
F
150 mm
W
Solution: The reaction at the obstacle acts through the center of
the disk (see sketch) Denote the contact point by B. When the moment
F
150 mm
Problem 5.126 Use the fact that the horizontal bar is
a three-force member to determine the angle ˛and the
magnitudes of the reactions at Aand B. Assume that
0˛90°.
2 m
B
a30⬚
60⬚
364
Problem 5.127 The suspended load weighs 600 lb.
Use the fact that ABC is a three-force member to
determine the magnitudes of the reactions at Aand B.
45⬚
30⬚
3 ft
4.5 ft
A
B
C
Solution: All of the forces must intersect at a point.
Problem 5.128 (a) Is the L-shaped bar a three-force
member?
(b) Determine the magnitudes of the reactions at Aand B.
(c) Are the three forces acting on the L-shaped bar
concurrent?
250
mm
500
mm
700 mm
300 mm
150 mm
2 kN
3 kN-m
A
B
Solution: (a) No. The reaction at Bis one-force, and the reaction
at Ais two-force. The couple keeps the L-shaped bar from being a three
force member.(b) The angle of the member at Bwith the horizontal is
366
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.
