Problem 6.124 The truss supports a 400-N load at G.
Determine the axial forces in members AC,CD, and CF.400 N
300 mm
B
D
F
H
G
ECA
300 mm300 mm300 mm
600 mm
from which AyD400 N.
The method of joints: The angle from the horizontal of element BD is
˛CF D90 tan1300
600⊲1tan ⊳ D53.13°.
Joint B:
FxDBCBD cos D0,
Joint DJoint C
Joint BJoint A
BD
DF
AC CE
CD
CD
AD
CF
θ
AD
α
CF
α
from which CD D240 N ⊲C⊳
Joint C :
Problem 6.125 Consider the truss in Problem 6.124.
Determine the axial forces in members CE,EF,
and EH.
Solution: Use the results of the solution of Problem 6.124:
AC CE CE EG
EF
CF
CF
α
α
Problem 6.126 Consider the truss in Problem 6.124.
Which members have the largest tensile and compressive
forces, and what are their values?
Problems 6.124 and 6.125 except for members EG and GH. These are:
Joint E:
CE
EG
EG
400 N
EH
α
500
Problem 6.127 The Howe truss helps support a roof.
Model the supports at Aand Gas roller supports. Use
the method of joints to determine the axial forces in
members BC,CD,CI, and CJ.
2 m 2 m 2 m 2 m 2 m 2 m
6 kN
4 m
A
B
C
G
F
E
D
HI JKL
4 kN
4 kN
2 kN2 kN
Solution: The free body diagrams for the entire truss and the
required joints are shown.
Joint A:
6 kN
E
D
y
4 kN
4 kN
Joint C:
Problem 6.128 For the roof truss in Problem 6.127,
use the method of sections to determine the axial forces
Problem 6.139. The equations of equilibrium for the section are
FXDTCDuCDX CTCJuCJX CTIJ D0,
TIJ D12.0kN,
Problem 6.139.
2 kN
4 kN
C
D
TCD
TCJ
502
Problem 6.129 A speaker system is suspended from
the truss by cables attached at Dand E. The mass of
the speaker system is 130 kg, and its weight acts at G.
Determine the axial forces in members BC and CD.
G
E
C
A
BD
0.5 m 0.5 m 0.5 m0.5 m 1 m
1 m
Solution: The speaker as a free body: The weight of the speaker
from which DD425.1N.
The structure as a free body: The sum of the moments about Cis
1 m
0.5 m1 m
2 m
Cy
CY
CE
DE
DE
AC
CE
CD
BD
E
βββ
α
α
Problem 6.130 The mass of the suspended object is
900 kg. Determine the axial forces in the bars AB
and AC.
Strategy: Draw the free-body diagram of joint A.
y
z
x
D (0, 4, 0) m
A (3, 4, 4) m
C (4, 0, 0) m
B (0, 0, 3) m
Solution: The free-body diagram of joint Ais.
The position vectors from pt Ato pts B,C, and Dare
From the equilibrium equation
504
Problem 6.131 Determine the forces on member ABC,
presenting your answers as shown in Fig. 6.25. Obtain
the answer in two ways:
(a) When you draw the free-body diagrams of the
individual members, place the 400-lb load on the
free-body diagram of member ABC.
(b) When you draw the free-body diagrams of the
individual members, place the 400-lb load on the
free-body diagram of member CD.
C
B
200 lb
400 lb
1 ft
1 ft
1 ft
D
E
F
1 ft 1 ft
Solution: The angle of element BE relative to the horizontal is
˛Dtan11
2D26.57°.
MFD3DxCBcos ˛D0,
from which DxDBcos ˛
3.
MAD2Bcos ˛3⊲400⊳3CxD0.
The sum of the forces
FxD400 CCxCBcos ˛CAxD0,
from which AxD400 lb .
Element DEF : No changes. The changes in the solution for
Element ABC CxD800 lb when the external load is removed,
Problem 6.132 The mass mD120 kg. Determine the
forces on member ABC.ABC
D
E
m
200 mm200 mm
300 mm
Solution: The weight of the hanging mass is given by
Ay
Cy
506
c
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Problem 6.133 Determine the reactions on member
ABC at Band C.
0.2 m 0.2 m
A
B
C
D
E
4 kN
0.2 m
0.2 m 2 kN-m
Solution: We draw free-body diagrams for the entire structure, and
for members BD and ABC.
From the entire structure:
Problem 6.134 The truck and trailer are parked on a
10°slope. The 14,000-lb weight of the truck and the
8000-lb weight of the trailer act at the points shown.
The truck’s brakes prevent its rear wheels at Bfrom
turning. The truck’s front wheels at Cand the trailer’s
wheels at Acan turn freely, which means they do not
exert friction forces on the road. The trailer hitch at D
behaves like a pin support. Determine the forces exerted
on the truck at B,C, and D.
14 ft
ABC
D
y
x
4 ft 3 ft
5 ft 6 in
6 ft 8 kip
14 kip
10⬚
3 ft
2 ft
9 ft
Solution: We separate the two vehicles and draw a free-body
diagram of each. Starting with the trailer we have
508
Problem 6.135 The 600-lb weight of the scoop acts at
a point 1 ft 6 in to the right of the vertical line CE. The
line ADE is horizontal. The hydraulic actuator AB can
be treated as a two-force member. Determine the axial
force in the hydraulic actuator AB and the forces exerted
on the scoop at Cand E.
1 ft 6 in
2 ft 6 in
1 ft
5 ft
2 ft
C
B
DE
A
Scoop
Solution: The free body diagrams are shown at the right. Place the
coordinate origin at Awith the xaxis horizontal. The coordinates (in
ft) of the points necessary to write the needed unit vectors are A(0,
0), B(6, 2), C(8.5, 1.5), and D(5, 0). The unit vectors needed for
uBC D0.981i0.196j,
and uBD D0.447i0.894j.
The scoop: The equilibrium equations for the scoop are
TCB
C
EY
1.5 ft 1.5 ft
600 lb
y
Problem 6.136 Determine the force exerted on the bolt
by the bolt cutters.
100 N
55 mm
75
mm
40 mm
A
C
D
B
Solving the equations simultaneously (we have extra (but compatible)
equations, we get FD1051 N, AXD695 N, AYD1586 N, BXD
695 N, BYD435 N, CXD695 N, CYD535 N, DXD695 N,
and DyD535 N
DX
DY
D
Problem 6.137 For the bolt cutters in Problem 6.136,
determine the magnitude of the force the members exert
on each other at the pin connection Band the axial force
in the two-force member CD.
510