1. The toggle pliers are used for a variety of clamping purposes. For the handle
position given by a=10o and for a handle grip P=150 N, calculate the clamping
force C produced. Note that pins A and D are symmetric about the horizontal
centerline of the tool.
2. The elements of a floor jack are shown in the figure. CDFE forms a
parallelogram. Calculate the force in the hydraulic cylinder AB corresponding to
the 10 kN load. What is the force in link EF?
3. The elements of a rear suspension for a front-wheel-drive car are shown in
the figure. Determine the magnitude of the force at each joint if the normal
force F exerted on the tire has a magnitude of 3600 N.
4. Determine the horizontal and vertical components of force that the pins at A,
B, and D exert on the A-frame.
1200 N
0.5 m 0.5 m 0.5 m 0.5 m
1 m
1 m
0.75 m
0.25 m
A E
B D
C
1600 N
5. The motion of the backhoe bucket shown is controlled by the hydraulic cylinders AD,
CG and EF. As a result of an attempt to dislodge a portion of a slab, a 13-kN force P is
exerted on the bucket teeth at J. Determine the force exerted by each cylinder.
6. A basketball hoop whose rim
height is adjustable is shown. The
supporting post ABCD weighs 400
N with the center of gravity at
point C, and backboard-hoop
assembly weighs 220 N with the
center of gravity at point G. The
height of the rim is adjustable by
means of the screw and hand crank
IJ, where the screw is vertical. If a
person with 800 N weight hangs on
the rim, determine the support
reactions at D and the forces
supported by all members.
Hint: Member IJ is a two-force
member.
7. The mechanism in the figure is used to raise the bucket of a bulldozer. The bucket and its contents
weigh 10 kN and have a center of gravity at H. Arm ABCD has a weight of 2 kN and a center of gravity
at B, arm DEFG has a weight of 1 kN and a center of gravity at E. The weights of the hydraulic cylinders
can be neglected. Determine the forces in the hydraulic cylinders CJ, BF and EI and also determine all
the forces acting at arm DEFG.
Two-force members: CJ, BF, EI.
FBD of bucket
FEI
Gx
Gy
10 kN
kNGGF
kN.GFGF
kN.Fcos.F.M
yyy
xEIxx
EIEIG
100100
88200
8820302130100
FBD of hydraulic cylinder EI
FEI FEI I
E
FBD of DEFG
kN.DsinFDGF
kN.DFcosFDGF
kN.F
sin.Gcos.Gsin.cosFcos.sinFsin.cos.F
M
yBFyyy
xEIBFxxx
BF
y
.
xBFBF
.
EI
D
61701910
8590190
4210
030813081302171302171306013060
0
10882882
30o
Gx Gy
1 kN
FEI
Dx
Dy
FBF
30o 30o
19o
G
F
E
D
D
B
F
oo ,sin
.
sin
.
sin
.
m.BFcos....BF
794160
5918121
59160218122181 222
aa
a
FBD of entire system
kNAAF
kNAAFF
kNF
sin.sin.sin.sin.sin..sin.F
M
yyy
xxCJx
CJ
CJ
A
13010120
1800
18
0309023072306013072308130106081
0
Ax
10 kN
P=FCJ
Ay
1 kN
2 kN
8. The figure shows a special rig
designed to erect vertical sections of a
construction tower. The assembly A has
a weight of 15 kN and is elevated by the
platform B, which itself has a weight of
20 kN. The platform is guided up the
fixed vertical column by rollers and is
activated by the hydraulic cylinder CD
and links EDF and FH. For the
particular position shown, calculate the
force exerted by the hydraulic cylinder
at D and the magnitude of the force
supported by the pin at E.
15 kN
20 kN
15 kN
20 kN
FBD of HF
FHF
Two-force members: CD, HF.
(FHF) y=35 kN
(FHF) x
1.25 m
3 m
a
kN.F.F
.
FHFH
o
91637253
335
61922
a
E
FBD of EDF
1 m
3 m
kN.F,. FHo 9163761922 a
Ey
Ex
FHF
a
FCD
0.7
5 m
D
C
o..
sin 4318163
1
kN.EcosFcosFEF
kN.EsinFsinFEF
kN.F.sinFcosF.sinFcosF
M
yCDFHyy
xCDFHxx
CDFHFHCDCD
E
742200
833300
8760025237501
0
a
a
aa
9. The elements of a stump grinder with a total mass (exclusive of the hydraulic cylinder DF and arm
CE) of 300 kg with mass center at G are shown in the figure. The mechanism for articulation about a
vertical axis is omitted, and the wheels at B are free to turn. For the nominal position shown, link CE is
horizontal and the teeth of the cutting wheel are even with the ground. If the magnitude of the force F
exerted by the cutter on the stump is 400 N, determine the force P in the hydraulic cylinder and the
magnitude of the force supported by the pin at C.
FBD of entire machine
Two-force member: DF.
Ey
Ex
By F
20o
N.EsinF.BEF
N.EcosFEF
N.B...B.sinF.cosF
M
yyyy
xxx
yy
E
597150208193000
873750200
7835210052819300551252206020
0
400400
FBD of EDC
Ey
Ex
P D
N.CsinPCEF
N.CcosPECF
N.P.sinP.cosP.EM
y.
y
.
yy
x..
xxx
.
yC
0516300
15274800
5431720901503510
54317259715
54317287375
59715
a
a
aa
E C
Cy
Cx
a
D
F
a
1300 mm
23
0 m
m
o.tan 03101300
230 aa
10. If the forces shown in the figure are applied to the digger at point G, find the
forces in the hydraulic cylinders HB and CD.
(1) FBD of entire digger
Ay
Ax
FHB
kN.Fcos.sinFsin.cosF.
M
kN.F.sinF.cosF.
M
CDCDCD
K
HBHBHB
A
492407025015702501531325
0
423208135123590355
0
(1)
(2)
(2) FBD of bucket+FK+EC+CD+CJ
Kx
Ky
FCD
15o
11. The linkage shown is used on a garbage truck to lift a 9000 N dumpster. Points A-G are
pins, and member ABC is horizontal. For the position shown, when the dumpster just fully
lifts off the ground, determine the force hydraulic cylinder CF. Roller at G is contact with the
dumpster.
Ax
Ay
FG
FBD of dumpster
NAFAF
NAFAF
NFFM
yGyy
xGxx
GGA
6910045sin90000
15910045cos0
225000)600()1500(90000
Ex
Ey
FCD
FG
FBD of member EDG
NEEF
NEEF
NFFM
yyy
xxx
GCDE
5.79549045sin2250045cos1125000
63640045cos11250045cos225000
1125000)300()1500(225000
Bx
By
FCF
Ex
Ey
From equilibrium of whole system;
NF
FEE
M
CF
CFyx
E
68.118724
0)450(45cos)45cos900900(450)45cos900()1950(900
0
12. A hydraulic lift-table is used to raise a 1000 kg crate. It consists of a platform and two
identical linkages on which hydraulic cylinders exert equal forces. In the figure only one
linkage and one cylinder are shown. Members EDB and CG are each of length 2a and
member AD is pinned to the midpoint of EDB. If the crate is placed on the table, so that
half of its weight is supported by the system shown, determine the force exerted by each
cylinder in raising the crate for q=60o, a=0.70 m and L=3.20 m. Show that the result
obtained is independent of the distance d.
FBD of Platform
Two-Force Members: AD, BC, CG, DH.
30o
q=60o
FAD
W/2
FC FB
A B C
BCCBCBy
ADADx
FW
FW
FFW
FFF
FFFx
220
20
000
Pulley C FC
C
FCG
30o
FBC
300300
0300
cosFFFcosFF
FsinFF
CGCCCGy
BCCGx
30o
q=60o
FBD of CG
C
G
FCG
q=60o 30
30
cos
FF
sin
FF
CCG
BCCG
FBD of BC
FCG
FBC FBC
30tanF
F
C
BC
30o
q=60o
N.F
FF.F.
.tanF.FF.
.F.F..sinF..cosFM
DH
/W
CBDH
F
CBDH
BCBDHDHE
BC
635124
070670
02113070670
021170350021261002120
2
FBD of EDB
B
E
FB
q=60o
Ex
o.sin
.
sin
.
m.DHcosEHDEEHDEDH
021260
91270
912602222
aa
D
H
a
Ey
FBC
FDH
0.7 m
0.61 m
1.4 m
Cosine theorem:
Sine theorem:
13. The mechanism is designed to keep its load level while raising it. A pin on the rim of the 80 cm
diameter pulley fits in a slot on arm ABC. Arm ABC and DE are each 80 cm long and the package
being lifted weighs 80 kN. The mechanism is raised by pulling on the rope that is wrapped around the
pulley. Determine the force P applied to the rope and all the forces acting on arm ABC when the
package has been lifted 80 cm as shown.
SOLUTION
FBD of Platform Two force member: ED
69.28 mm
FBD of DE
FBD of ABC
FBD of the pulley
14. A simple folding chair comprised of two identical frames, one on each side, is shown in
the figure. The half frame shown carries half the weight of a 70 kg person. Determine all
the forces acting on member BF. Neglect the weights of the connecting elements (not
shown) and the seat, and also the friction at points A, B and F.
Detail of contact at F
Dimensions in “mm”
Geometry of the Chair
SOLUTION
Two force member: EG
Equilibrium of the whole system:
FBD of DC
GE two force member
FBD of BF
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