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