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.

FBD of Upper Handle

FBD of Lower Handle

P=150 N

P=150 N

Ay

By

By

Dy

Symmetry axis

NA

A

M

y

y

B

46.4103

010sin2010sin2010cos100150

0

10o

FBD of Upper Jaw

Ay

Oy

C

NCC

MO

82.13670602046.4103

0

+

Mechanical Advantage:

+

12.9150

82.1367

B

2. 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.

Two-force member: CD

FCD

Bx

By

a

x

y

09.12280

60tan 1

a

NF

FF

M

CD

CDCD

B

1898

0)70(sin)305(cos)165(3600

0

aa

NB 441040001856 22

FBD of CD FCD

FCD

FBD of Wheel + BC

NBFBF

NBFBF

yCDyy

xCDxx

400003600sin0

18560cos0

a

a

C

D

+

NF

F

M

EF

EF

A

5920

0)350(4000)260(sin

0

Bx

By

30.65200

435tan 1

FEF

Ax

Ay

FBD of member AEBx

y

NAFAF

NAFAF

yEFyy

xEFxx

138404000sin0

433001856cos0

NA 455013844330 22

+

3. Calculate the x- and y-components of all forces acting on each member of the

loaded frame.

4. The motion of the backhoe bucket is controlled by the hydraulic cylinders AB, DE, and FI.

Determine the forces exerted by pin C and cylinders AB and IF in supporting the 7.5 kN load shown.

5. The device shown is an overload prevention mechanism. When the force acting on the

smooth peg at D reaches 1 kN, the peg will be sheared, allowing the jaws at C to open and

thereby releasing the eye-bolt. Determine the maximum value of the tension P that can be

applied without causing the eye-bolt to be released. Neglect friction.

6. The design of a hoisting

mechanism for the dump truck

is shown in the enlarged view.

Determine the compression P in

the hydraulic cylinder BE and

the magnitude of the force

supported by the pin at A for the

particular position shown,

where BA is perpendicular to

OAE and link DC is

perpendicular to AC. The dump

and its load together have a

mass of 9 Mg with center of

mass at G.

FCD

W=9000(9.81) N

Ox

Oy

Two-force members : DC and BEFBD of Dump

kNNF

FF

M

CD

CDCD

O

797.78856.78796

02225.69cos7475.69sin9605.24sin81.9900012005.24cos81.99000

0

+

69.5o

Two-force members : DC and BEFBD of ABC

kNFFFM BEBECDA 865.1140300312

1223000

22

+

FCD

69.524.5=45o

Ax

Ay

123

FBE300 mm

300 mm

kNAAFFF xxBECDx 717.550312

1245cos0

22

kNAAFFF yyBECDy 859.270312

345sin0

22

7. 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

o87.36

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

1200 N

FBD of Entire Frame

Ay

Ax o87.36

N

NA

NA

F

x

x

x

1500

0sin1200

0

NA

NA

F

y

y

y

800

0cos16001200

0

NN

N

M A

4500

02cos25.2160075.0120025.21200

0

+

Ay

38005.0

05.0112

0

yx

yxyx

C

BB

BBAA

M

FBD of ABC

Ay=800 N

Ax=1500 N

FBD of BD

Bx

By

Cy1

Cx1

A

B

C

B DBx

By

Dx

Dy

1600 N

1

+

2

NB

BM

y

yD

1200

0175.016000

+

3

ND

DBF

y

yyy

2800

016000

1 → NBx 3200

4

NBD

DBF

xx

xxx

3200

00

8. 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.

800 N24

7

220 N

400 N

Dy

MD

800 N

24

7

224 N

768 N

FBD of Entire Structure

ND

DF

x

xx

224

02240

x

y

ND

D

F

y

y

y

1388

0768220400

0

mNM

M

M

D

D

D

2.1867

035.1768322472.0220

0

+

Dx

Dx

Two-force members : IJ and AE

NF

F

M

AE

AE

F

19.1607

0)75.0(768)12.0(22048.0cos

0

220 N

800 N

24 7

224 N

768 N

Fx

Fy

FAE

FBD of member EFH

(backboard-hoop assembly)

E

A

0.6 m

0.4

8 m

38.66o

NF

FF

F

x

xAE

x

1031

0224cos

0

19.1607

F

+

NF

FF

F

y

yAE

y

1992

0768220sin

0

NF 2243)1992()1031( 22

FAE

FIJ

FIJ

NF

FFFM

IJ

yxIJB

4669

0)60.0()48.0(15.00

FBD of member IBF

FxF

B

I

Fy

Bx

By

FIJ

NB

NBFBFF

NFBFBF

yyyIJy

xxxxx

6740)6661()1031(

666100

103100

22

+

IJ: Two-force member

9. 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

FEIFEI IE

+

E

FBD of DEFG

kNF

GGFFF

M

BF

yxBFBFEI

D

42.10

030sin8.130cos8.130sin2.171cos30cos2.171sin30sin6.0130cos6.0

0

1088.288.2

30o

GxGy

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

+

kNDFDGF

kNDFFDGF

yBFyyy

xEIBFxxx

61.7019sin10

85.9019cos0

FBD of entire system

kNF

F

M

CJ

CJ

A

18

030sin9.0230sin7.230sin6.0130sin7.230sin8.13.01060sin8.1

0

Ax

10 kN

P=FCJ

Ay

1 kN

2 kN

+

kNAAF

kNAAFF

yyy

xxCJx

13010120

1800

10. 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

FHF

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

+

11. 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

PD

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

12. 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