1. For a gripping force of 90 N, determine the normal force N

exerted on the round stock by each jaw. Also determine the

magnitude of the force supported by the pin at O. (4/83)

N

FBD of lower piece

NN

N

MO

360

0)100()400(90

0

+

NO

NOONF

NOONF

yyy

xxx

400

13.21309020sin0

29.338020cos0

360

360

normal force N exerted on the round stock by each jaw and force supported by the pin at O

Ox

Oy

20°

2. The toggle pliers are used for a variety of clamping

purposes. For the handle position given by a = 10° and for a

handle grip P = 150 N, calculate the clamping force FC

produced.

Note that pins

A and D are

symmetric

about the

horizontal

centerline of

the tool.

a = 10°, P = 150 N, calculate the clamping force FC produced. All loading is vertical and symmetric.

FBD of lower handle

B

a

Bcos10 Bsin10

Dy

Dx

NBBMD 46.42530)20(10sin)100(10cos1500 +

NDDBF

DDF

yyy

xxx

46.410301500

000

a = 10°, P = 150 N, calculate the clamping force FC produced. All loading is vertical and symmetric.

FBD of DOC

Dy FC

NFFDM CCyO 82.13670)60()20(0 +

O

C

3. Compute the force

in link AB of the

lifting tongs which

cross without

touching.

force in link AB, tongs cross without touching

FBD of Joint E

FED

400(9.81)=3924 N

FEC

DAF

E

D 360

540

33.69° 33.69° 33.69°

NFFFF

NFFAFF

kNNA

AsnFF

M

yyEDyy

xEDxxxx

x

xEDED

F

19620

18.515000

09.814.8093

0)720()990(69.33)1320(69.33cos

0

NFFF

FFF

EDEDy

EDECx

3537392469.33sin20

0

FED

FED

FED

Ax

ED

Fx

Fy

4. 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. (4/122)

magnitudes of forces at each joint

FCD

Bx

By

a

x

y

09.12280

60tan 1

a

NBFBF

NBFBF

NF

FF

M

yCDyy

xCDxx

CD

CDCD

B

400003600sin0

18560cos0

1898

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

0

a

a

aa

NB 441040001856 22

CD

FCD

FCD

FBD of wheel + BC

NA

NAFAF

NAFAF

NF

F

M

yEFyy

xEFxx

EF

EF

A

455013844330

138404000sin0

433001856cos0

5920

0)350(4000)260(sin

0

22

magnitudes of forces at each joint

Bx

By

x

y

30.65200

435tan 1

FEF

EF

FEF

FEF

Ax

Ay

FBD of AEB

5. The figure shows the sideview

of the upper portion of a wooden

attic ladder. The ladder is held in

place by aluminum support

members and springs on both

sides and by a piano hinge at O

that runs the width of the

ladder. If the mass of the ladder

is 30 kg with mass center at G,

determine the magnitude of the

force in each of the two

supporting springs EF. Note that

bar DE rests on top of the

smooth support at C in this

configuration but is not

constrained horizontally. (4/111)

Piano hinge (or continuous hinge) m = 30 kg, determine magnitude of the force in each of the two supporting springs EF.

GAO

FAB

W=30 (9.81)=294.3 N

Ox

Oy

NFF

FF

M

ABAB

ABAB

O

21.261,2.13066924.500

0)444(3.294

)600(18.71cos)324(18.71sin

0

+

264

90

a

A

B

a 71.18°

a

600

AB

FAB

FAB

A

B

m = 30 kg, determine magnitude of the force in each of the two supporting springs EF.

BCD

Cx

Cy

xy

yx

yx

C

DD

DD

DD

M

219.026.9

1924224.1777

0)192()42()426(18.71cos21.261)138(18.71sin21.261

0

+

FAB

a

Dx

Dy

m = 30 kg, determine magnitude of the force in each of the two supporting springs EF.

DCE

2Fsp

NC

NFFD

NDNDF

FDFDF

FNFNM

spsp

F

x

CyCyy

spxspxx

spCspCD

sp

68.29,75.0219.026.9

,00

2,020

75.0,0)72(2)192(0

2

+

Dx

Dy

xy DD 219.026.9

6. Determine the cutting force F applied by the blade DE to the

bar at G in terms of the force P applied to the handle of the

heavy-duty cutter.

Assume that

the cutting

force F is

perpendicular

to the lower

edge of the

blade DE.

cutting force F applied by the blade DE to the bar at G in terms of the force P applied to the handle, force F is perpendicular to the lower edge of the blade DE.

FBE

a

x

y

FBD of ABC

Ax

Ay

4

17

CPF

FFPM

BE

BEBEA

7.23

0)4(sin)17(cos)413(0

12.1028

5tan 1

aa

a

cutting force F applied by the blade DE to the bar at G in terms of the force P applied to the handle, force F is perpendicular to the lower edge of the blade DE.

FBE=23.7P

a

x

y

FBD of Blade DE

Dx

Dy

86

14°

PF

FFPPM D

4.45

0)5(14sin)48(14cos)13(sin7.23)86(cos7.230

aa

F

7. 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.

Determine the compression P in the hydraulic cylinder BE and the magnitude of the force supported by the pin at A.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.

Dumper

DC Two force member

Determine the compression P in the hydraulic cylinder BE and the magnitude of the force supported by the pin at A.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.

ABC

BE Two force member

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 N

24

7

mABCD= 400 N (at C), mhoop= 220 N (at G) IJ= two-force member, if a person with 800 N weight hangs on the rim, determine the support reactions at D and the forces supported by all members.

Dx

mNM

M

M

ND

D

F

ND

DF

D

D

D

y

y

y

x

xx

2.1867

0)35.1(768

)3(224)72.0(220

0

1388

0768220400

0

224

02240

220 N

400 N

Dy

MD

800 N

24

7

224 N

768 N

FBD of Entire frame

x

y

mABCD= 400 N (at C), mhoop= 220 N (at G) IJ= two-force member, if a person with 800 N weight hangs on the rim, determine the support reactions at D and the forces supported by all members.

FAE

NF

NF

FFF

NF

FFF

NF

NFF

F

NF

F

M

y

yAEyy

x

xAExx

AE

AEy

AEx

AEy

AEx

AEx

F

2243)1992()1031(

1992

07682200

1031

02240

1584)1004()1225(

1004,5

4

60.0

48.0

1255

0)75.0(768)12.0(22048.0

0

22

22

FAEx

220 N

FAEy

800 N

24

7

224 N

768 N

Fx

Fy

FBD of EFGH

x

y

mABCD= 400 N (at C), mhoop= 220 N (at G) IJ= two-force member, if a person with 800 N weight hangs on the rim, determine the support reactions at D and the forces supported by all members.

FIJ

NB

NBFBFF

NFBFBF

NF

FFF

M

yyyIJy

xxxxx

IJ

yxIJ

B

6740)6661()1031(

666100

103100

4669

0)60.0()48.0(15.0

0

22

Fx

Fy

FBD of IBF

Bx

By

IJ two force member

FIJ

FIJ

x

y

9. The figure shows a wheel puller which is designed to remove a V-belt pulley P from its tight-fitting shaft S by tightening of the central screw. If the pulley starts to slide off the shaft when the compression in the screw has reached 1.2 kN, calculate the magnitude of the force supported by the each jaw at A. The adjusting screws D support horizontal force and keep the side arms parallel with the central screw.

compression in the screw =1.2 kN, calculate the magnitude of the force supported by the each jaw at A, adjusting screws D support horizontal force

FCB

x

y

kNFF

kNFFFF

F

F

FFF

FF

FFF

EHCB

EHxCBxEHyCBy

CBx

CBy

EHxCBxx

EHyCBy

EHyCByy

85.0

6.0

145tan

0

2.1

02.10

1.2 kN

FEH

E

H

FEHx

FEHy

FCBx

FCBy

45°

45°

FCBy

FCBy

FCBx

FCBx

FCB

FCB

CB

EC

compression in the screw =1.2 kN, calculate the magnitude of the force supported by the each jaw at A, adjusting screws D support horizontal force

x

y

kNDF

kNAAA

kNA

A

AAFM

kNFF

kNFAF

xx

yx

x

x

xyCBxD

CBxCBy

CByyy

42.00

626.0

18.0

,)6.0(15)6.0(60150

0)150()15()60(0

6.0

6.00

22

FCBy

FCBy

FCBx

FCBx

FCB

FCB

Ax

Ay

FCBy

FCBx

FCB

45°

CB

Dx

ABD

10. The pruning mechanism of a pole saw is shown as it cuts a branch S. For

the particular position drawn, the actuating cord is parallel to the pole and

carries a tension of 120 N. Determine the shearing force P applied to the

branch by the cutter and the total force supported by the pin at E. The

force exerted by the light return spring at C is small and may be neglected.

actuating cord is parallel to the pole and carries a tension of 120 N. Determine the shearing force P applied to the branch by the cutter and the total force supported by the pin at E. The force exerted by the light return spring at C is small and may be neglected.

FDE

DC two force member

actuating cord is parallel to the pole and carries a tension of 120 N. Determine the shearing force P applied to the branch by the cutter and the total force supported by the pin at E. The force exerted by the light return spring at C is small and may be neglected.

ABC

11. The figure shows a simplified sketch of the mechanism used to raise the bucket of a bulldozer. The bucket and its contents weigh 10 kN and have acenter 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 weight of the hydraulic cylinders can be ignored. Calculate the forces in the hydraulic cylinders and all forces acting on arm DEFG for the position shown.

mbucket = 10 kN, mABCD = 2 kN (B), mDEFG = 1 kN (E). Calculate the forces in the hydraulic cylinders and all forces acting on arm DEFG.