1. The pole is subjected to the two forces shown. Determine the components of

reaction of A assuming it to be a ball-and-socket joint.

“Ball-and-socketjoint”

2. The shaft assembly (consisting of welded pieces AB, ED and CD) is supported

by a thrust bearing at A and a radial bearing at B. The assembly is subjected to a

force at C and a couple . If it is known that the y component of the reaction

at bearing B is (N), determine the vector expressions of the force , couple

and the bearing reactions at A and B. Link ED lies in the yz plane. ED=250 mm.

F

M

j104

F

M

Ax

Az

Ay

Bz

By

3. Structure ABCD is supported by a collar at D that can rotate and slide along bar EF which is

fixed and is frictionless. Structure ABCD makes contact with smooth surfaces at A and C

where normal direction to the surface at A lies in a plane that is parallel to the xy plane.

Force P is parallel to the y axis. If P=10 kN, determine the reactions at A, C and D. Note that

collar D acts like a wide radial bearing.

n

Dy

Dz

NC

NA

My

Mz

4. Because of friction a very large force T is required to raise the 10 kg body when the cable

is wound around a fixed rough cylinder as shown. For the static equilibrium values shown,

determine all of the reactions at point O. The 0.3 m dimension refers to the point A where

the cable loses contact with the cylinder. Neglect the weight of the fixed cylinder.

Ox

Mx

OyMy Oz

Mz

W=10(9.81) N=0.098 kN

Ox

Tyz

OyMy Oz

Mz

W=10(9.81) N=0.098 kN

Reactions at O:

y

20o

z

y

x Txy

30o

30o

A

Mx

A

kMjMiMM

kOjOiOR

zyxo

zyxo

[kN]098.0[N]1.9881.910

89.841.2415

20sin30cos20cos30cos30sin

iiiW

kjiT

kTjTiTT

Ox

OyMy

Oz

Mz

W=10(9.81) N=0.098 kN

y

x Txy

30o

30o

A

30cos3026 mm

30sin3015mm

Mx

mmkN.MmmkN.MmmkN.M

kMjMiM

k.j.ikjii.jk

M

zyx

zyx

O

7285634121236519394

0

89841241580015260980301100

0

kN.OTOF

kN.OTOF

kN.O.TOF

zzzz

yyyy

xxxx

89800

412400

90214009800

5. The uniform panel door has a mass of 30 kg

and is prevented from opening by the strut C,

which is a light two-force member whose

upper end is secured under the door knob and

whose lower end is attached to a rubber cup

which does not slip on the floor. Of the door

hinges A and B, only B can support force in the

vertical z-direction. Calculate the compression

C in the strut and the horizontal components

of the forces supported by hinges A and B

when a horizontal force P = 200 N is applied

normal to the plane of the door as shown.

NjP

Nk.k).(W

kBjBiBB

jAiAA

k.j.FnFF

k.j..

kj.n

zyx

yx

CDHCC

DH

200

329481930

8060

8060251

750

O (0, 0, 0) A (1.2, 0, 1.8) B (1.2, 0, 0.2)

G (0.6, 0, 1) E (0, 0, 2) D (0.2, 0, 1)

H (0.2, 0.75, 0)

jAiAkjki

kjFkikkiM

M

yx

CB

B

6.12008.12.1

8.06.08.03.2948.06.0

0

E

G

W

AxAy

Bx

By

BzFC

D

H

O

P

NA

NAFAj

NAAFi

NFFk

xy

xCx

yyC

CC

06.138

64.8958.1768.06.1

1056.136048.0

40002406.0

NB

NBFBAF

NABBAF

xy

yCyyy

xxxxx

47.170

14506.02000

64.8900

E

G

W

AxAy

Bx

By

BzFC

D

H

O

P

02406.058.1766.18.03606.148.0 kFjAFiAFM CxCyCB

6. The control surface of an aircraft is supported by a thrust bearing at point C and is

actuated by a bar connected to point A. The 1 kN force acts in the negative z direction, and

the line connecting points A and B is parallel to the z axis. Determine the value of force F

needed for equilibrium and all support reactions.

F Cx

Cy

Cz

Mx

Mz

F

1 kN

kF,kFjFiFF D

1

15

5

15

2

15

14

kMiMM,kCjCiCR zxCzyxC

Reactions at C:

06.03.015

5

15

2

15

141.02.0

0

kMiMkjikjiFkj

M

zx

C

kNMMFk

mkNMkNFFj

MFFi

kMiMijiFjFiFkF

zz

x

x

zx

6.0015

8.2

343.021.303.015

4.1

06.015

2.0

15

1

06.03.015

2.0

15

4.1

15

1

15

8.2

F Cx

Cy

Cz

Mx

Mz

F

1 kN

kF,kFjFiFF D

1

15

5

15

2

15

14

kMiMM,kCjCiCR zxCzyxC

Reactions at C:

kN.CCFF

kN.CCFF

kN.CCFF

zzz

yyy

xxx

0720115

50

4280015

20

9962015

140

7. The shaft, lever and handle are welded together and constitute a single rigid

body. Their combined mass is 28 kg with mass center at G. The assembly is

mounted in bearings A and B, and rotation is prevented by link CD. Determine the

forces exerted on the shaft by bearings A and B while the 30 N.m couple is applied

to the handle as shown.

Ax

Ay

Bx

By

W

F

C

F

x

y 600 mm

45

0 m

m

jF.iF.F

8060

06.06.03043.604.8236.008.006.0

06.03081.92822.03.08.06.01.06.0

0

iBjBkkjkFiFjF

jBiBkkijkjiFkj

M

yx

yx

A

NBBFj

NBBFi

NFFk

xx

yy

2.11206.04.8206.0

49.3306.008.0

2.251043.9036.0

Ax

Ay

Bx

By

W

F

C

F

x

y 600 mm

45

0 m

m

jF.iF.F

8060

NABAFF

NAAFF

yyyy

xxx

46.16708.00

78.1107.2742.1126.00

NB

NB

NF

x

y

2.112

49.33

2.251

8. The electric sander has a mass of 3 kg with

mass center at G and is held in a slightly tilted

position (z-axis vertical) so that the sanding disk

makes contact at its top with the surface being

sanded. The sander is gripped by its handles at

B and C. If the normal force R against the disk

is maintained at 20 N and is due entirely to the

force component Bx (i.e., Cx = 0), and if the

friction force F acting on the disk is 60 percent

of R, determine the components of the couple M

which must be applied to the handle at C to hold

the sander in position. Assume that half of the

weight is supported at C.

Cy

Cz

29.43 N

My Mx

Mz

Nm.MNm.MNm.M

kMjMiM

k.k.i.

k.ik.j.i.

jik.i.

M

zyx

zyx

C

56241118571

0

432930040

71514203020040

122040120

0

N

WCC

NFNR

zx 715.142

81.93

20

12100

602020

NCB

CBWF

NC

CF

NB

BF

zz

zzz

y

yy

x

xx

715.14

00

12

0120

20

0200