1) a) Reduce the force system to a force-couple system at O. b)Replace the force couple system obtained by a

wrench and determine the coordinates of the point in the yz plane through which the line of action of the

wrench passes.

60 mm

2) A 77 N force 𝐹1 and a 30 N.m couple 𝑀1 are applied to corner E of the bent plate shown. If 𝐹1 and 𝑀1are to

replaced with an equivalent force-couple system ( 𝐹2, 𝑀2) at corner B and if (M2)z=0 determine, (a) distance d,

(b) 𝐹2 and 𝑀2.

(F1=77 N, M1=30 N.m) at point E

(M2)z=0 at point B

Coordinates: E(250, 0, 70) K(250-d, 30, 0) H(310, 60,0)

mNkjiFr

kjikjiFr

--

07.1419.15225.1

49424207.0085.025.0

1

1

kjikji

F

494242110

706060771 -

-

112 FrMM

(at point E)

5800

703030

7030

703030

22221

--

--

d

kjid

d

kjidM

(at point B)

(F1=77 N, M1=30 N.m) at point E

(M2)z=0 at point B

mNkjiFr

07.1419.15225.11kjiF

4942421 - 112 FrMM

5800

703030

21

--

d

kjidM

(M2)z=0

254.1495800

007.145800

7030

2

2

-

d

d

d=128.36 mm

kjiM

kjiM

07.1403.68.25

580036.128

703036.12830

1

21

--

--

112 FrMM

kiM

22.21575.242 -

kjiF

4942422 -

3) The mass center of 15-N link OC is located at G, and the spring constant of

k=25 N/m is unstretched length when q=0. Calculate the tension T and the

reactions at O for q=45o.

Ox

Oy

T

Fspring

W=mg O

A

B

C

D

OAC : isosceles triangle

45o

b

45o

mmODOCOA 480

67.5o

67.5o

67.5ob

a mmAC 37.3675.22sin4802

b

a

sin

240

sin

480

45sin

AB

Sine theorem:

33.10667.28 ab

k=25 N/m, unstretched length when q=0.

Ox

Oy

T

Fspring

W=mg=15 NO

A

B

C

D

45o

b

67.5o

67.5ob

a

33.10667.28 ab

Unstretched length of spring:

Length of spring when q=45o:

mmBClo 240

mmlAB

AB

spring 67.353

45cos2404802240480 222

-

NllklkF ospringspring 84.21024067.35325 3 -- -

NT

T

MO

302.5

045sin1601548067.28sin84.24805.67sin

0

--

+

NOTOF xxx 667.005.67cos67.28sin84.20 --

NOTOF yyy 593.1205.67sin1567.28cos84.20 --

Ox

67.5o

67.5o

4) Total mass of the member ABC and pulley C is 30 kg and its mass center is at point G. Detemine the reaction

acting on pin A.

DB

A

G

C

600 mm

100 mm

R=125 mm

R

1 kN

125 mm

250mm

30o

225 mm

FBDB

A

G

C

600 mm

100 mm

R=125 mm

R

1 kN

125 mm

250 mm

30o

W=mg=30(9.81) N

Ax

Ay

30o

NA

A

F

y

y

y

325.1160

030cos100081.930

0

--

NF

F

M

BD

BD

A

33.3593

025.030sin125.0375.030sin100030cos125.06.030cos1000225.081.930

0

-

NAAF xxx 33.3093030sin100033.35930 --

+

225 mm

5) Uniform rod AB has a mass of 25 kg and a length L = 1 m.What must be q for equilibrium? Also what are the normalforces acting from the surfaces to the rod at points A and B?Assume all surfaces are smooth.

FBD of rod AB

RA

RB45o

W=25(9.81) N

0sincos707.0cos625.122

0sin145sincos145coscos5.081.925

0

qq-q

q-q-q

B

BB

A

R

RR

M

BABAx RRRRF 414.1045sin30sin0 -

081.92545cos30cos0

414.1

- B

R

Ay RRF

B

NRNR AB 51.17995.126

1

1 qqqq-q sin75.89cos87.320sincos95.126707.0cos625.122

11.2075.89

87.32tan

cos

sinqq

q

q

+

6) A 450 mm long uniform rod AB has a weight of 304 N and is attached to a ball-and-socket joint at A. The end B of the rod

rests against an inclined frictionless surface and is held in the equilibrium position shown by cord BC. Knowing that cord BC

is 450 mm long, determine the tension in the cord and the reactions at A and B.

7) The shaft AB is supported by a thrust bearing at A and a radial bearing at B. Determine the force in cable CD, and the

bearing reactions at A and B caused by the 90 N vertical force applied at E. Neglect weights.

Radial bearing at B:

Thrust bearing at A:

Tension in cable CD:

0447.0894.0657060

9070125110

---

-

jiTkji

kjikBiBjM zxA

kBiBB zx

kAjAiAA zyx

jiTji

TT

447.0894.03060

3060

22--

--

kF

90-

082.26055.29

11.5858.62112506300110110

---

kTiT

jTkTjiiBkBM zxA

0055.296300110 - TBi z

011.5811250 - Tj

082.2658.62110 - TTBk x

T=193.598 NBx=157.34 NBz=6.136 N

NAAF

NAAF

NAAF

z

B

zz

y

T

yy

x

TB

xx

z

x

864.830136.6900

538.860598.193447.00

736.150598.193894.034.1570

-

-

--

1

2

3

8) The lever AB is welded to the bent rod BCD which is supported by bearing E and cable DG. Assuming that the

bearing can exert an axial thrust and couples about axes parallel to the x and z axes, determine the tension in cable

DG and the reaction at E under the action of the 220 N force. The mass of ABCD is neglected.

220 N

240 mm

60 mm

250 mm

225 mm

160 mm

120 mm

GD

C

E

B

y

z

x

A

Reactions at E (Single thrust bearing) :

kEjEiER zyxE

kjTkj

TT

iF

88.047.0225120

225120

220

22--

--

-

kMiMM EzExE

220 N

240 mm

60 mm

250 mm

225 mm

160 mm

120 mm

GD

C

E

B

y

z

x

A

kEjEiER zyxE

kjTTiF

88.047.0220 ---kMiMM EzExE

088.047.019016022024060 ---- kMiMkjTjiikjM EzExE

065.167 ExMTi

018.14152800 - Tj

029.7513200 - EzMTk

T=374 NMEx=-62701.1 N.mmMEz=14958.46 N.mm

065.16718.14129.755280013200 -- kMiMiTjTkTjkM EzExE

1

2

NEEF

NEEF

NEEF

z

T

zz

y

T

yy

xxx

12.329037488.00

78.175037447.00

22002200

-

-

-

9) Two rods are welded to form a T-shaped structure. The end D of the structure rests against a frictionless vertical wall,

while ends A and B are supported by radial bearings. When a 600 N magnitude vertical force P is applied to the midpoint E

of the part DC of the structure, determine the reactions at D.

25 cm15 cm

15 cm

40 cm

50 cm

50 cm

Coordinates: A(25, 0, 0) B(55, 0, -40) C(40, 0,-20) D(0, 50,-50) E(20, 25,-35)

Dx

Ay

By

Axz

Bxz

0 ABAAB nMM

BrPriDrM ABAExADA

///

Brjkji

iDkjiM

AB

xA

---

--

/60035255

505025

BrikjDkDM ABxxA

---- /2100030005050

kinAB

8.06.0 -

06.0210008.030008.050 ----- xAB DM

ND

D

x

x

255

01020040

-