1. A force F of magnitude 50 N is exerted onthe automobile parking-brake lever at theposition x = 250 mm. Replace the force by anequivalent force–couple system at the pivotpoint O.

EQUIVALENT FORCE–COUPLE SYSTEM

10sin25.025sin1.020sin5010cos25.025cos1.020cos50 xy FF

oM

+

jijiF

98.461.1720cos5020sin50

mNM o 29.17

20o

F

Mo=C=17.29 N.mResultants 1

2. Replace the force system acting on the post by a resultant force, and specify where its

line of action intersects the post AB measured from point B.

Resultants 2

3. The device shown is part of an automobile seat-back-release mechanism. The part is subjected

to the 4 N force exerted at A and a 300 Nmm restoring moment exerted by a hidden torsional

spring. Determine the y-intercept of the line of action of the single equivalent force.

jiR

jiFR

035.186.3

15sin415cos4

Total Moment about O (resultant couple):

Resultant Force:

mmNM

M

O

O

8.155

3004015cos41015sin4+

y

xO

d jiR

035.186.3

xR

yR

Ox MdR

mmy

mmd

d

3.40

3.40

8.15586.3

3

4. A machine component is subjected to the forces and couples shown. The component is

to be held in place by a single rivet that can resist a force but not a couple. Determine the

location of the rivet hole if it is to be located (a) on line FG, (b) on line GH.

120 N

80 N

60 N

40 N.m

42 N.m

200 N

4

Determine the location of the rivet hole if it is to be

located (a) on line FG, (b) on line GH.

ij

ji

ji

FR

6080

70sin12070cos120

15sin20015cos200

jiFR

53.24414.212 Total Moment about G (resultant couple):

Resultant Force:

47.015cos20005.015sin200

19.070sin12047.070cos12013.06059.0804042

GM

mNMG 34.63

Resultants 5

Determine the location of the rivet hole if it is to be

located (a) on line FG, (b) on line GH.

Equivalent Force-Couple System at point G:

jiR

53.24414.212

mNMG 34.63

GGM

jiR

53.24414.212

(a) (b)

G

y

Equivalent Force-Couple System at G:

my 298.014.212

34.63

G

jiR

53.24414.212

jiR

53.24414.212

x

mx 259.053.244

34.63

F

H

Resultants 6

5. Replace the parallel force system acting on the plate by a resultant force and specify its

location on the x–z plane.

jjjjFR

10352

SOLUTION:

Resultant force

jik

jikjikM

CFrM

O

O

35.15.0

25.15.255.05.1

Sum of the moments about point O

kiM

kikikiM

O

O

1014

5.45.1355.25.7

O

kiizkx

kijkzix

RrMO

10141010

101410

jR

10

r

mzmx 4.1,1

Location on the x-z plane which the resultant force acts.

Resultants 7

6. Represent the resultant of the force system acting on the pipe

assembly by a single force and a couple at A .

kjiFR

100180120

SOLUTION:

j

A

A

jk

kiikM

CFrM

40

25.016050

1003.012025.0

kjM A

50100

Equivalent force-couple system at A

A

y

x

z

kjiR

100180120

kjM A

50100

Sum of the moments about point A

Resultant force

Resultants 8

7. The pulley wheels are subjected to the loads shown.

Determine the equivalent force-couple system at point O.

kjFR

5001000

jkijki

kjikjiM

CFrM

o

o

60018.02.140018.02.1

30024.06.020024.06.0

kjiM o

120030012

Equivalent force-couple system at point O

O

x

y

z

kjR

5001000

kjiM o

120030012

SOLUTION:

Sum of the moments about point O

Resultant force

Resultants 9

8. Two forces and a couple (applied to portion AB) are applied to the car’ s exhaust system as

shown. Replace the given force system with an equivalent force-couple system at D.

25 N.m

115 N

100 N

kjR

kjjFR

FF

5039.28

30sin10030cos100115

21

SOLUTION:

Resultant force

Resultants 10

25 N.m

115 N

100 N

CFrM D

kjiM

kjiM

D

D

868.179825.8

207.43568.4115243.94475.85

kjjkikikFr

505.86225.03.079.046.031.036.022

jikikFr

11569.046.031.036.011

ikFr

3.947.4311

kjiFr

568.4124745.8522

y

z

C

kjkj

C

2015375.0

3.0225.025

Equivalent force-couple system at D

kjiM

kjR

D

868.179825.8

5039.28

Sum of the moments about point D

B

A

11

9. Replace the three forces acting on the plate by a wrench. Specify the magnitude of theforce and couple moment for the wrench and the point P(x, y) where its line of actionintersects the plate.

kjiFR

800300500

Resultant force

Equivalent force-couple system at A

jjikjM

CFrM

A

A

300468004

kiM A

18003200

Sum of the moments about point

xy

z

AM

R

A

kjiFR

800300500

Equivalent force-couple system at A

kiM A

18003200

Reduction to wrench:

Unit Vector of Resultant force: kjikji

nR

808.0303.0505.0

95.989

800300500

13

0304000018003200800300500 kikji

(Couple is not perpendicular to the force. Force-couple system can be reduced to a wrench – positive wrench.)

Parallel component of the couple to resultant force:

mNkjikiM R 4.3070808.0303.0505.018003200//

Parallel component of the couple to resultant force in vector form:

kjikjiM R

64.568833.93055.1550808.0303.0505.04.3070//

(Wrench Resultants)

kjiM R

64.568833.93055.1550//

xy

z

RM

R

A

//RM

kjiR

800300500

Positive Wrench

Resultants 14

Normal component of the couple to resultant force in vector form

kjikiMMM RAR

64.568833.93055.155018003200//

kjiMMM RAR

64.388833.93045.1649//

Coordinates of the point in the xy plane through which the line of action of the wrench passes

xy

z

RM

R

A

//RM Positive Wrench

kjikjijyixMRr R

64.388833.93045.1649800300500

kjiiykyjxkx

64.388833.93045.1649800500800300

mxxj

myyi

16.133.930800

06.245.1649800

x

y

z

R

A

//RM

(Positive Wrench)

(1.16,2.06,0)

10. 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.

Resultants 15

11. The force-couple system acting at O is equivalent to the wrench acting at A.

If and , determine . NkjiR

7001400600 mNM //R 1200

oM

kjikji

R

RnR

418.0835.0358.07001400600

7001400600

222

(Unit vector of resultant force)

kjikjinMM RRR

6.50110026.429418.0835.0358.01200////

kjikjijiRrM R

300021001400700140060023

kjiMMM RRO

6.350110986.1829//

SOLUTION:

Resultants 16

jiR

jiijiFR

jijiF

iF

jijiF

35201040

800150050027202040

800150017

81700

17

151700

500

272020405

43400

5

33400

3

2

1

kjiCFrM

kC

kjijikjFr

kijFr

kjiFr

jikjijijikFr

CFrM

A

A

6.11391770960

800

75075040080015005.05.0

1505003.0

6.48910201360

272020405.016.03.02720204017

8

17

1534.05.0

33

22

11

11

17

12. The pulleys and the gear are

subjected to the loads shown. For these

forces, determine the equivalent force-

couple system at point A.