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4.5 Equivalent Force-Couple Systems

4.5 Equivalent Force-Couple Systems Example 1, page 1 of 1

1. Replace the force at A by an equivalent

force and couple moment at point O.

y

x

4 m

6 m

20 N

A4

35

1 Express the force in rectangular components.

A

20 N

6 m

4 m

x

y

34

5

0

0

(20 N)5

16 N = 4

3 = 12 N5

(20 N)

Calculate the moment about O.2

MO = (16 N)(4 m) + (12 N)(6 m)

= 136 N·m

3 Display the equivalent force and couple moment

at point O.

0

53

4

A

20 N

x

y

ns.

136 N·m


2. The 60-N force acts at point A on the lever as shown.

Replace the force at A by a force and couple moment

acting at point O that will have an equivalent effect.

A

1 Calculate the moment about O.

MO = (60 N)(200 mm) cos 60°

= 6000 N·mm = 6 N·m

Display the equivalent force

and couple moment at point O.

2

60 N

60°

60°

60 N

A

A

60 N

6 N m ns.

(200 mm) cos 60°

O

O

O

200 mm200 mm


3. Replace the 2-lb force acting on the end of the bottle

opener by an equivalent force and couple moment

acting on the underside of the bottle cap at A. Use your

results to explain how a bottle opener works.

1 Calculate the couple moment about point A.

MA = (2 lb)(3 in.)

= 6 lb·in.

Display the equivalent force

and couple moment at point A.2

ns.

A

2 lb

B

B

3 The bottle opener works by pulling (with a 2-lb

force) on the edge of the cap while

simultaneously twisting (with a 6 lb·in. moment)

the entire cap away from the top of the bottle.

6 lb·in. ns.

3 in.

2 lb


4. Replace the given forces and couple moment by a

resultant force and couple moment at A.

2 Calculate the resultant force

A

120 lb260 lb

800 lb·ft

7 ft4 ft2 ft

Express the inclined force in rectangular components.

1

2 ft 4 ft 7 ft

800 lb·ft

120 lb

1312

5 A

Rx = Fx: Rx = 100 lb

Ry = Fy: Ry = 240 lb 20 lb

= 360 lb

= 360 lb

5

12 13

513

(260 lb) =100 lb

240 lb=(260 lb)1312

Calculate the resultant couple moment about A.3

MRA = MA

= (240 lb) (2 ft + 4 ft + 7 ft)

+ (120 lb) (4 ft + 7 ft) 800 lb·ft

= 3640 lb·ft


Determine the magnitude and direction of the

resultant force.4

ns.

R = (360)2 + (100)2

= 374 lb

= tan-1

100360

= 74.5°

ns.360 lb

100 lb

5 Display the equivalent force and couple moment at A.

A

374 lb

3640 lb·ft

ns.

74.5°


5. Replace the force F = 3 kN acting on corner A of the

block by an equivalent force and couple moment acting

at the center C of the block.

A

y

xC

z

F = 3 kN

150 mm

150 mm

100 mm 100 mm


Calculate the couple moment about C.

A

y

xC

z

F = 3 kN

150 mm

150 mm

1

M C = rCA F

= {100i 150k}mm 3j}kN

= 100( i j 150( k j

= k = i

= { 450i 300k}kN·mm

= { 450i 300k}N m Ans.

ij

k

rCA

100 mm100 mm


Display the equivalent force and couple moment at C.

A

y

xC

z

150 mm

150 mm

2

F = { 3j}kN Ans.

MC = { 450i 300k}N·m

100 mm 100 mm


B

C

x

y

z

4 lb

7 lb

A

48 in.

6. Replace the forces acting on the ice auger by an

equivalent force and couple moment acting at A.

6 in.


B

C

x

4 lb

7 lb

A

48 in.

Rx = Fx: Rx = 0

Ry = Fy: Ry = 4 lb

Rz = Fz: Rz = 7 lb

Calculate the resultant force1

z

6 in.

y


B

C

x

4 lb

7 lb

A

48 in.

MAR = MA

= rAC { 4j} lb + rAB {7k} lb

2

k

ji

rAC

rAB

z

6 in.

Calculate the resultant couple moment about A.

= 0, because rAC and { 4j} are parallel,

or, what amounts to the same thing,

the line of action of the { 4j} lb force

passes throughout point A.

= 0 + { 6i + 48j} in. {7k} lb

= 6(7)i k + 48(7)j k

= j = i

= {336i + 42j} lb·in.

y


B

C

x

y

A

MAR = {336i + 42j} lb·in.

R = { j + 7k} lb

ns.

z

3


7. Replace the forces and couple moment by a single force and

specify where it acts.

A B C D

2 ft 8 ft 4 ft 3 ft

3 kip 4 kip

20 kip·ft

40°


A B C D

2 ft 8 ft 4 ft 3 ft

3 kip

20 kip ft

Rx = Fx: Rx = 3.064 kip = 3.064 kip

Ry = Fy: Ry = 3 kip 2.571 kip = 5.571 kip

R = (3.064)2 + (5.571)2

= 6.358 kip

Calculate the resultant force2

1 Resolve the inclined force into

rectangular components.

y

(4 kip) cos 40° = 3.064 kip

(4 kip) sin 40° = 2.571 kip

40°

3.064 kip

5.571 kip

= tan-1

3.0645.571 = 61.2°

R

4 kip

3

ns.

ns.


4

A B C D

2 ft 8 ft 4 ft 3 ft

3 kip

4 kip

20 kip·ft

We equate the moment of the new

system, about point A, to the moment

of the original force-couple system

(The choice of point A for summing

moments is arbitrary; any other point

would work as well, except that we

must use the same point for both the

original system and the new system.)

6

5 This is a new force-couple system that we want to make

equivalent to the original force-couple system. We

already know that the forces are equivalent because R

is the resultant of the forces in the original system.

Now we have to make sure that the moment is also

equivalent. We do this by placing R at some unknown

distance d from the left end and then choosing d so that

the moment of this new system is the same as that of

the original system.

3.064 kip

2.571 kip

40° = 61.2°

7

5.571 kip

3.064 kip

DCA

MAR = MA

or,

(5.571 kip)d = 20 kip·ft (3 kip)(2 ft) (2.571 kip)(2 ft + 8 ft)

Solving gives

d = 2.10 ft Ans.

d

R = 6.358 kip

This is the original force-couple moment system.


8. Replace the forces acting on the frame by a single force

and specify where its line of action intersects

a) member BCD and b) member AB.

1 Determine the resultant force.

Rx = Fx: Rx = 800 N = 800 N Ry = Fy: Ry = 400 N 600 N 300 N

= 1300 N

= 1300 N

400 N

800 N

600 N 300 N

x

y

A

B C D

E

2 m

2 m

R = (800)2 + (1300)2 = 1526 N

= tan-1 = 58.4°1300 N

800 N

1300 800

Ans.

Ans.

4 m 4 m


Choose d so that the moment about B of the resultant

force equals the moment of the original force system.

(The choice of point B was arbitrary.)

2 Part a) To determine where the line of

action of the resultant force intersects

member BCD, place the force on BCD,

an unknown distance d from point B.

400 N

800 N

600 N 300 N

xA

B C D

E

3

MBR = MB

or

(1300 N)d = (600 N)(4 m) (300 N)(4 m + 4 m) (800 N)(2 m)

Solving gives

d = 4.92 m Ans.

E

D

C

B

A

1300 N

d

R = 1526 N

800 N 58.4°

2 m

4 m4 m

2 m


Choose d' so that the moment about B of the resultant

force equals the moment of the original force system.

4 Part b) To determine where the line of

action of the resultant force intersects

member AB, place the force a distance

d' from point B.

400 N

800 N

600 N 300 N

xA

B C D

E

5

MBR = MB

or

(800 N)d' = (600 N)(4 m) (300 N)(4 m + 4 m) (800 N)(2 m)

Solving gives

d' = 8.0 m Ans.

E

DB

A

1300 N

R = 1526 N

800 N 58.4°d'

2 m

2 m

4 m4 m


The intersection of the line of action with a line drawn

through A and B occurs below point B.

A

B D

E

6

800 N

58.4°1300 N

R = 1526 N

4 m

d' = 8 m


7 Alternative solution for part b. Once we know where

the line of action intersects member BCD, we can

use geometry to find the intersection with AB.

E

DCB

A

1300 N

d = 4.92 m

line of action of

resultant forceG

d' 4.92

tan 58.4° =

Solving gives,

d' = 8.0 m

(same result as before)

From triangle CBG,8

58.4°

d'


6 ft6 ft

9. Determine the value of force P such that the line of

action of the resultant of the forces acting on the truss

passes through the support at H. Also determine the

magnitude of the resultant.

A B C D E

I JHGF

8 ft

160 lb 200 lb 180 lb

260 lb

P

30°

6 ft6 ft


30°

AB C D E

I JHGF

6 ft 6 ft 6 ft 6 ft

8 ft

160 lb 200 lb 180 lb

260 lb

P

Rx = Fx: Rx = 130 lb P

Ry = Fy: Ry = 225.2 lb 160 lb 200 lb 180 lb

= 765.2 lb

= 765.2 lb (2)

Determine the resultant force. 1

(260 lb) cos 30°

= 225.2 lb

(260 lb) sin 30°

= 130 lb


MHR = MH :

0 = (225.2 lb)(6 ft + 6 ft) (130 lb)(8 ft)

+ (160 lb)(6 ft) (180 lb)(6 ft) + P(8)

Solving gives

P = 192.8 lb

= 192.8 lb

F G HJ

I

EDCB

Resultant R

225.2 lb

Rx = 130 lb P

= 130 lb ( 192.8 lb)

= 322.8 lb

Ry = 765.2 lb

ns.

ns.

AB C D E

IJ

HGF

6 ft 6 ft 6 ft 6 ft

8 ft

160 lb 200 lb 180 lb

P

Choose force P so that the

moment about H of the

resultant force equals the

moment of the original force

system about H. Any other

point besides H could be

used, but H has the advantage

that the moment of the

resultant R is zero, since R is

known (as part of the

statement of the problem) to

pass through H.

2

130 lb

3


Magnitude of resultant.

From Eqs. 1 and 2,

R = (322.8)2 + ( 765.2)2 = 831 lb ns.

4


10. A machine part is loaded as shown. The part is to be attached to

a supporting structure by a single bolt. Determine the equation of

the line defining possible positions of the bolt for which the given

loading would not cause the part to rotate. Also, determine the

magnitude and direction of the resultant force.

x

y

0.3 m

0.5 m

90 N·m

20 N·mO

60 N80 N

0.6 m0.4 m

60°


x

y

90 N·m

20 N·mO

60 N

Rx = Fx: Rx = 40 N = 40 N Ry = Fy: Ry = 69.28 N 60 N

= 129.28 N

= 129.28 N

Determine the resultant force. 1

Ans.

Ans.

129.28 40

40 N

129.28 N

R = (40)2 + (129.28)2 = 135.33 N

= tan-1 = 72.8°

80 N

60°

(80 N) cos 60° = 40 N

(80 N) sin 60° = 69.28 N


x

0.5 my

x

y

90 N·m

20 N·mO

60 N

Rotation is caused by moment.

If the given force-couple

system is replaced by a single

equivalent force (the resultant)

and a specified line of action,

then the moment would be zero

about any point on the line of

action. So the line of action is

the line on which the bolt

should be placed to prevent rotation.

2

80 N

60°

40 N

3 To find the equation of the line of action, place the resultant at a

general point (x, y). Then equate moments about, say, point O for the

resultant (the figure on the right) and the original loading (the figure

on the left):

MOR = MO

or

(40 N)y (129.28 N)x = 90 N·m 20 N·m + (40 N)(0.3 m + 0.5 m)

(69.28 N)(0.4 m) (60 N)(0.4 m + 0.6 m)

40 N

O

y

x

69.28 N

0.4 m

0.3 m (x, y)

129.28 NR

72.8°

0.6 m


Solving for y gives the equation of a line

y = 3.232 x + 0.357 Ans.

This line defines the possible locations of

the bolt.

0

y

x

0.4 m

0.3 m

All points at the top of the machine part have a y coordinate of

y = 0.3 m + 0.5 m

= 0.8 m.

Substituting y = 0.8 m into the equation of the line for the bolt

locations,

y = 3.32 x + 0.357

and solving for x gives

x = 0.133 m

4 5

0.5 m

0.357 m

0.6 m

0.133 m


Ry = Fy: Ry = 30 kip 15 kip 20 kip 11 kip

= 76 kip

= 76 kip

11. The rectangular foundation mat supports the four

column loads shown. Determine the magnitude,

direction, and point of application of a single force that

would be equivalent to the given system of forces.

1

A

Determine the resultant.

B

C D

y

20 kip

15 kip

11 kip

z

x

30 kip

10 ft

5 ft8 ft

3 ft


3 First, equate moments about the x axis. We can use

either the scalar definition of moment, M = Fd, or the

vector product definition. Let's use the scalar definition.

2

x

z

y

A

B

C

D(x, 0, z)

x z

R = 76 kip

z

30 kip11 kip

15 kip

C, B x, A

8 ft 3 ft

zC, B

yR = 76 kip

x, A

zView from positive x axis

20 kipy

To make the resultant

force R equivalent to

the original system of

forces, place R at the

point (x, 0, z) and then

determine values of x

and z such that R

produces the same

moments about the x

and z axes as the given

forces produce.

DD

3 ft

8 ft5 ft

10 ft

30 kip

x

z

11 kip

15 kip

20 kip

y

DC

B

A

MxR = Fd: (76 kip)z = (20 kip)(8 ft + 3 ft) + (15 kip)(8 ft + 3 ft)

+ (11 kip)(3 ft ) + (30 kip)(0) (1)4


R = 76 kip

zx

(x, 0, z) D

C

B

A

y

z

x

x

y

Cz, B, ADC

30 kip

View from positive z axis

x

R = 76 kipy

D

5 ft10 ft

z, B, A

15 kip20 kip 11 kip

Use two-dimensional views. 6

M zR = Fd: (76 kip)x = (20 kip)(10 ft) (11 kip)(10 ft + 5 ft) (2)7

5 Solving Eq. 1 gives

z = 5.5 ft Ans.

Next, equate moments about the z axis.

A

B

C D

y

20 kip

15 kip

11 kip

z

x

30 kip

10 ft

5 ft8 ft

3 ft


8 Solving Eq. 2 gives

x = 4.80 ft Ans.


Rz = Fz: Rz = 300 N 850 N 400 N = 1550 N

= 1.55 kN

12. Three signs are supported by an arch over a highway and are

acted upon by the wind forces shown. Replace the forces by an

equivalent force and specify its point of application.

Ans.

Determine the resultant. 1

6 m

C

4 m

B

850 N

400 N

A

O

3.5 m

y

300 N

4 m1 mz

5 m

x


A

B

C

(x, y, 0)

1550 N

y

z

y

y

Ox

5 m

z 1 m4 m

300 N

y

3.5 m

O

A

400 N

850 N

B

4 m

C

6 m

x

x

4 MxR = Fd: (1550 N)y = (850 N)(6 m) (400 N)(4 m) (300 N)(3.5 m) (1)

Place the resultant R at the arbitrary point (x, y, 0) and then

determine values of x and y such that R produces the same

moments about the x and y axes as the given forces produce.

2

3 First equate moments about the x axis.

B

C

A

850 N

400 N

300 N

3.5 m4 m

6 m

zx,O

y

x,O

y

R = 1550 N

zView from the

positive x axis


Solving Eq. 1 gives

y = 5.0 m Ans.

5

x

=

1 m

R = 1550 N

y, Oxy,O

400 N850 N300 N

CBA

View from the

positive y axis

Next, equate moments about the y axis.6

Ans.

MxR = Fd: (1550 N)x = (300 N)(1 m) + (850 N)(1 m + 5 m) + (400 N)(1 m + 5 m + 4 m)

Solving gives

x = 6.06 m

7

x5 m 4 m


13. Three forces act on the pipe assembly. Determine the

magnitude of forces P and Q if the resultant of all three

forces is to act on point A. Also determine the magnitude

of the resultant.

B

A

C

D

O

P

Q

y

x

z

200 N3 m

2 m

1.5 m

1 m

1 Express the resultant R in terms of P and Q.

R = Fy: R = P + Q + 200 N (1)


Place the resultant R at point A and then determine values of

P and Q such that R produces the same moment about the x

and z axes as the given forces, P, Q, and 200 N, produce.

B

C

D

P

Q

x

z

200 N

rOD

rOC

rOB

y

2

R = ( P Q 200 N)j

We can use either the scalar definition of moment M = Fd,

or the vector definition, M = r F. Let's use the vector

definition and calculate moments with respect to point O.

rOA

z

D

A

C

3

Introduce position vectors, all with tails at point O.

rOB = {3i} m

rOC = {5i + 1.5k} m

rOD = {5i + 2.5k} m

rOA = {3i + 1.5k} m

x

4

y

A

3 m

2 m1 m

1.5 m

3 m

1.5 m

O

O


For equivalence, equate the moment of the resultant R

about point O to the moment of the given forces about O.5

MOR = MO: rOA R = rOB { Pj} + rOC { Qj} + rOD { 200j}

or

{3i + 1.5k} ( P Q 200)j = {3i} { Pj} + {5i + 1.5k} { Qj} + {5i + 2.5k} { 200j}

3( P Q 200)i j + 1.5( P Q 200)k j = 3( P)i j + 5( Q)i j + 1.5( Q)k j + 5( 200)i j + 2.5( 200)k j

= k = i = k = k = i = k = i

k

ji

Collecting coefficients of i, j, and k gives

1.5(P + Q + 200)i 3(P + Q + 200)k = (1.5Q + 500)i + ( 3P 5Q 1000)k

Equating coefficients of i gives

1.5(P + Q + 200) = 1.5Q + 500 (2)

Equating coefficients of k gives

3(P + Q + 200) = 3P 5Q 1000 (3)

6


P and Q were defined as

downward directed

forces.

8

Using these values in Eq. 1 gives

R = P + Q + 200 N

= 133.3 N

Solving Eqs. 2 and 3 gives,7

ns. P = 133.3 N

Q = 200 N = 200 N

133.3 200

ns.

ns.


Alternative solution: the computation are simplified

somewhat if we sum moments about point A instead

of point O.

Equate the moment of R about point A to

the moment of the given forces about A.11

200 N

rAC

D

A

rAD C

Q

9

rAB

P

Introduce position vectors, all with

tails at point A.

rAB = { 1.5k} m

rAC = {2i} m

rAD = {2i + k} m

10

x

R

z

y

3 m

2 m1 m

1.5 m

O

B

A

C

D

O

y

B



0 = 1.5P + 200 (4)


0 = 2Q 400 (5)

Solving Eqs. 4 and 5 gives,

P =

Q = N

Same result as before.

MAR = MA: 0 = rAB { Pj} + rAC { Qj} + rAD { 200j}

or

0 = { 1.5k} { Pj} + {2i} { Qj} + {2i + k} { 200j}

0 = 5( P)k j + 2( Q)i j + 2( 200)i j + ( 200)k j

= i = k = k = i

0i + 0j = ( 1.5P + 200)i + ( 2Q 400)k

Because R passes

through point A, the

moment is zero.

12

k

ji


14. The end plate of a pressurized tank is held in place by forces from three bolts.

Determine the required value of bolt-force P and angle if the resultant of the

three bolt forces is to act through the center of the plate at O.

y

x

z

P 500 NB

A

700 N

O

0.5 m

40°

C


1 Place the resultant, R, at point O and then equate the moment of R about O (which is

zero because R passes through O) to the moment of the given forces about O.

2

y

x

z

P 500 NB

A

700 N

O

0.5 m

C

rOC

rOB

rOA

Introduce position vectors,

all with tails at point O.

rOA = {0.5j} m

rOB = {0.5 cos 40°i 0.5 sin 40°j} m = {0.3830i 0.3214j} m

rOC = { 0.5 cos i 0.5 sin j} m

y

z

B

A

O0.5 m

CR

x

40°


Equate moments about O.3

MOR = MO: 0 = rOA { 700k} N + rOB { 500k} N + rOC { Pk} N

or

0 = {0.5j} { 700k} + {0.3830i 0.3214j} { 500k} + { 0.5 cos i 0.5 sin j} { Pk}

0 = 0.5( 700)j k + 0.3830( 500)i k 0.3214( 500)j k 0.5 cos ( P)i k 0.5 sin ( P)j k

= i = j = i = j = i

k

ji

Collecting coefficients of i and j gives

0i + 0j = [0.5( 700) 0.3214( 500) 0.5 sin ( P)]i + [0.3830(500) 0.5 cos (P)]j (1)


0 = [0.5( 700) 0.3214( 500) 0.5 sin ( P)]

or, after some arithmetic,

P sin = 378.6 (2)

4

R passes through O, so

produces zero moment.


5 Equating coefficients of j in Eq. 1 gives

0 = 0.3830(500) 0.5 cos (P)

or, after some arithmetic,

P cos = 383.0 (3)

Eqs. 2 and 3 are best solved by using a calculator capable of solving simultaneous

nonlinear equations. Alternatively, dividing Eq. 2 by Eq. 3 gives

6

P sin 378.6

P cos 383.0

tan

Solving for gives,

= 44.7°

=

Substituting for in Eq. 3 gives,

P cos = 383.0

44.7°

Solving gives

P = 539 N

ns.

ns.


z

6 m

A

O

y

x

B

60 N

40 N

140 N

15. Replace the system of forces by a wrench.

Determine the pitch and axis of the wrench.

12 m

4 m


14

r AB

r AB

Thus the vector form of the 140-N force is

F AB = (140 N)

= (140 N)

= {40i + 120j 60k} N (1)

42 + 122 + ( 6)2

4i + 12j 6k

4 m

12 mr

AB

B

x

y

A 6 m

z

2

1 Express the 140-N force in terms of

rectangular components. First introduce a

position vector r ABfrom A to B.

r AB {4i + 12j 6k} m

4 Forces along x and z axes

3 The resultant of all three forces is

R F

{40i + 120j 60k} N {40i} N + {60k} N

120j (2)

140 N

40 N

60 N


60 N

z

6 m

A

O

40 N

y

x

B

C

D

Moment arm is zero.

(40 N force passes through O)

MRO r

OA F AB + r

OA (60k) + 0 ( 40i)

(6k) (40i +120j 60k) + (6k) (60k)

{ 720i + 240j} N m

6

Moment resultant about O:5

F AB

r OA {6k} m

12 m

6(40) k i + 6(120) k j + 6( 60) k k + 6(60) k k

j i 0 0


Express MRO in terms of components parallel M and

perpendicular M to R.

7

By definition,

pitch of wrench, p,

M = { 720i} N m

R {120j} N

M {240j} N m

MRO { 720i +240j} N m

x

z

y

8 M

R

240120

2 m Ans.

||

||

||


10

Old line of action of R

(Intersection of new line

of action with x-z plane)

New position of R

New line of action of R

Old position of R

R

r OP {xi + zk} m

O

r OP

P(x, 0, z)x(120) i j + z(120) k j 720i

k i

(120z)i + (120x)k 720i + 0k


120z = 720

z = = 6 m


120x = 0

x = 0

720120

Now move R to a new line of action such that in the

new location R will produce a moment (about O)

equal to M .

Equate the moment of R in the new position to M

rOP R M

or

{xi + zk} {120j} 720i

M = { 720i} N m

R {120j} Nx

z

y

9


6 m

Ans.

Ans.

||

R {120j} N

M {240j} N m

x

z

y

O

Because a couple moment has the same value

about all points it can be moved to the new

line of action of R to form the wrench.

The axis of the wrench is a vertical line

(same direction as R {120j}) passing

through the point (0, 0, 6 m)

12

11


F 1 4 lb

M 1 60 lb in.

1.5 in.

M 2 80 lb in.

30°F

2 6 lb

z

6 in.

1 in.

3 in.

Dx

16. In a machining operation, holes are simultaneously drilled at points

A and B of the wedge. The drill at A produces a force and couple

moment perpendicular to the planar surface at A. The force and couple

moment at B are similarly perpendicular to the planar surface at B.

Replace the forces and couple moments by a wrench. Determine

a) the magnitude R of the resultant force,

b) the pitch of the wrench,

c) the axis of the wrench, and

d) the point where the axis intersects the x-z plane.

E

B

A

C

y

4 in. 4 in.


Express the forces and couple moments in rectangular components.

90° 30°

View from positive x axis

1

Equal

2

E

x, DCz

30°

A30°

E

C

A

x, Dz

Geometry

M 2 60 lb in.

F 1 4 lb

60°

yy60°

60°

F 1 (4 lb) sin 60°j (4 lb) cos 60°k

{ 3.464j 2k} lb (1)

M 1 (60 lb in.) sin 60°j (60 lb in.) cos 60°k

{ 51.962j 30k} lb in. (2)

3


z

B

y

M 2 80 lb in.

F 2 6 lb

x

4 F 2 { 6i} lb (3)

M 2 { 80i} lb in. (4)


4 in.

z

C

4 in.

E

A

B

y

B

30°

E

(3 in.) cos 30° 2.598 in.

A1.5 in.

x, D1 in.

6 in.

3 in.

y

Cz

x

Next, determine the coordinates of points A and B.

A y (3 in.) sin 30° 1.5 in.

A z (6 in. + 1 in.) 2.598 in.

4.402 in.

6

B y 1.5 in.

B z 1 in.

A x 4 in.

B x 4 in. + 4 in.

8 in.

8

7

5


z

A

B

y

xO

Introduce position vectors from O, the origin of coordinates, to A and B.

r OA A

xi + A yj + A

zk

{4i + 1.5j + 4.402k} in. (5)

9

10

r OB B

xi + B yj + B

zk

{8i + 1.5j + k} in. (6)

11


Resultant force

R F: R { 3.464j 2k} + { 6i}

{ 6i 3.464j 2k} lb (7)

12

Resultant moment about point O

MRO M

O: MRO r

OA F 1 + r

OB F 2

13

Carrying out the cross products and simplifying gives

MRO {12.249i + 2j 4.856k} (8)

i j k i j k

MRO = 4 1.5 4.402 + 8 1.5 1

0 3.464 2 6 0 0


Display R and MO with respect to the xyz axes.14

x

y

z

O

R

MRO {12.249i + 2j 4.856k}

R { 6i 3.464j 2k} lb


O

MRO

R

Two intersecting vectors define a plane. Display the plane defined by R and MRO.15

P Q

RS


O

MRO

R

P Q

RS

M

u

16 Unit vector in R direction

By Eq. 7 for R,

( 6)2 + ( 3.464)2 + ( 2)2

17

0.832i 0.480j 0.277k (10)

7.211 lb Ans. (9)

u RR

R

u RR

6i 3.464j 2k

7.211

Thus

M

Express MRO in terms of components

parallel M and perpendicular M to R.

||

||

M 9.806 lb in. (11)

{12.249i + 2j 4.856k}·{ 0.832i 0.480j 0.277k}

Performing the multiplications and

then simplifying gives

18 M component of MRO in direction of u

MRO· u

||

||

19

M M u

( 9.806){ 0.832i 0.480j 0.277k}

{8.159i + 4.707j + 2.716k} lb in. (12)

In vector form, since M lies in the

direction of the unit vector u,||

||||


The component of MRO perpendicular to R can now be

computed by starting with the vector sum

{12.249i + 2j 4.856k} {8.159i + 4.707j + 2.716k}

{4.090i 2.707j 7.572k} lb in. (13)

and rearranging to get

20

M = MRO M

MRO M M

by Eq.8 by Eq.12

||

||


Now we move the force R to a new line of

action such that R will produce a moment

about O exactly equal to M

21

Old position of R



New position of R

(below plane SPQR)

O

R

P Q

RS

M

||

M

O

R Parallel


View in terms of xyz axes

O

R



y

x

22

P(x, 0, z)-intersection of new line

of action with the x-z plane

New position of R

Old position of R

Parallel

r OP

zBecause the force R, in its new position, is to create a

moment about O equal to M , we can write

r OP R M

23

{xi + zk} { 6i 3.464j 2k} 4.090i 2.707j 7.572k

3.464zi + (2x 6z)j 3.464xk 4.090i 2.707j 7.572k

Performing the multiplications and simplifying gives

or,R

O

by Eq.7

by Eq. 13



3.464z 4.090 (14)

Similarly for j and k

2x 6z 2.707 (15)

3.464x 7.572 (16)

These are three equations in only two unknowns, but the

equations are not all independent. Eqs 14 and 16 imply

Substituting these values into the left hand side of Eq.15 gives

2x 6z 2(2.186) 6(1.181) 2.714

Thus Eq.15 is satisfied (Round-off, error leads to 2.714 rather than 2.707).

24

z 4.0903.464

1.181 (17)

x 7.5723.464

2.186 (18)


Summary: The effect of the drilling operations on the two

surfaces of the wedge is to push the wedge down the

wrench axis (direction of u) while simultaneously causing

the wedge to tend to rotate counterclockwise (as viewed

from a position above the x-z plane) about the wrench axis.

Because a couple moment is the same about all

points, we can move M to point P. Similarly

we can slide R to P along R's line of action, by

the principle of transmissibility.

2528

The axis of the wrench is a line passing through the

point (2.186 in., 0, 1.181 in.) with direction, by Eq. 10,

u 0.832i 0.480j 0.277k

26

R Ru

(7.211 lb)u

by Eq. 9

M M u

( 9.805 lb in.)u

Pitch of wrench M

R

9.806 lb in.

7.211 lb

1.360 in.

27by Eq. 11

y

z

x

2.186 in.

1.181 in.

Pu

Ans.

Ans.

|| ||

||

||

4.5 equivalent force-couple systems - civil … equivalent force-couple systems example 1, ... { 4j...

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