equivalent forces - dynamics

7/29/2019 Equivalent Forces - Dynamics

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4.5 Equivalent Force-Couple Systems


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4.5 Equi valent Force-Couple Systems Example 1, page 1 of 1

1. Replace the force at A by an equivalentforce 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

3 45

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 Nm

3 Display the equivalent force and couple momentat point O.

0

53

4

A

20 N

x

y

ns.

136 Nm


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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 Nmm= 6 Nm

Display the equivalent forceand 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


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

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 whilesimultaneously twisting (with a 6 lbin. moment)

the entire cap away from the top of the bottle.

6 lbin. ns.

3 in.

2 lb


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4. Replace the given forces and couple moment by aresultant force and couple moment at A.

2 Calculate the resultant force

A

120 lb260 lb

800 lbft

7 ft4 ft2 ft

Express the inclined force in rectangularcomponents.

1

2 ft 4 ft 7 ft

800 lbft

120 lb

13125 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 lbft

= 3640 lbft


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Determine the magnitude and direction of theresultant 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 lbft

ns.

74.5


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5. Replace the force F = 3 kN acting on corner A of theblock by an equivalent force and couple moment actingat the center C of the block.

A

y

xC

z

F = 3 kN

150 mm

150 mm

100 mm 100 mm


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Calculate the couple moment about C.

A

y

xC

z

F = 3 kN

150 mm

150 mm

1

MC = rCA F

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

= 100( i j 150( k j

= k = i

= { 450i 300k}kNmm

= { 450i 300k}N m Ans.

ij

k

rCA

100 mm100 mm


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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}Nm

100 mm 100 mm


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B

C

x

y

z

4 lb

7 lb

A

48 in.

6. Replace the forces acting on the ice auger by anequivalent force and couple moment acting at A.

6 in.


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B

C

x

4 lb

7 lb

A

48 in.

Rx = Fx: Rx = 0Ry = Fy: Ry = 4 lb

Rz = Fz: Rz = 7 lb

Calculate the resultant force1

z

6 in.

y


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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 forcepasses throughout point A.

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

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

= j = i

= {336i + 42j} lbin.

y


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B

C

x

y

A

MAR= {336i + 42j} lbin.

R= { j + 7k} lb

ns.

z

3


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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 kipft

40


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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.


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4

A B C D

2 ft 8 ft 4 ft 3 ft

3 kip

4 kip

20 kipft

We equate the moment of the newsystem, about point A, to the moment

of the original force-couple system(The choice of point A for summing

moments is arbitrary; any other pointwould work as well, except that we

must use thesame point for both theoriginal system and the new system.)

6

5 This is a new force-couple system that we want to makeequivalent to the original force-couple system. We

already know that theforces are equivalent because Ris the resultantof 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 unknowndistance d from the left end and then choosing d so that

the moment of this new system is the same as that ofthe 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 kipft (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 originalforce-couple momentsystem.


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8. Replace the forces acting on the frame by a single forceand specify where its line of action intersectsa) member BCD and b) member AB.

1 Determine the resultant force.

Rx = Fx: Rx = 800 N = 800 NRy = 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 = 58.41300 N

800 N

1300800

Ans.

Ans.

4 m 4 m


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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 ofaction of the resultant force intersectsmember 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


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Choose d' so that the moment about B of the resultantforce equals the moment of the original force system.

4 Part b) To determine where the line ofaction of the resultant force intersects

member AB, place the force a distanced' 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 58.4d'

2 m

2 m

4 m4 m


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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


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7 Alternative solution for part b. Once we know wherethe 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 ofresultant forceG

d'4.92

tan 58.4 =

Solving gives,

d' = 8.0 m

(same result as before)

From triangle CBG,8

58.4

d'


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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 themagnitude 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


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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


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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 H JI

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 theresultant force equals themoment of the original force

system about H. Any otherpoint besides H could beused, but H has the advantage

that the moment of theresultant R is zero, since R is

known (as part of thestatement of the problem) to

pass through H.

2

130 lb

3


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Magnitude of resultant.

From Eqs. 1 and 2,

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

4


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10. A machine part is loaded as shown. The part is to be attached toa 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 themagnitude and direction of the resultant force.

x

y

0.3 m

0.5 m

90 Nm

20 NmO

60 N80 N

0.6 m0.4 m

60


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x

y

90 Nm

20 NmO

60 N

Rx = Fx: Rx = 40 N = 40 N

Ry

= Fy: R

y= 69.28 N 60 N

= 129.28 N

= 129.28 N

Determine the resultant force.1

Ans.

Ans.

129.2840

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


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x

0.5 my

x

y

90 Nm

20 Nm O

60 N

Rotation is caused by moment.

If the given force-couplesystem 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 ofaction. So the line of action is

the line on which the boltshould be placed to preventrotation.

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 theresultant (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 Nm 20 Nm + (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


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Solving for y gives the equation of a line

y = 3.232 x + 0.357 Ans.

This line defines the possible locations ofthe 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


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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


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3 First, equate moments about the x axis. We can useeither 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 kipC, 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 tothe original system of

forces, place R at thepoint (x, 0, z) and thendetermine values of x

and z such that Rproduces the same

moments about the xand 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


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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

MzR= 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


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8 Solving Eq. 2 gives

x = 4.80 ft Ans.


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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 areacted upon by the wind forces shown. Replace the forces by anequivalent 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


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A

B

C

(x, y, 0)

1550 N

y

z

y

y

Ox

5 mz 1 m

4 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 samemoments 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 m6 m

zx,O

y

x,O

y

R = 1550 N

zView from the

positive x axis


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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 thepositive 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


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

magnitude of forces P and Q if the resultant of all threeforces 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)


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

P and Q such that R produces the same moment about the xand z axes as the given forces, P, Q, and 200 N, produce.

B

C

D

P

Q

xz

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

A3 m

2 m1 m

1.5 m

3 m

1.5 m

O

O


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40/67


P and Q were defined asdownward directedforces.

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.


41/67


Alternative solution: the computation are simplified

somewhat if we sum moments about point A insteadof point O.

Equate the moment of R about point A tothe moment of the given forces about A.

11

200 N

rAC

D

A

rAD C

Q

9

rAB

P

Introduce position vectors, all withtails 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


42/67


Equating coefficients ofi gives

0 = 1.5P + 200 (4)

Equating coefficients ofkgives

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 passesthrough point A, the

moment is zero.

12

k

ji


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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 thethree 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


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

rOCrOB

rOA

Introduce position vectors,all with tails at point O.

rOA = {0.5j}m

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

rOC = { 0.5 cos i 0.5 sin j} m

y

z

B

A

O0.5m

CR

x

40


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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 ofi andj gives

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

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.


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5 Equating coefficients ofj 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 simultaneousnonlinear equations. Alternatively, dividing Eq. 2 by Eq. 3 gives

6

P sin 378.6P cos 383.0

tan

Solving for gives,

= 44.7

=

Substituting for in Eq. 3 gives,

P cos = 383.0

44.7Solving gives

P = 539 N

ns.

ns.


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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


48/67


14

rAB

rAB

Thus the vector form of the 140-N force is

FAB = (140 N)

= (140 N)

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

42 + 122 + ( 6)2

4i + 12j 6k

4 m

12 mrAB

B

x

y

A 6 m

z

2

1 Express the 140-N force in terms ofrectangular components. First introduce aposition vectorrABfrom A to B.

rAB {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


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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 rOA F

AB + r

OA (60k) + 0 ( 40i)

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

{ 720i + 240j} N m

6

Moment resultant about O:5

FAB

rOA {6k} m

12 m

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

j i 00


50/67


Express MRO in terms of components parallel M andperpendicularM 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.

||

||

||


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10

Old line of action ofR

(Intersection of new lineof action with x-z plane)

New position ofR

New line of action ofR

Old position ofR

R

rOP {xi + zk} m

O

rOPP(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

Equating coefficients ofkgives

120x = 0

x = 0

720

120

Now move Rto a new line of action such that in thenew location Rwill produce a moment (about O)

equal to M .

Equate the moment ofRin the new position to M

rOP R M

or

{xi + zk} {120j} 720i

M = { 720i} N m

R {120j} Nx

z

y

9


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6 m

Ans.

Ans.

||

R {120j} N

M {240j} N m

x

z

y

O

Because a couple moment has the same valueabout all points it can be moved to the newline of action ofRto 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


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F1 4 lb

M1 60 lb in.1.5 in.

M2 80 lb in.

30

F2 6 lb

z

6 in.

1 in.

3 in.

Dx

16. In a machining operation, holes are simultaneously drilled at pointsA 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, andd) the point where the axis intersects the x-z plane.

E

B

A

C

y

4 in. 4 in.


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Express the forces and couple moments in rectangular components.

90 30

View from positive x axis

1

Equal

2

E

x, DCz

30

A

30

E

C

A

x, Dz

Geometry

M2 60 lb in.

F1 4 lb

60

yy60

60

F1 (4 lb) sin 60j (4 lb) cos 60k

{ 3.464j 2k} lb (1)

M1 (60 lb in.) sin 60j (60 lb in.) cos 60k

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

3


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z

B

y

M2 80 lb in.

F2 6 lbx

4 F2 { 6i} lb (3)

M2 { 80i} lb in. (4)


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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.

Ay (3 in.) sin 30 1.5 in.

Az (6 in. + 1 in.) 2.598 in.

4.402 in.

6

By 1.5 in.

Bz 1 in.

Ax 4 in.

Bx 4 in. + 4 in.

8 in.

8

7

5


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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

rOB Bxi + B

yj + B

zk

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

11


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Resultant force

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

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

12

Resultant moment about point O

MRO MO: M

RO 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


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Display Rand 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


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O

MRO

R

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

P Q

RS


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O

MRO

R

P Q

RS

M

u

16 Unit vector in Rdirection

By Eq. 7 forR,

( 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 componentsparallel M and perpendicularM to R.

||

||

M 9.806 lb in. (11)

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

Performing the multiplications andthen simplifying gives

18 M component ofMRO in direction ofu

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 thedirection of the unit vectoru,

||

||||


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The component ofMRO perpendicular to Rcan now becomputed 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

||

||


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Now we move the force Rto a new line ofaction such that Rwill produce a moment

about O exactly equal to M

21

Old position ofR



New position ofR

(below plane SPQR)

O

R

P Q

RS

M||

M

O

R Parallel


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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 ofR

Old position ofR

Parallel

rOP

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

moment about O equal to M , we can write

rOP 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


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3.464z 4.090 (14)

Similarly forj 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)


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equivalent forces - dynamics

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