4 - moment of a force

8/17/2019 4 - Moment of a Force

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ES 11 Lecture 4


2/63

➔Moment of a force is the measure of

the tendency of a force F to make

the rigid body rotate about a fixed

axis perpendicular to the plane ofthe force F .


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F

r

Mo

o

d

A

pt. O – fixed point/axis on the plane of the force and pt.

r - position vector of F acting at pt. A relative to the fixed

pt. O. It is a vector from the point to the line of action of F

pt. A – pt. of application of force F


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F

r

Mo

o

d

A

(vector)

(scalar or magnitude)

The moment of a force about a given point is the same even if it is

computed using a different displacement vector as long as?

0=

0 =

0 = ×

d – perpendicular distance from the point to the line of action of

the force


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F o

A

'

F

r

r'

B

Two forces F and F’ are equivalent if, and only if, they are

equal (i.e. have the same magnitude and same direction)

and have equal moments about a given point O.

F = F’ Mo = M’ o


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The moment about a given

point O of the resultant of

several concurrent forces is

equal to the sum of themoments of the various

moments about the same

point O.

2121 F r F r F F r


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Varignon’s Theorem makes it possible to replacethe direct determination of the moment of a

force F by the moments of two or more

component forces of F .

A

O

A

O

R

r

r

1 F

2 F

3 F

...321 F F F xr R xr


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The moment of F about B(not from the origin),

F r M B A B

/

k z z j y yi x x

r r r

B A B A B A

B A B A

/

r - position vector from the point O to the force

k F j F i F F z y x


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z y x

B A B A B A B

F F F

z z y y x xk ji

M


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z B A y B A

Z B

z B A y B A B

F y y F x x

M M

k F y y F x x M

For two-dimensionalsystems,

0 = ×

+Positive sign

convention in 2

dimensions


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a) What is the

moment of F aboutpoint A?

b) What is the

perpendiculardistance from point

A to the line of

action of F ?

Example: The line of action of the 90-N

force F passes through points B and C.

z

y

x

B Ar /

F

A(5,0,6)m

B(4,11,0)m

C(0,7,7)m

=90N

BC r /

BC l


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z

y

x

B Ar /

F

A(5,0,6)m

B(4,11,0)m

C(0,7,7)m

=90N

BC r /

BC l


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14

a) To evaluate the cross-product, we need thecomponents of F. The vector from B to C is (0-4)i +

(7-11)j + (7-0)k = 4i - 4j + 7k. Dividing this vector by

its magnitude, we obtain a unit vector eBC

has the

same direction as F .

.

k jie BC 9

7

9

4

9

4

k N j N i N e F BC 70404090


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.

Choose the vector r , let say The position vector from A to B

k jir A B 6001154/

mk jir A B 611/ Evaluating r B/A x F (moment of F about A)

F xr M A B A /

jik k ji 7024044040240770

40

11

40

1

704040

6111

jik ji

Nmk ji M A 480310530


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.

b) The perpendicular distance is

m F

M A

66.8704040

480310530

222

222


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The cables AB and AC extend from an

attachment point A on

the floor to attachment

points B and C in thewalls. The tension in

cable AB is 10 kN and the

tension in cable AC is 20

kN. What is the sum of

the moments about O of

the forces exerted on A

by the two cables?

B(8,0,4)m

x

z

y

C(0,6,3)m

A(6,4,0)m


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The cables AB and AC extend from an

attachment point A on the floor to

attachment points B and C in the walls. The

tension in cable AB is 10 kN and the tension

in cable AC is 20kN. What is the sum of themoments about O of to the forces exerted

on A by the two cables?

B(8,0,4)m

x

z

y

C(0,6,3)m

A(6,4,0)m


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Let F AB and F AC be the forces exerted on the attachment point Aby the two cables. To express F AB in terms of its components,

we determine the position vector from A to B,

B(8,0,4)m

x

z

y

C(0,6,3)m

A(6,4,0)m

k jik ji 442044068

and divide by its magnitude to the

yield the unit vector e AB

k ji

k ji

e AB 3

2

3

2

3

1

6

442

kN k jie F AB AB

67.667.633.310

kN k ji F AC 57.871.514.17


20/63

Choose the vector r such that the lines of action of both forces

pass through point A, we can use the vector from O to A to

determine the moment of both forces about point O.

jir O A 46/

AC O A ABO A F xr F xr Mo //

57.871.514.17046

67.667.633.3046

k jik ji

jk k i jk k i 42.5156.6826.3428.3402.4032.1302.4068.26

kNmk jio M 48.4944.9196.60

x

z

y

C

A

B O

O Ar / AC F

AB F


21/63

A force of 800N acts on a bracket as

shown, determine a) the moment of theforce about B b) perpendicular distance

from B to the line of action of the force

B

A

200mm

160mm

800N060


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B

A

200mm

160mm

800N060


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B

A

200mm

160mm

800N060

jmimr B A 16.02.0/

j N i N F 00 60sin80060cos800

j N i N F 693400

j N i N x jmim F xr M B A B 69340016.02.0/

k Nmk Nm 646.138

k Nm M B 6.202

b) Perpendicular distance is 253.2 mm


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➔In 2 dimensional problems, it may be easier to

decompose the force into components➔Using the principle of transmissibility, the

perpendicular distances can be determined.

➔Determine the sign of the moments using theright hand rule.


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Consider a line L and a force F . Let MP be the

moment of F about an arbitrary point P on L.

L

P

R xr M p

r F

L

P

p M

L M

F xr M P

L P L e M M

➔measure of the tendency of a force F to make the

rigid body rotate about a line


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In vector form, ML is given by

L L P L ee M M

L L L ee F xr M

Where e is the unit vector along L


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The mixed triple product can be expressed in

determinant as

z y x

z y x

z y x

L

F F F

r r r

eee

F xr e M

Notice that the value of the above formula is a scalar which

determines the magnitude and direction of M L, relative to e.

Note: If ML is +

then component isthe same direction

as e, negative

otherwise.


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30/63

Determine the moment of a force

about an arbitrary line AB.

y

z

x

A(2,0,4)m

(8,6,4)m

B(-7,6,2)m

r k ji F 206010


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y

z

x

A(2,0,4)m

(8,6,4)m

B(-7,6,2)m

r k ji F 206010


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y

z

x

A(2,0,4)m

(8,6,4)m

B(-7,6,2)m

r k ji F 206010

Solution: If we chose point A, the vector r from A to the pointof application of F is

k jir 440628

m jir 66


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y

z

x

A(2,0,4)m

(8,6,4)m

B(-7,6,2)m

r k ji F 206010

The moment of F

about A is

206010066

k ji

F xr M A

Nmk ji M A 300120120

mk jir A B 269/

mr A B 0.11/ k ji

r

r e

A B

A B

A B11

2

11

6

11

9

/

/

/


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y

z

x

A(2,0,4)m

B(-7,6,2)m

F

A Bl /

A B A A B L e M e M //

A Be /30011

2120

11

6120

11

9

Nmk ji M L

11

2

11

6

11

9109

Nme M A B L /109


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36/63

A pair of forces of:

• Equal magnitudes

• Opposite directions

• Parallel LOAs

d

A couple tends to

cause rotation of anobject even though

the vector sum of the

forces is zero.


37/63

d

If you can't turn the nut ofa wheel, what do you doto turn it?• You can call a stronger

person to turn it,thereby increasing F, or

• You can place yourhands farther apart onthe wrench, thereby

increasing d.

Why does either of thesework?


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F1

F2

F3

F4

F5

F7

F6


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Oo na! Ako na! Ako nang

mag-isa!


41/63

A couple can onlycause rotation of anobject.

The resultant force ofa couple is zero.

Its resultant moment

is not zero.


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43/63

.

y

C

AB

D

1 N

1 N

3 m

5 m

1 m

1 m

E

1 m

Nmk N jmi N jmi M C

2)15()13( Nmk N jmi N jmi M D

2)11()13( Nmk N jmi N jmi M E

2)11()11( The moment of a couple about any point is

constant.

3 m


44/63

F

F

1r

2r P

)(21 F xr F xr M p

F xr r 21

21 r r r

F xr M p The displacement vector r is from any point on the line of action

of the force to any point on the line of action of the other force.

F should be the vector where r terminates!


45/63

The force F is 10i – 4 j (N). Determine the moment of

the couple.

(6, 6, 0) m

(8, 3, 0) m

F

F

Th f F i 10i 4j (N)


46/63

The force F is 10i – 4 j (N).

Determine the moment of the

couple.

(6, 6, 0) m

(8, 3, 0) m

F

F


47/63

(6, 6, 0) m

(8, 3, 0) m r

F

F

jir ˆ3ˆ2

ji x ji F xr M ˆ4ˆ10ˆ3ˆ2

m N k M ˆ22

(6, 6, 0) m (8, 3, 0) m

P (10, 7, 3) m

1r

2r

F

F

F xr F xr M

21

0410

314

ˆˆˆ

0410

342

ˆˆˆ

k jik ji

M

m N k M ˆ

22


48/63

From the definition of the moment of a couple, itfollows that two couples, one consisting of the forces

F 1 and F 2, will have equal moments if

2211 d F d F Magnitude

d1 d2

1 F

1 F

2 F

2 F


49/63

Two couples are said to be equivalent if and only iftheir moments have the same magnitude and

direction. In addition, since M = r x F , the orientation

of the planes of r and F should be the same.

Characteristics of a Couple

• Magnitude, M = Fd

• Direction (clockwise or counterclockwise)• Aspect or Orientation of the plane where force

(that make up the couple) are.


50/63

1) The magnitude of the forces that make up the

couple can be changed with an inverse change in

the perpendicular distance without changing the

effect (magnitude) of the couple.

1.0m

10N

10N 0.5m

20N

20N


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2) The directions of the forces can be changed aslong as the magnitude of the couple and the

orientation of the plane where the forces are

acting remains the same.

1.0m

10N

10N

1.0m

10N

10N


52/63

3) Two couples having the same moment M, areequivalent even if they are at different planes as

long as those 2 planes are parallel.

x

10N

1.0m

10N

10N

1.0m y

z

M

' M


53/63

The sum of two couples M1 and M2 is a couple with

M equal to the vector sum of M1 and M2.

21

M M M

+ =

1 M

2 M M


54/63

The couple vector M may

be resolved into component

vectors Mx, My , and Mz

directed along the axes of

coordinates and

representing couples acting,respectively in the yz, xz,

and xy planes.

The moment of a couple or couple vector obeys the

law of vector addition.

y

x

z

z

y x

M

X M

Z M

Y M


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Couple Vector Force Vector

y

x

z

M

M


56/63

➔Determine the sum of the moments exerted

on the pipe by the two couples.

60O

60O

30N

30N 20N 20N

y

x

z

4m

4m

2m


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30sin60O

20N 20N

y

x

z

30sin60O

30cos60O

30cos60O

M1

M2

M3

m N iim N M ˆ40ˆ2201

m N j jm M o ˆ9.103ˆ460sin302

m N k k m M o ˆ60ˆ460cos303

m N k ji M M M M ˆ60ˆ9.103ˆ40321


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➔Find the resultant of the three couples.

y

x

z

270Nm

220Nm

200Nm

0.9m

1.2m

60O


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y

x

z

270Nm

220N

m

200Nm

0.9m

1.2m

60O

m N k M ˆ2701

m N k jk j M oo ˆ110ˆ5.190ˆ30sin220ˆ30cos2202

m N k ik i M ˆ120ˆ160ˆ2005

3ˆ200

5

43

m N ii M M X X ˆ160ˆ

m N j j M M Y Y ˆ5.190ˆ

m N k k k M M Z Z ˆ40ˆ120110270ˆ


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60

m N k ji M ˆ40ˆ5.190ˆ160

m N M M M M Z Y X 0.252405.190160 222222

y

x

z M

Z M

X M

Y M

o X

X M

M 4.129

252

160coscos

11

oY

Y M

M 9.138

252

5.190coscos

11

o Z

Z M

M 9.80

252

40coscos 11


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1. Beer, Vector Mechanics for Engineers: Statics,

9th Ed


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62

What is the

moment of theforce F about

the bar BC? y

z

x

A(4,2,2)m

B(0,0,3)m

C(0,4,0)m

kN k ji F 362


63/63

The mixed triple product is

362

124

6.08.00

//

F xr e B A BC

kNm F xr e B A BC 8.24//

mkN ee F xr e M BC BC B A BC BC //// 8.24

BC l /

BC r /

B BC e /

4 - moment of a force

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