4.2 Moments in Three-Dimensional Force · PDF file4.2 Moments in Three-Dimensional Force Systems Example 1, page 1 of 3 200 mm 200 mm A F = 50 N 1 1. Use the cross-product definition

4.2 Moments in Three-Dimensional Force Systems

4.2 Moments in Three-Dimensional Force Systems Example 1, page 1 of 3

200 mm

200 mm

A

F = 50 N

1

1. Use the cross-product definition of the moment of a force to determine the

moment of the force about point A. Also, compare the sign of the result with

that obtained from the scalar definition of positive moment, M = Fd.

B

HEAVY DUTY 250MM CHINA

Introduce a position vector rAB

with tail at A and head at B.

B

A

rAB = { 200i} mm

x

y

+

2 Express the force in rectangular components.

z

y

x

F = 50 N

30°

(50 N) sin 30° = 25 N

(50 N) cos 30° = 43.30 NF = {43.30i 25j} N (1)


y

x

z

z

30°


6 MA = ( 200 mm)( 25 N)k

= {5000k} N·mm

= {5k} N·m Ans.

3 Use the vector cross product to compute the moment

about point A.

MA = rAB F

= { 200i} mm {43.30i 25j} N

= [ 200(43.30) i i + ( 200)( 25) i j ] N·mm

= 0, because cross

product of parallel

vectors is zero

= k, by the right-hand rule

4

j

k

i

A memory aid for cross products:5

Assign a plus sign if the product is in

the order indicated by the arrowheads;

minus sign otherwise

k

i

j

i j = +k

i k = j

from i to j

thumb points out

of i-j plane

fingers of right hand

curl from i to j

ik

j

Apply the

right-hand rule.


k

50 N

8 The 50-N force tends to rotate the

wrench counterclockwise about

the axis (through A) defined by

the unit vector k. By the

right-hand rule, this definition of

positive moment reduces to the

usual sign convention for positive

moment in coplanar problems.

7 Display the MA vector (a double-headed arrow)

z

y

x

MA = {+5k} N·m

y

50 N

MA = {+5k} N·m

x

+

direction of fingersthumb points

out-of-plane

Ans.



A

A

z

30°

30°


1

2. A force F = 20 N is applied to the end of a string of length L. The other end of

the string is tied to the handle of a wrench as shown. Use the cross-product

definition of the moment to determine the moment of F about point A. Discuss the

effect of distance L on your answer.

Introduce a position vector rAC

with head at C and tail at A.

F = 20 N

B

A

C

rAC

z

y

x


30°x

z

y

C

L


B

F = 20 N

A

200 mm

30°


200 mm

A

2

B

Determine the rectangular components of rAC.

C

y

z

x

rAC

L cos 30°

rAC = (L cos 30° + 200 mm)i (L sin 30°)j (1)53

L sin 30°4

L

x

y

z

C

F = 20 N

30°

Determine the rectangular

components of the force F

6

F = (20 N) cos 30° i (20 N) sin 30° j (2)


30°


7

MA = rAC F

= {( L cos 30° 200)i L sin 30° j} mm { 20 cos 30° i 20 sin 30° j} N

= [( L cos 30° 00)( 20 cos 30°)(i i)

= 0

+ ( L cos 30° 00)( 20 sin 30°)(i j)

= k

+ ( L sin 30° )( 20 cos 30°)(j i)

= k

+ ( L sin 30°)( 20 sin 30°)(j j) ] N·mm

= 0

j

k

i

= k[( L cos 30° 200)( 20 sin 30°) ( L sin 30°)( 20 cos 30°)] N·mm

= k[( L cos 30°)( 20 sin 30°) 200( 20 sin 30°)

( L sin 30°)( 20 cos 30°)] N·mm

The terms involving L drop out

= {2000k} N·mm = {2k} N·m Ans.

Use the cross product to compute the moment.


Ans.

30°

30°


Note: All position vectors

and force vectors are in

the xy plane.

z

x

y

3

2

1

F

rA3

rA1rA2

rAC

C

Discussion: Distance L does not appear in the answer

for MA. Thus the result is valid for all values of L,

and so the head of the position vector can be located

anywhere (for example, points 1, 2, 3, or C in the

figure) on the line of action of the force.

B

8

A


3. A shower/bathtub grab bar is being pulled by a force F = 30 lb as

shown. Determine the moment of F about the support A. Also determine

the coordinate direction angles of the moment vector and interpret the

result.

16 in.

8 in.

5 in.

A

B

z

x

y

F = 30 lb

40°

60°


16 in.

8 in.

5 in.

A

B

z

x

y

40°

60°

Introduce a position vector with tail at A and head at B.

rAB = { 16i + 8j + 5k} in. (1)

1

rAB

F = 30 lb


Determine the rectangular

components of the force F.

2

F y = (30 lb) cos 60°

= 15 lb (2)

3

(30 lb) sin 60°4

F z = (30 lb)(sin 60°)(sin 40°)

= 16.70 lb (3)

5

F x = (30 lb)(sin 60°)(cos 40°)

= 19.90 lb (4)

6

Component form of F,

from Eqs. 2-4:

F = { 19.90i 15j + 16.70k} lb (5)

7

40°

60°

x

y

z

F = 30 lb


8

MA = rAB F

= { 16i + 8j + 5k } { 19.90i 15j + 16.70k}

Because both vectors each have three non-zero

components, evaluating the cross products is

easier if we use the determinant form.

9

i j k

MA = 16 8 5

19.90 15 16.70

Expand the determinant in terms of the first row,

remembering to insert the minus sign for the j term.

10

8 5 16 5 16 8MA = i j + k

15 16.70 19.90 16.70 19.90 15

= i[8(16.70) 5( 15)] j[ 16(16.70) 5( 19.90)] + k[ 16( 15) 8( 19.90)]

= {208.6i +167.7j + 399.2k} lb·in. Ans.

Calculate the moment.


Observation: The above procedure is tedious and error prone. A much

better approach is to use a calculator with a built-in function for evaluating

the cross product of two vectors. If such a calculator is available, all you

need to do is enter the components of the vectors and then let the calculator

perform the arithmetic. As a result, typical errors such as missing a minus

sign or mis-copying a number from one line to the next are avoided.

11

To determine the coordinate direction angles, first determine the

magnitude of the moment.

M A = (208.6)2 + (167.7)2 + (399.2)2

= 480.6 lb·in.

480.6

399.2

480.6

167.7

480.6

208.6

Coordinate direction angles

= cos-1 = cos-1 = 64.3° Ans.

= cos-1 = cos-1 = 69.6° Ans.

= cos-1 = cos-1 = 33.8° Ans.zM

A

M A

M A

M A

M Ay

M Ax


16 in.

5 in.

A

B

z

x

y

40°

60°

64.3°

69.6°

33.8°

M A = 480.6 lb·in.

Interpretation: The force F = 30 lb

produces a moment of 480.6 lb·in. about

point A. This moment tends to rotate the

grab bar about an axis defined by the moment vector.

12

Axis of rotation

Ans.

F = 30 lb


4. A force F = 15 N acting parallel to the z axis is applied to the

handle of a socket wrench to turn a bolt at A. Determine the

moment of the force about the point A. Also, state which

component of the moment tends to turn the bolt.

100 mm

80 mm

A

B

x

y

z

F = 15 N

C


Introduce a position vector

with head at B and tail at A.

rAB = {80i 100j} mm

1

F = 15 N

80 mm

z

A

y

100 mm

B

x


MA = rAB F

= (80i 100j) { 15k}

= 80( 15)(i k) + ( 100)( 15)(j k)

= j = i

= ( 1200)( j) + (1500)i

= {1500i + 1200j} N·mm

= {1.5i + 1.2j} N·m Ans.

2

i

k

j

(F points in negative z-direction.)

rAB

C


Display the moment vector.3

The component that tends to

rotate the shaft AC of the

wrench about the x axis

(and thus turn the bolt) is

M A = 1.5 N·m Ans.

4

100 mm

15 N

B

x

y

z

A

80 mm

M A = 1.5 N·m

x

M A = 1.2 N·m

y

M A

x

C


5. Pulley B is used to drive pulley C. Determine the resultant

moment about bearing A produced by the belt forces acting on

pulley B. Also, interpret your result.

z

x

y

30°

Belt forces

A

BC

D

E

F

Q = 55 N

P = 30 N

40 mm

Radius = 70 mm


z

x

y

30°

Belt forces

A

BC

D

E

F

Q = 55 N

P = 30 N

40 mm

Radius = 70 mm

1 Introduce position vectors with

tails at A and heads at E and F.

rAF

rAE

2 From the figure,

rAE = {40i 70j} mm (1)


30°

z

x

y

30°

Belt forces

ABC D

E

F

Q = 55 N

P = 30 N

40 mm

Radius = 70 mm

3 To determine the components of r AF, consider a

view of the pulley from the positive x axis:

rAF

rAE

7 In component form, from Eqs. 2 and 3,

rAF

= {40i + 60.62j 35k} mm (4)

Q = 55 N

P = 30 N

z

y

B

F

6

4

5

rAFz = (70 mm) cos 60° = 35 mm (3)

Angle = 90° 30° = 60°

rAFy

= (70 mm) sin 60°

= 60.62 mm (2)

Radius = 70 mm


z

x

y

30°

Belt forces

AB

CD

E

F

Q = 55 N

P = 30 N

40 mm

Radius = 70 mmrAF

rAE

8 Express the forces in rectangular components.

Q = (55 N) sin 30°j + (55 N) cos 30°k

= {27.5j + 47.63k} N (5)

P = {30k} N (6)


MA= rAF × Q + rAE × P

9

i j k i j k

MA = 40 60.62 35 + 40 70 0

0 27.5 47.63 0 0 30

60.62 35 40 35 40 60.62 = i j + k

27.5 47.63 0 47.63 0 27.5

= i[60.62(47.63) ( 35)(27.5)] j[40(47.63) ( 35)(0)] + k[40(27.5) 60.62(0)]

+ i[ 70(30) 0(0)] j[40(30) 0(0)] + k[40(0) 70)(0)]

= {1750i 3105j + 1100k} N·mm

= {1.750i 3.105j + 1.100k} N·m Ans.

70 0 40 0 40 70 + i j + k

0 30 0 30 0 0

Calculate the resultant moment.


40 mm

z

x

y

30°

Belt forces

AB

C

55 N

30 N

M x = 1.750 N·m

Interpretation: The belt forces acting on pulley B tend to

rotate the entire structure with a 3.73 N·m moment about an

axis defined by the direction of the moment vector MA.

The 1.750 N·m x component of MA is the component of

moment that rotates the shaft and drives pulley C.

11

Ans.

The magnitude of the moment is

M A = (1.750)2 + ( 3.105)2 + (1.100)2

= 3.73 N·m

10


6. A child on a bicycle collides with a mailbox and

exerts the force F shown. If the base of the pole at O

will fail if the magnitude of the moment there exceeds

60 N·m, determine if the mailbox will fall over.

250 mm75 mm

A

O

x

y

z

F = {80i + 12j 10k} N

900 mm


Introduce a position vector with tail at O and

head at A.

rOA = {250i + 900j + 75k} mm

1

i j k

= 250 900 75

80 12 10

MO = rOA × F

2

rOA

= i[900( 10) (12)(75)] j[250( 10) 80(75)] + k[250(12) 80(900)]

= { 9900i + 8500j 69000k} N·mm

= { 9.9i + 8.5j 69.0k} N·m

900 75 250 75 250 900 = i j + k

12 10 80 10 80 12


250 mm75 mm

A

O

x

y

z

F = {80i + 12j 10k} N

900 mm


Magnitude of moment

M O = ( 9.9)2 + (8.5)2 + ( 69.0)2

= 70.2 N·m

3

Because 70.2 N·m exceeds the 60 N·m

maximum allowable moment, the

mailbox will fall over.

4

Ans.


7. Copper tubing emerges from the wall at A and is subjected to a

force F at its free end B. The tubing will fail if the magnitude of

the moment at A exceeds 3 N·m. Determine the largest value of

the force F that can be applied to the free end of the tubing.

30°

40°250 mm

300 mm

A

Bx

z

y

F

35°

200 mm


250 mm

200 mm

300 mm35°A

B

x

z

y

F

1 Introduce a position vector rAB

with tail at A and head at B.

r AB

30°

40°

35°


250 mm

200 mm

300 mm

A

B

x

z

y

F

2 Determine the

components of r AB.

r AB

3 r ABx = 200 mm + 250 mm = 450 mm (1)

250 mm

D

C

30°

40°

35°


7 In component form, from Eqs. 1-3,

r AB = {450i 172.1j + 245.7k} mm (4)

r ABz = (300 mm) cos 35° = 245.7 mm (3)6

4

r ABy

= (300 mm) sin 35°

= 172.1 mm (2)

5

To determine the y and z components of rAB,

consider a view from the positive x axis.

r AB

35°

300 mm

y

z

x, B, D

C, A


8 Determine the components of F.

F sin 40°10

F z = F sin 40° sin 30°

= 0.3214F (6)

11B x

z

y

F

F y = F cos 40°

= 0.7660F (5)

9

F x = F sin 40° cos 30°

= 0.5567F (7)

12

13 In component form, from Eqs. 5-7,

F = F{0.5567i 0.7660j + 0.3214k} (8)

30°

40°


MA = rAB × F

14

i j k

= 450 172.1 245.7

0.5567F 0.7660F 0.3214F

172.1 245.7 450 245.7 450 172.1 = i j + k

0.7660F 0.3214F 0.5567F 0.3214F 0.5567F 0.7660F

= (132.9F)i (7.8F)j (248.9F)k

M A = (132.9F)2 + ( 7.8F)2 + ( 248.9F)2

= F (132.9)2 + ( 7.8)2 + ( 248.9)2

= 282.3F N·mm

= 0.2823F N·m

15

Equate M A to the largest allowable moment, 3 N·m:

M A = 3 N·m

0.2823F N·m

Solving gives

F = 10.6 N Ans.

16

Calculate the resultant moment. Compute the magnitude.


8. Two forces, P = 60 N and Q = 80 N act on the vertices of a cube as

shown. Determine the moment of each force about point O, if the length

of each edge of the cube is 2 m. Also, determine the shortest distance

from O to the line BF.

x

z

y

A B

CD

E F

G

O

P = 60 N Q = 80 N

Each edge is

2 m long.


x

z

y

A B

CD

E F

G

O

P = 60 N Q = 80 N

Each edge is

2 m long.

To determine the

moment of the force

P about point O, we

have several choices

of position vector

(Recall that the head

of the vector can lie

anywhere on the line

of action of P).

1

Select the position vector with

the simpler form.

rOD = {2i + 2j +2k} m

rOE = {2j} m (1)

2

Simpler (only one component)

rOD

rOE


x

z

y

A B

CD

E F

G

O

P = 60 NQ = 80 N

Each edge is

2 m long.

Determine the components of

the force P by introducing a

position vector rDE:

rDE = { 2i 2k} m

3

P = (60 N)

= (60 N)

= { 42.43i 42.43k} N (2)

4

rDE

2i 2k

( 2)2 + ( 2)2

rDE

r DE

The force P has magnitude 60 N and

points in the direction from D to E, so


5 Calculate the moment.

MOP = rOE P

= {2j} { 42.43i 42.43k}

= (2)( 42.43)(j i) (2)(42.43)(j k)

= { 84.86i + 84.86k} N·m Ans.

j

k

i= k = i

Since rOE has only one component, it is easier to

calculate cross products of base vectors individually

rather than use the determinant approach to calculating

the cross product.



M OQ

= r OB × Q

8

i j k

= 2 0 2

0 56.57 56.57

0 2 2 2 2 0 = i j + k

56.57 56.57 0 56.57 0 56.57

= {113.14i 113.14j 113.14k} N·m (3) Ans.

To determine the moment of the force Q

about point O, we could introduce either

position vector rOB or rOF. Let's

arbitrarily choose rOB.

rOB = {2i + 2k} m

6

Determine the components of

force Q by noting it has

magnitude 80 N and points from F to B so

Q = (80 N)

= (80 N)

= { 56.57j + 56.57k} N

7

2j + 2k

( 2)2 + (2)2

rFB

r FB

x

z

y

A B

CD

E F

GO

P = 60 N Q = 80 N

rFB

rOB

rOF

Each edge is

2 m long.


x

z

y

A B

CD

E F

G

O

P = 60 N Q = 80 N

Each edge is

2 m long.

Finally, to determine the

shortest distance from point O

to line FB, consider the plane

defined by O and FB:

9

From the figure it is clear that of the various distances

d 1, d

2, d

3, ..., from O to line FB, the perpendicular

distance d is the shortest. But we also know that, by

the scalar definition of moment, the moment of Q

about point O equals the perpendicular distance from O

to the line of action of Q times Q. That is,

M QO

= Qd

Thus,

d =

= = 2.45 m Ans.

10

Q = 80 N

B

F

O

d 1

d 2

d 3

d

(113.14)2 + ( 113.14)2 + ( 113.14)2

80

M QO

Q

d


9. Determine the moment about the screw at A of the force F = 2 N

applied to the sheet-metal bracket shown. Also, determine the shortest

distance from A to the line connecting B and C.

x

y

z

60 mm

100 mm

70 mm

80 mm

30 mm

60 mm50 mm

A

B

C

F = 2 N

25°


To calculate the moment

we could use either the

position vector rAB or rAC

.

1

rAC has a single component,

rAC = (50 mm + 60 mm)i

= { 110i} mm (1)

while rAB has two components, so let's

choose the simpler, rAC.

2

x

y

z

AC

F = 2 N

rAC

rAB

B 25°

50 mm 60 mm


25°

B 25°

x

30 mm

B

A

Next, we need to calculate the components

of the force, F. To do this, we first need to

calculate the coordinates of point B. By

inspection, the x coordinate of B is 60 mm.

3

To find the y and z

coordinates of point B,

consider a view from

the positive x-axis.

4

y coordinate of B

= 30 mm 70 mm 42.26 mm

= 142.26 mm

5

z coordinate of B

= 90.63 mm + 80 mm

= 170.63 mm

6

y

z

60 mm

100 mm

70 mm

80 mm

60 mm50 mm

AC

F = 2 N

100 mm

80 mm

30 mm

70 mm

(100 mm) sin 25° = 42.26 mm

(100 mm) cos 25°

= 90.63 mm

z

y


To determine the components of the force, introduce the

position vector rBC.

rBC = ( 50 mm 60 mm)i

+ [0 ( 142.26 mm)]j

+ (0 170.63 mm)k

= { 110i + 142.26j 170.63k} mm (2)

7

F = (2 N)

= (2 N)

= { 0.887i + 1.148j 1.377k} N

8

110i + 142.26j 170.63k

( 110)2 + (142.26)2 + ( 170.63)2

rBC

r BC

y

z

60 mm

100 mm

70 mm

80 mm

60 mm50 mm

AC

F = 2 N

C coordinates

( 50 mm, 0, 0)

B coordinates (60, 142.26 mm, 170.63 mm)

rBC

B 25°The force F has magnitude 2 N and

points from B to C so

x

30 mm


9

MA = rAC F

= { 110i} { 0.887i + 1.148j 1.377k}

= ( 110)( 0.887)(i i) + ( 110)(1.148)(i j) + ( 110)( 1.377)(i k)

= { 151.4j 126.3k} N·mm (3) Ans.

= j= k= 0

k

j

i



To determine the shortest distance

between point A and line BC, consider the

plane formed by A and BC.

10

Magnitude of moment about A = Fd

or,

( 151.4)2 + ( 126.3)2 N mm = (2 N)d

Solving gives,

d = 98.6 mm Ans.

F = 2 N

d

A

B

C

11

by Eq. 3

y

z

60 mm

100 mm

70 mm

80 mm

60 mm50 mm

AC

F = 2 N

B 25°

x

30 mm


10. If the tension in the cable BC is T = 80 lb,

determine the moment about point A of the cable force

acting on the frame at point B. Also, determine the

shortest distance from A to the line through B and C.

32 in.

18 in. 10 in.

AB

x

y

z

T = 80 lb

C

12 in.


To calculate the

moment, we could

use either position

vector rAB or rAC.

1

Since rAC has only one

component while r AB has three,

let's use rAC.

rAC = {12j} in. (1)

2

32 in.

18 in.10 in.

AB

x

y

z

T = 80 lb

C

rAC

rAB

12 in.


To determine the components of the force, T,

introduce the position vector rBC.

3

T = (80 lb)

= (80 lb)

= { 69.62i + 4.35j 39.16k} lb

4

32i + 2j 18k

( 32)2 + 22 + ( 18)2

rBC

r BC

rBC = (0 32 in.)i + (12 in. 10 in.)j + (0 18 in.)k

= { 32i + 2j 18k} in.

32 in.

18 in. 10 in.

AB

x

y

z

T = 80 lb

C

rBC

The force T has magnitude 80 lb and

points from B to C so

12 in.


5 Calculate the moment.

MA = rAC T

= 12j { 69.62i + 4.35j 39.16k}

= (12)( 69.62)(j i) + (12)(4.35)(j j) + (12)( 39.16)(j k)

= { 469.92i + 835.44k} lb·in. Ans. (2)

= k = i= 0

k

j

i

by Eq. 1


Finally, to determine the shortest distance

between point A and line BC, consider the

plane formed by A and BC.

6

Magnitude of moment about A = Td

or,

( 469.92)2 + (835.44)2 = (80 lb)d

Solving gives,

d = 11.98 in. Ans.

7

32 in.

18 in. 10 in.

AB

x

y

z

T = 80 lb

C

T = 80 lb

d

Extension of BC

A

B

C

by Eq. 2

12 in.

Documents

4.2 Moments in Three-Dimensional Force · PDF file4.2 Moments in Three-Dimensional Force Systems Example 1, page 1 of 3 200 mm 200 mm A F = 50 N 1 1. Use the cross-product definition