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4.3 Moment of a Couple

4.3 Moment of a Couple Example 1, page 1 of 3 1. Two swimmers on opposite sides of a boat attempt to turn the boat by pushing as shown. Determine the couple moment about a) point A on the bow, b) point B on the stern, and c) point C.

A 1m 50 N

Also state what general principles your results demonstrate.

2m C

50 N

1.5 m

B

4.3 Moment of a Couple Example 1, page 2 of 3 1

Calculate the moment about A. 2 MA = (50 N)(1 m)

Calculate the moment about B.

(50 N)(1 m + 2 m) MB =

= 100 N·m

(50 N)(2 m + 1.5 m)

(50 N)(1.5 m)

ns. = 100 N·m

ns.

A 1m 50 N

50 N

2m C

2m C

50 N

50 N

1.5 m

B

4.3 Moment of a Couple Example 1, page 3 of 3 3

Moment about C MC = (50 N)(2 m)

50 N = 100 N·m

ns.

2m C

50 N

4

General principles demonstrated: 1) The couple moment is the same about every point, and ns. 2) The calculation of the couple moment is simplified if the moment is calculated about a point on the line of action of one of the forces making up the couple.

4.3 Moment of a Couple Example 2, page 1 of 3 2. Determine the magnitude and sense of the resultant couple moment acting on the rectangular plate.

30 N 20 N 8m

30° A

1 Resolve the inclined forces into rectangular components.

D 4m

B

30 N (20 N) sin 30° = 10 N

C 30° 20 N

30 N

20 N

30° A

D

(20 N) cos 30° = 17.32 N

(20 N) cos 30° = 17.32 N B

C 30° (20 N) sin30° = 10 N

30 N

20 N

4.3 Moment of a Couple Example 2, page 2 of 3 2 Calculate the moments 3 Moment of the couple formed by the forces at B and D. 10 N

30 N

MD = (30 N)(8 m)

8m 17.32 N

= 240 N·m

A

D

B

C

This moment value would be the same for any other point besides D.

4m

30 N

17.32 N

10 N

4 Moment of the couple formed by the forces at A and C. MC = (17.32 N)(4 m) = 10.72 N·m

(1)

(10 N)(8 m) (2)

This moment value would be the same for any other point besides C.

4.3 Moment of a Couple Example 2, page 3 of 3 5 Since both couple moments are the same about all points, we can move them to any arbitrary point we choose and then add them to get the resultant couple moment. M = MC + MD = 10.72 + 240 = 229 N·m

ns.

No subscript because valid for all points

A

D 229 N·m

B

6

Arbitrary location

C

4.3 Moment of a Couple Example 3, page 1 of 2 3. Determine the value of the force P such that the resultant couple moment of the two couples acting on the beam is 900 lb·ft clockwise.

5 ft

4 ft

7 ft

P

P

B

D

200 lb E

A C 5 4 3 200 lb

0.5 ft

5 4 3

4.3 Moment of a Couple Example 3, page 2 of 2 5 ft

7 ft

P

P

B

D

160 lb 0.5 ft 120 lb

E

A 120 lb

C 1 160 lb

3

MD =

P(5 ft)

(1)

Since the couple moments are the same about all points, we can move them both to any arbitrary point we choose and then add them to get the resultant couple moment: M = P(5) + 1860

4

Compute the moment of the couple at B and D.

(3)

2

Compute the moment of the couple at C and E. ME = (120 lb)(0.5 ft) + (160 lb)(5 ft + 7 ft)

Since the resultant couple moment is specified to be 900 lb·ft clockwise, Eq. 3 becomes 900 lb·ft = P(5) + 1860 Solving gives P = 552 lb

ns.

= 1860 lb·ft

(2)

4.3 Moment of a Couple Example 4, page 1 of 2 4. The wrench applies a 10 N·m couple moment to the bolt. To prevent the plate from rotating, two 2-N forces are applied as shown. Determine the distance s such that the resultant couple moment acting on the plate and bolt is zero.

10 N·m  

C A

B

2N

2N s

4.3 Moment of a Couple Example 4, page 2 of 2

1

To calculate the moment of the couple, sum the moments of the two forces at A and B with respect to the point A: MA = (2 N)s

10 N·m  

2

C A

B

Since couple moments are the same about any point, the two couple moments of (2 N)s and 10 N·m can be considered to act at the same point and thus added to give the resultant moment, M (Note that no subscript is needed on M, since M is independent of where the moment is calculated): M = (2 N)s

2N

10 N·m

2N s

3

Because the resultant moment, M, is to be zero, the latter equation becomes 0 = (2 N)s

10

Solving gives s=5m

ns.

4.3 Moment of a Couple Example 5, page 1 of 5 5. Two cords are wrapped around pegs attached to a board as shown. Determine the value of such that P is as small as possible while still producing a resultant couple moment of zero. Also determine the value of P corresponding to this value of . Neglect the size of the pegs.

80 N A

B P

P D

C 80 N

4m

2m

4.3 Moment of a Couple Example 5, page 2 of 5

80 N B

1 2m

Moment of the couple at B and C. MC = (80 N)(2 m) = 160 N·m

(1)

C

80 N

2 A P

P D

To balance the 160 N·m couple moment computed in Eq. 1, the moment of the couple at A and D must be 160 N·m. To achieve this value of couple moment with the smallest possible value of force P, P must be perpendicular to line AD

4.3 Moment of a Couple Example 5, page 3 of 5

B

A 2m

P

P C

3

Now use geometry to find .

4

Equal angles

5

tan-1 4 m 2m

D

4m

= 63.4°

(2)

A 6 2m P

distance AD =

= 4.472 m

P 7 C

D 4m

(2 m)2 + (4 m)2

Moment of couple A and D MD = P

distance AD

= P(4.472 m)

(3)

ns.

4.3 Moment of a Couple Example 5, page 4 of 5 8

The resultant couple moment must be zero: M = MC + MD = 0

(4)

or, 160 + P(4.472 m) = 0 Solving gives P = 35.8 N 9

ns.

Alternative solution. Do not assume that force P is perpendicular to line AD. Calculate the couple moment of P about point D in terms of the unknown angle . MD = (P cos )(2 m) + (P sin )(4 m)

by Eq. 1 160

80 + 2 sin

2m

P

P sin

P D

(Eq. 4 repeated) by Eq. 5 (P cos )(2) + (P sin )(4)

4m

Solving for P gives P = cos

A

(5)

Now substitute the expressions for MC and MD into Eq. 4: MC + MD = 0

P cos

(6)

4.3 Moment of a Couple Example 5, page 5 of 5 10 To find the minimum value of P, use dP = 0 d From Eq. 6, dP = d d d

cos

80 + 2 sin

=0

so 80 ( sin + 2 cos ) =0 (cos + 2 sin )2 Thus sin tan

+ 2 cos

=0

=2

= tan-1 2 = 63.4° This is the same result as Eq. 2.

4.3 Moment of a Couple Example 6, page 1 of 3 6. A plumber uses two pipe wrenches so that he can loosen pipe BC from pipe AB without also loosening pipe AB from the connection at the wall, A. Determine the moment of the forces about a) A and b) D. Also state what general principles your results demonstrate. y 130 mm

250 mm

B

C x

A 80 N 175 mm

z D

80 N

E

4.3 Moment of a Couple Example 6, page 2 of 3 y

130 mm

250 mm 1

B

C

Part a): Determine the moment about point A. Introduce position vectors from point A to points D and E respectively. x

A

rBD

z

175j } mm

rAE = {130 mm + 250 mm}i

80 N

rAD

rAD = {130i

= {380i

175j } mm

(1) (175 mm}j (2)

175 mm D

E

80 N 2

Use the cross product definition of moment.

MA = rAD = {130i

{80k} N + rAE 175j }

= 130(80)i

{ 80k} N

{80k} + {380i

k 175(80)j = j

= [130(80) + 380( 80)]( j )

175j}

k + 380( 80)i

Equal magnitude, opposite sign, so cancel out

{ 80k}

k 175( 80)j = j

k i

j

= {20 000j} N·mm

k = {20j} N·m

4.3 Moment of a Couple Example 6, page 3 of 3 y 3

250 mm B

C

Part b): To determine the moment about point D, first introduce a position vector with tail at D and head at E.

rDE = {250i} mm

x

A

MD

80 N

z

175 mm D 5

6

rDE

4 E

80 N The two forces (the couple) produce a moment that tends to rotate the entire pipe assembly about the vertical axis (j component). The force applied at E also produces a moment on pipe BC about the x axis, while the force applied at D produces a moment on the other pipe, AB, about the x axis. Since the forces have opposite senses, the moments have opposite senses and thus the total moment applied (to the entire pipe) about the x axis adds to zero (no i component in MD). General principles demonstrated: 1) The couple moment is the same about every point, and 2) the calculation of the couple moment is simplified, if the moment is calculated about a point on the line of action of one of the forces making up the couple.

Calculate the moment.

MD = rDE

{ 80k} N

= {250i}

{ 80k}

= 250( 80)i

k = j

= {20000j } N mm

i j

= {20j } N m

k Same result as in Part a).

4.3 Moment of a Couple Example 7, page 1 of 3 7. Determine the couple moment produced by the two forces applied to the handle of the crank as shown. y 40° F = 20 N 300 mm 25°

A

B

x

25°

40° G = 20 N z

4.3 Moment of a Couple Example 7, page 2 of 3 1 Express the force F in rectangular components. 2

Fy = (20 N) cos 40° 40°

= 15.32 N

y

F = 20 N

3

25°

(20 N) sin 40°

A

= 12.86 N

z

5 4

x

Fx = (12.86 N) cos 25°

Fz = (12.86 N) sin 25° = 11.66 N = 5.43 N 6

In component form,

F = {11.66i

15.32j + 5.43k} N

4.3 Moment of a Couple Example 7, page 3 of 3 7

Introduce a position vector with tail at B and head at A 8

Calculate the couple moment.

rBA = { 300i} mm

M = rBA

F

= { 300i} 40°

{11.66i

15.32j + 5.43k}

y = 300(11.66)i

i 300( 15.32)i

j 300( 5.43)i

k

F = 20 N =0

300 mm 25°

A

rBA B

= j

= {1629j + 4596k} N·mm

x

25°

=k

i

j = {1.63j + 4.60k} N·m

ns.

k 40° G = 20 N z

9

Because the couple has both j and k components, it tends to cause rotation about both the y and z axes.

4.3 Moment of a Couple Example 8, page 1 of 6 8. Determine the resultant couple moment of the two couples acting on the block. y

A

B 6 lb

20 in.

5 in. x

6 lb C

D

10 lb

30° E z

10 lb

F

4.3 Moment of a Couple Example 8, page 2 of 6 y 1 6 lb A

20 in.

5 in.

rAC

C

B

x

Introduce a position vector from a point, A, on the line of action of the { 6i} lb force, to a point C, on the line of action of the {+6i} lb force.

rAC = {20k} in.

6 lb

10 lb

D 2 Calculate the moment of the 6-lb couple. 30°

z

E

10 lb

F

M6 = rAC

{6i} lb

= {20k}

{6i}

= 20(6) k

i j

= {120j } lb·in.

i

j k

4.3 Moment of a Couple Example 8, page 3 of 6 y 3 6 lb

B

A

20 in.

5 in. x

C

rDF = { 5j + 8.660k} in.

6 lb

10 lb

Introduce a position vector from a point, D, on the line of action of the { 10i} lb force, to a point, F, on the line of action of the {+10i} lb force.

D

D, C

rDF

30° E

10 lb

F

F, E

z 4

5 in.

30°

Calculate the moment of the 10-lb couple.

M10 = rDF

5 in. = 8.660 in. tan 30°

{10i}

= { 5j + 8.660k} = 5(10)j

i + 8.660(10)k

= k = {86.60j + 50k} lb·in.

i

j

{10i}

i =j

k

View from positive x axis looking back on yz plane

4.3 Moment of a Couple Example 8, page 4 of 6 5

MR

Compute the resultant couple moment from Eqs. 1 and 2,

y

B

A

= {120j} + {86.60j + 50k} = {206.6j + 50k} lb·in.

M6

M10

MR = M6 + M10

ns.

x

The two couples combine to produce a rotation of the block about the axis defined by the MR vector. Since MR is the same for all points, it can be considered to act at an arbitrary point on top of the block.

C

D

30° E z

F

4.3 Moment of a Couple Example 8, page 5 of 6 y

6 6 lb

20 in.

C

B A 120 lb·in

x

7 30°

F

Using the right hand rule and the scalar equation M = Fd, we can see that the couple moment produced by the 6-lb couple has magnitude 6 lb 20 in. = 120 lb·in. and can be represented as a vector perpendicular to the plane of the forces (plane ABDC) and pointing up.

5 in.

D

6 lb

E

Alternative solution that does not use the cross product definition of a moment:

Similarly, the 10-lb couple produces a couple moment with magnitude 10 lb distance DF = 10 lb 10 in. = 100 lb·in and with direction perpendicular to the plane CDFE.

y

z

5 in. = 10 in. sin 30° 20 in.

100 lb·in

B

A

5 in.

100 lb·in

D, C x 5 in.

C

30°

D

10 lb

F, E View from positive x axis looking back on yz plane 30°

E z

10 lb

F

4.3 Moment of a Couple Example 8, page 6 of 6 8

To add the 100 lb·in. couple moment and the 120 lb·in. couple moment vectorially, we first have to express the 100 lb·in. couple moment, which is perpendicular to FD, in terms of its y and z components.

11

Moment vector perpendicular to top surface (plane ABDC) of block

100 lb·in (100 lb·in)(sin 60°) = 86.6 lb·in 10 90°

30° = 60° D, C

MR = 86.6j + 50.0k + 120j = {206.6j + 50k} lb·in.

30° (100 lb·in.)(cos 60°) = 50.0 lb·in. 30° F, E y z 9

12 In component form,

x

View from positive x axis looking back on yz plane

Same answer as before.

4.3 Moment of a Couple Example 9, page 1 of 5 9. Force F is applied to the doorknob as shown. Determine where on the surface of the door a force G = F should be applied so that the resulting couple moment has a y component of 2 N·m but no x or z components. y

F = {16i + 12k} N

70 mm 650 mm

x 700 mm z

4.3 Moment of a Couple Example 9, page 2 of 5 y

F = {16i + 12k} N 1 70 mm 650 mm

Introduce a force G.

G= F

A

rBA

= {16i + 12k} N

B(xB, yB, 0) = { 16i

G= F 700 mm

x

12k} N

G acts at a point B(xB, yB, 0) on the surface of the door. G and F form a couple.

z 2

Draw a position vector rBA from point B on the surface of the door to point A on the doorknob, and determine its rectangular components.

rBA = (700 mm

xB)i + (650 mm

yB)j + (70 mm

0)k

4.3 Moment of a Couple Example 9, page 3 of 5 3

Calculate the couple moment.

M = rBA

F i

=

j

(700

xB)

(650

16

k yB)

0

(650

y B)

70

= i

4

0

12

= i [(650

yB)(12)

70(0)]

= i [(650

yB)(12)]

70 12

j

(700

xB)

16

j[(700

xB)(12)

yB)(12) = 0

(2)

yB)(16) = 0

(3)

(700

xB)

(650

16 70(16)] + k[(700

j [( 12xB + 7280)] + k [ (650

and Mz = 0, so Mz = (650

+k

12

Use the fact that we know Mx = 0, so Mx = (650

70

yB)(16)]

y B)

0 xB)(0)

(650 (1)

yB)(16)]

4.3 Moment of a Couple Example 9, page 4 of 5 5 Solving Eq. 2 gives yB = 650 mm

ns.

This value of yB also satisfies Eq. 3: (650

yB)(16) = 0

(Eq. 3 repeated)

650 The y component of M, from Eq. 1, is My = ( 12xB + 7280) N·mm

(4)

Since My was specified at the beginning of the problem as My = 2 N · m =

2000 N·mm

Eq. 4 becomes 2000 = ( 12xB + 7280) Solving gives xB = 440 mm

ns.

4.3 Moment of a Couple Example 9, page 5 of 5 y

7

M = { 2j} N·m The couple moment tends to open the door.

F = {16i + 12k} N B

G

440 mm 650 mm 6 x 700 mm z

Thus forc...