# The roles of science in national development

Today’s Objectives:

Students will be able to

a) define a couple, and,

b) determine the moment of a couple.

APPLICATIONS

A torque or moment of 12 N · m is required to rotate the wheel. Which one of the two grips of the wheel above will require less force to rotate the wheel?

APPLICATIONS

(continued)

The crossbar lug wrench is being used to loosen a lug net. What is the effect of changing dimensions a, b, or c on the force that must be applied?

MOMENT OF A COUPLE

A couple is defined as two

parallel forces with the same

magnitude but opposite in

direction separated by a

perpendicular distance d.

The moment of a couple is defined as

MO = F d (using a scalar analysis) or as

MO = r × F (using a vector analysis).

Here r is any position vector from the line of action of –F to the line of action of F.

Couple moment only depends on F and d

F

d

A

rOA

rOB

-F

O

B

MB = + dF

What about point O, is MO = + dF too?

First note

=

=

=

MA = + dF

MOMENT OF A COUPLE

(continued)

The net external effect of a couple is that

the net force equals zero and the magnitude

of the net moment equals F d

Since the moment of a couple depends

only on the distance between the forces,

the moment of a couple is a free vector. It

can be moved anywhere on the body and

have the same external effect on the body.

Moments due to couples can be added using

the same rules as adding any vectors.

EXAMPLE - SCALAR APPROACH

Given: Two couples act on the

beam and d equals 8 ft.

Find: The resultant couple

Plan:

1) Resolve the forces in x and y directions so they can

be treated as couples.

2) Determine the net moment due to the two couples.

EXAMPLE - SCALAR APPROACH

The x and y components of the

top 60 lb force are:

(4/5)(60 lb) = 48 lb vertically up

(3/5)(60 lb) = 36 lb to the left

Similarly for the top 40 lb force:

(40 lb) (sin 30 ) up

The net moment equals to

(40 lb) (cos 30 ) to the left

+ ΣM = -(48 lb)(4 ft) + (40 lb)(cos 30º)(8ft)

= -192.0 + 277.1 = 85.1 ft·lb

VECTOR APPROACH

C

rCD

d

A

-F

F

d is not always easy to see in 3D.

B

Can use any point on one line

of action to the other as long

as you have the coordinates of

the points.

D

EXAMPLE – VECTOR APPROACH

Given: A force couple acting on

the rod.

Find: The couple moment

acting on the rod in

Cartesian vector

notation.

A

B

Plan:

1) Use M = r × F to find the couple moment.

2) Set r = rAB and F = {14 i – 8 j – 6 k} N .

3) Calculate the cross product to find M.

EXAMPLE – VECTOR APPROACH

rAB = {0.8 i + 1.5 j – 1 k} m

F = {14 i – 8 j – 6 k} N

A

B

M = rAB × F

=

i

0.8

14

j

k

1.5 -1

-8 -6

N·m

= {i (-9 – (8)) – j (- 4.8 – (-14)) + k (-6.4 – 21)} N·m = {-17 i – 9.2 j – 27.4 k} N·m

CONCEPT QUIZ

1. F1 and F2 form a couple. The moment

of the couple is given by ____ .

A) r1 × F1

B) r2 × F1

C) F2 × r1

D) r2 × F2

F1

r1

r2

F2

2. If three couples act on a body, the overall result is that

A) the net force is not equal to 0.

B) the net force and net moment are equal to 0.

C) the net moment equals 0 but the net force is not

necessarily equal to 0.

D) the net force equals 0 but the net moment is not

necessarily equal to 0 .

GROUP PROBLEM SOLVING – SCALAR APPROACH

Given: Two couples act on the

beam. The resultant

couple is zero.

Find: The magnitudes of the

forces P and F and the

distance d.

PLAN:

1) Use definition of a couple to find P and F.

2) Resolve the 300 N force in x and y directions.

3) Determine the net moment.

4) Equate the net moment to zero to find d.

GROUP PROBLEM SOLVING – SCALAR APPROACH

From the definition of a

couple:

P = 500 N and

F = 300 N.

Resolve the 300 N force into vertical and horizontal

components. The vertical component is (300 cos 30º) N and

the horizontal component is (300 sin 30º) N.

It was given that the net moment equals zero. So

+ ΣM = -...

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