The roles of science in national development

Topics: Force, Torque, Momentum Pages: 17 (951 words) Published: November 1, 2014
MOMENT OF A COUPLE
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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