CHAPTER 4
MOMENT
Today’s Objectives
a) Understand and define moment
b) Determine moments of a force in 2-D and 3-D
Moment of force
APPLICATIONS
What is the effect of 30 N
force on the lug nut?
MOMENT OF A FORCE (2-D)
The moment of a force about a point provides a measure
of the tendency for rotation.
MOMENT OF A FORCE (2-D)
In 2-D, the magnitude of the moment is
Mo = F d
As shown, d is the perpendicular distance from point O to the
line of action of the force.
In 2-D, the direction of MO is either clockwise or counter-
clockwise depending on the tendency for rotation.
MOMENT OF A FORCE (2-D)
Often it is easier to determine MO by using the
components of F as shown.
Fy F
Fx
b a
O
Using this approach, MO = (FY a) – (FX b)
Note the different signs on the terms!
The typical sign convention for a moment in 2-D is that counter-
clockwise is considered positive.
MOMENT OF A FORCE (3-D) - Vector Formulation
Moments in 3-D can be calculated using scalar (2-D) approach but
it can be difficult and time consuming. Thus, it is often easier to
use a mathematical approach called the Vector Cross Product.
Using the Vector Cross Product, MO = r F
r is the position vector from point O to any point on the line of
action of F.
CROSS PRODUCT
The cross product can be written as:
Each component can be determined using 2 2 determinants
MOMENT OF A FORCE (3-D)
So, using the cross product,
moment can be expressed as
By expanding the above equation using 2 2 determinants,
we get (sample units are N - m or lb - ft)
MO = (r y FZ - rZ Fy) i - (r x Fz - rz Fx ) j + (rx Fy - ry Fx ) k
Example #1
Given: 400 N force is applied
as shown and = 20°
Find: The moment of the
force at joint A.
Plan
1) Resolve the force along x and y axes.
2) Determine MA using scalar analysis.
Solution
Fx = 400 cos 20° N (←)
Fy = 400 sin 20° N (↓)
+ MA = {(400 cos 20°)(2) + (400 sin 20°)(3)} N·m
= 1160 N·m
Example #2
Given: a = 3 in , b = 6 in and c = 2 in
Find: Moment of F about point P
Plan
1) Find rPA
2) Determine MP = rPA x F
Solution r PA = { 3 i + 6 j - 2 k } in
i j k
MP = 3 6 -2 = { -2 i - 3 j - 12 k } lb · in
3 2 -1
MOMENT OF A COUPLE
Today’s Objectives
a) Define a couple
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
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 (Section 4.6)
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)
r is any position vector from the line of action of F to the
line of action of F.
MOMENT OF A COUPLE
The net external effect of a couple is that
the net force equals zero and the magnitude
of the net moment = 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.
Moments due to couples can be added using
the same rules as adding any vectors.
Example #1 - SCALAR APPROACH
Given: Two couples act on the
beam & d = 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.
Solution
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°) - vertically up
(40 lb) (cos 30°) - to the left
The net moment equals to
+ M = -(48 lb)(4 ft) + (40 lb)(cos 30º)(8ft)
= -192.0 + 277.1 = 85.1 ft·lb
Example #2 – 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 300 N force in x and y directions.
3) Determine the net moment.
4) Equate the net moment to zero to find d
Solution
From the definition of a couple
P = 500 N
F = 300 N
Resolve 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
+ M = - (500)(2) + (300 cos 30º)(d) - (300 sin 30º)(0.2) = 0
Now, solve this equation for d
d = (1000 + 60 sin 30º) / (300 cos 30º) = 3.96 m
Example #3 – VECTOR APPROACH
Given: A force couple acting on the rod.
A
Find: The couple moment acting on the
rod in Cartesian Vector.
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
Solution
rAB = {0.8 i + 1.5 j – 1 k} m
F = {14 i – 8 j – 6 k} N
A
M = rAB F
= i j k B
0.8Trial Edition
Created with the N·m 5.
1.5 -1of SmartDraw
14 -8 -6
= {i (-9 – (8)) – j (- 4.8 – (-14)) + k (-6.4 – 14(1.5))} N·m
= {-17 i + 9.2 j – 27.4 k} N·m
EQUIVALENT FORCE-COUPLE SYSTEMS
Today’s Objective
Find an equivalent system for a system of forces and couples
RESULTANT OF A FORCE AND COUPLE SYSTEM
If the force system lies in 2-D, the reduced equivalent system can
be obtained using the following three scalar equations.
FINDING THE RESULTANT OF A
FORCE AND COUPLE SYSTEM
When several forces and couple moments
act on a body, you can move each force
and its associated couple moment to a
common point O.
Now you can add all the forces and
couple moments together and find one
resultant force-couple moment pair.
Example #1
Given: 2-D force and couple system
as shown.
Find: The equivalent resultant force
and couple moment acting at
A, and then the equivalent
single force location along
the beam AB.
Plan
1) Sum all the x and y components of the forces to find FRA.
2) Find and sum all the moments resulting from moving each
force to A.
3) Shift the FRA to a distance d such that d = MRA/FRy
Solution
+ FRx = 25 + 35 sin 30° = 42.5 lb
+ ↑ FRy = -20 - 35 cos 30° = -50.31 lb
FR
+ MRA = -35 cos30° (2) - 20(6) + 25(3)
= -105.6 lb·ft
FR = ( 42.52 + (-50.31)2 )1/2 = 65.9 lb
= tan-1 ( 50.31/42.5) = 49.8 °
The equivalent single force FR can be located on the beam AB
at a distance d measured from A.
d = MRA/FRy = -105.6/-50.31 = 2.10 ft
Example #2
Given: 2-D force and couple
system as shown.
Find: The equivalent resultant
force and couple moment
acting at A.
Plan
1) Sum all X and Y components of the forces to find FRA
2) Find and sum all the moments resulting from moving each
force to A and add them to 500 lb-ft free moment to find MRA
Solution
Summing the force components
+ Fx = (4/5) 150 lb + 50 lb sin 30° = 145 lb
+ Fy = (3/5) 150 lb + 50 lb cos 30° = 133.3 lb
Now find the magnitude and direction of the resultant.
FRA = ( 145 2 + 133.3 2 )1/2 = 197 lb and = tan-1 ( 133.3/145)
= 42.6 °
+ MRA = { (4/5)(150)(2) – 50 cos30° (3) + 50 sin30° (6) + 500 }
= 760 lb·ft
DISTRIBUTED LOADING
In many situations a surface area
of a body is subjected to a
distributed load. Such forces are
caused by winds, fluids, or the
weight of items on the body’s
surface.
We will analyze the most common
case of a distributed pressure
loading. This is a uniform load
along one axis of a flat rectangular
body.
In such cases, w is a function of x
and has units of force per length.
EXAMPLES
We will consider only rectangular and triangular loading
diagrams whose centroids are well defined.
Rectangular loading, FR = 400 10 = 4,000 lb and x = 5 m
Triangular loading ,
FR = (0.5) (600) (6) = 1,800 N and x = 6 – (1/3) 6 = 4 m
Note: The centroid in a triangle is at a distance one third the
width of the triangle as measured from its base.
Example #1
Given: The loading on the beam as
shown.
Find: The equivalent force and its
location from point A
Plan
1) Consider the trapezoidal loading as two separate loads (one
rectangular and one triangular).
2) Find FR and x for each of these two distributed loads.
3) Determine the overall FR and x for the three point loadings.
Solution
For rectangular loading of height
0.5 kN/m and width 3 m,
FR1 = 0.5 kN/m 3 m = 1.5 kN
x1 = 1.5 m from A
For triangular loading of height 2 kN/m and width 3 m,
FR2 = (0.5) (2 kN/m) (3 m) = 3 kN
and its line of action is at x2 = 1 m from A
For combined loading of the three forces,
FR = 1.5 kN + 3 kN + 1.5 kN = 6 kN
+ MRA = (1.5) (1.5) + 3 (1) + (1.5) 4 = 11.25 kN • m
Now, FR x = 11.25 kN • m
Hence, x = (11.25) / (6) = 1.88 m from A
Tutorial
Moment
Problem #1
Determine:
1. The resultant force (FR)
2. The intersection of (FR) with member BC
Solution
Problem #2
• The cable CED supports the shown platform through a
frictionless pulley at E
Determine:
• The moment about O of the force exerted by cable at C
TCED = 1349 N
Solution
Problem #3
• Two cables BD & FE hold the shown concrete wall.
Determine:
1. The moment about O of the force exerted by cable FE
2. The perpendicular distance from C to cable BD
TFE = 675 N
TBD = 900 N
Solution
1. The moment about O of the force exerted by cable FE
Solution
2. The perpendicular distance from C to cable BD
