# Force resultant Lecture by alicejenny

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```									                                                     University of Palestine
College of Engineering & Urban Planning

ENGINEERING MECHANICS
STATICS & DYNAMICS

Instructor: Eng. Eman Al.Swaity
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Lecture 5                                        Chapter 4: Force resultant
Chapter(4)
Force System Resultants
Chapter Objectives
•   To discuss the concept of the moment of a force and show
how to calculate it in two and three dimensions.
•   To provide a method for finding the moment of a force
about a specified axis.
•   To define the moment of a couple.
•   To present methods for determining the resultants of
nonconcurrent force systems.
•   To indicate how to reduce a simple distributed loading to
a resultant force having a specified location.

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Lecture 5                                             Chapter 4: Force resultant
Today’s Objectives :
Students will be able to:
a) understand and define moment, and,
b) determine moments of a force in 2-D
and 3-D cases.

The moment (Torque) of a force about a point or axis provides a
measure of the tendency of the force to cause a body to rotate
about the point or axis. For example, consider the horizontal
force Fx, which acts perpendicular to the handle of the wrench
and is located a distance dy from point 0, is seen that this force
tends to cause the pipe to turn about the Z axis.
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Lecture 5                                         Chapter 4: Force resultant
Magnitude: The magnitude of Mo is

where d is referred to as the moment arm or perpendicular
distance from the axis at point 0 to the line of action of the force.
Units of moment magnitude consist of force times distance, e.g.,
N. m or lb. ft.
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Lecture 5                                           Chapter 4: Force resultant
Direction:
The direction of Mo will be specified
by using the "right-hand rule." To do
this, the fingers of the right hand are
curled such that they follow the sense
of rotation, which would occur if the
force could rotate about point 0, The
thumb then points along the moment
axis so that it gives the direction and
sense of the moment vector, which
is upward and perpendicular to the
shaded plane containing F and d.

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Lecture 5                                      Chapter 4: Force resultant
READING
F = 10 N
1. What is the moment of the 10 N force
about point A (MA)?
A) 10 N·m       B) 30 N·m    C) 13 N·m
d=3m
D) (10/3) N·m   E) 7 N·m                   • A

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Lecture 5                                      Chapter 4: Force resultant
APPLICATIONS (continued)

What is the effect of the 30 N
force on the lug nut?

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Lecture 5                        Chapter 4: Force resultant
MOMENT IN 2-D

The moment of a force about a point provides a measure of the
tendency for rotation (sometimes called a torque).

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Lecture 5                                       Chapter 4: Force resultant
MOMENT IN 2-D (continued)

In the 2-D case, 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.
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Lecture 5                                         Chapter 4: Force resultant
MOMENT IN 2-D (continued)
F
a
b                            For example, MO = F d and the
O                           direction is counter-clockwise.
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. We can
determine the direction of rotation by imagining the body pinned
at O and deciding which way the body would rotate because of
the force.
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Lecture 5                                       Chapter 4: Force resultant
EXAMPLE 1
Given: A 400 N force is
applied to the frame
and  = 20°.
Find: The moment of the
force at A.

Plan:
1) Resolve the force along x and y axes.
2) Determine MA using scalar analysis.

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Lecture 5                                       Chapter 4: Force resultant
EXAMPLE 1 (continued)

Solution
+  Fy = -400 cos 20° N
+  Fx = -400 sin 20° N
+ MA = {(400 cos 20°)(2) + (400 sin 20°)(3)} N·m
= 1160 N·m

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Lecture 5                                       Chapter 4: Force resultant
EXAMPLE 2
 Find    the moment of F1 at point o

F1=100kN         F2=200kN

+                          O

4.5m

M = F.d
M = 100  4.5 = 450 kN.m

Find the moment of F2 at point o
M = F.d
M = 200  0 = 0 kN.m
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Lecture 5                                       Chapter 4: Force resultant
      Example 3
Fine the moment of all forces at point A
50N

1m         100N             100N
50N
1m                 A               50N
-
Don’t forget
the signs
3m     3m        2m    2m

Moment at A = F1d1 + F2d2 + F3d3 +…. =  Fi .d i
Five forces means five moments
MA= (-)500 + (-)1002 + 1003 + (-)506 + 501
MA= -150 N.m

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Lecture 5                                           Chapter 4: Force resultant
      Moment theorem (Varignon’s theorem)
The moment of force F at point O is
equal to the sum of moments of the
force components at the same point O
Example 4
Fined the magnitude of the moment
about point O in four deferent ways
Technique I
the moment arm d
d = 4cos40o + 2sin40o
d = 4.35m
the moment = Fd                                     -
M = -6004.35 = -2610 N.m
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Lecture 5                                         Chapter 4: Force resultant
Example 4
Technique II
Replace the force by its components
F1 = 600cos40o = 460N                                       -
F2 = 600sin40o = 386N
Moment theorem
M0= (-)4604 + (-)3862
M0 = -2610N.m
Technique III
the principle of transmissibility
d1 = 4 + 2tan40o = 5.68m
M0= (-)4605.68 + 3860                             -
M0 = -2610N.m
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Lecture 5                                      Chapter 4: Force resultant
Example 4
Technique IV
the principle of transmissibility
d1 = 2 + 4tan50o = 6.77m
M0= 4600 + (-)3866.77 = -2610N.m

-

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Lecture 5                                    Chapter 4: Force resultant
GROUP PROBLEM SOLVING
Given: A 40 N force is
applied to the wrench.
Find: The moment of the
force at O.
Plan: 1) Resolve the force
along x and y axes.
2) Determine MO using
scalar analysis.
Solution: +  Fy = - 40 cos 20° N
+  Fx = - 40 sin 20° N
+ MO = {-(40 cos 20°)(200) + (40 sin 20°)(30)}N·mm
= -7107 N·mm = - 7.11 N·m
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Lecture 5                                     Chapter 4: Force resultant
For each case illustrated in Fig. 4-4, determine the moment of the
force about point O.

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Lecture 5                                         Chapter 4: Force resultant
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Lecture 5   Chapter 4: Force resultant
In general, the cross product of two vectors A and B results in
another vector C , i.e., C = A  B. The magnitude and direction
of the resulting vector can be written as
C = A  B = A B sin  UC
Here UC is the unit vector perpendicular to both A and B
vectors as shown (or to the plane containing the
A and B vectors).                                               21
Lecture 5                                        Chapter 4: Force resultant
CROSS PRODUCT

The right hand rule is a useful tool for determining the direction of
the vector resulting from a cross product.
For example: i  j = k
Note that a vector crossed into itself is zero, e.g., i  i = 0

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Lecture 5                                          Chapter 4: Force resultant
CROSS PRODUCT (continued)
Of even more utility, the cross product can be written as

Each component can be determined using 2  2 determinants.

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Lecture 5                                        Chapter 4: Force resultant
The moment of a force F about point 0, or actually about the
moment axis passing through 0 and perpendicular to the plane
containing 0 and F, Fig. 4-12a, can be expressed using the vector
cross product, namely,

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Lecture 5                                        Chapter 4: Force resultant
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Lecture 5   Chapter 4: Force resultant
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Lecture 5   Chapter 4: Force resultant
      Moment
M=rF
r = xi + yj + zk
F = FXi + FYj + FZk

i       j      k
M  rx      ry      rz
FX      FY      FZ

M = (ryFz – rzFy)i + (rzFx – rxFz)j + (rxFy – ryFx)k
Mx = ryFz – rzFy
My = rzFx – rxFz
Mz = rxFy – ryFx

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Lecture 5                                                      Chapter 4: Force resultant
EXAMPLE 2

Given: a = 3 in, b = 6 in and c = 2 in.
Find: Moment of F about point O.
o              Plan:

1) Find rOA.
2) Determine MO = rOA  F .

Solution rOA = {3 i + 6 j – 0 k} in
MO =     i j k
3 6 0    = [{6(-1) – 0(2)} i – {3(-1) – 0(3)} j +
3 2 -1      {3(2) – 6(3)} k] lb·in
Lecture 5                = {-6 i + 3 j – 12 k} lb·in 4: Force resultant
Chapter
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Lecture 5   Chapter 4: Force resultant
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Lecture 5   Chapter 4: Force resultant
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Lecture 5   Chapter 4: Force resultant
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Lecture 5   Chapter 4: Force resultant
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Lecture 5   Chapter 4: Force resultant
Recall that when the moment of a force is computed about a point, the
moment and its axis are always perpendicular to the plane containing
the force and the moment arm.
In some problems it is important to find the component of this moment
along a specified axis that passes through the point. To solve this problem
either a scalar or vector analysis can be used.

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Lecture 5                                                  Chapter 4: Force resultant
APPLICATIONS

With the force F, a person is
creating the moment MA.
What portion of MA is used in
turning the socket?

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Lecture 5                                Chapter 4: Force resultant
SCALAR ANALYSIS
Recall that the moment of a force about any point A is MA= F dA
where dA is the perpendicular (or shortest) distance from the point
to the force’s line of action. This concept can be extended to find
the moment of a force about an axis.
In the figure above, the
moment about the y-axis
would be My= 20 (0.3) =
6 N·m. However this
calculation is not always
trivial and vector analysis
may be preferable.

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Lecture 5                                          Chapter 4: Force resultant
VECTOR ANALYSIS

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Lecture 5                     Chapter 4: Force resultant
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Lecture 5   Chapter 4: Force resultant
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Lecture 5   Chapter 4: Force resultant
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Lecture 5   Chapter 4: Force resultant
Example
    Determine the moment MZ of T about z-axis passing O
 Technique I
AB      (12  0)i  (0  15) j  (9  0)k
T  T .n F  T     10
AB        (12  0) 2  (0  15) 2  (9  0) 2
T  10  0.566i  0.707 j  0.424k 

T  5.66i  7.07 j  4.24k
MO  r F
M O  15 j  (5.66i  7.07 j  4.24k )

M O  84.9k  63.6i

 M Z  84.9kN .m
Minus sign indicate that the Mz is
in the negative Z direction
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Lecture 5                                                Chapter 4: Force resultant
 Technique II
The force T resolved into two components
Tz & Txy
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 X Y  tan 1             25.1O
122  152
T XY  T cos xy  9.06kN
T Z  T sin  xy  4.24kN
TZ is parallel to the Z-axis. That’s means no
moment comes from TZ about Z-axis.
MZ is due only to TXY
M Z  T XY .d
12
d                  .15  9.37
12  15
2      2

 M Z  9.06  9.37  84.9kN .m
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Lecture 5                                                    Chapter 4: Force resultant
EXAMPLE

Given: A force is applied to
A                           the tool to open a gas valve.
Find: The magnitude of the
B                    moment of this force about
the z axis of the value.

Plan:
1) We need to use Mz = u • (r  F).
2) Note that u = 1 k.
3) The vector r is the position vector from A to B.
4) Force F is already given in Cartesian vector form.
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EXAMPLE (continued)

A                   rAB = {0.25 sin 30° i + 0.25 cos30° j} m

B
= {0.125 i + 0.2165 j} m
F = {-60 i + 20 j + 15 k} N
Mz = (rAB  F)

0       0         K1
= with the 0.2165 SmartDraw 5.
MzCreated0.125 Trial Edition of0
-60      20         15

= 1{0.125(20) – 0.2165(-60)} N·m
= 15.5 N·m

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Lecture 5                                           Chapter 4: Force resultant
University of Palestine
College of Engineering & Urban Planning

ENGINEERING MECHANICS
STATICS & DYNAMICS

45
Lecture 5                             Chapter 4: Force resultant

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