Bài giảng giáo trình Engineering Mechanics Statics 13th edition Hibbeler (Lecture Notes Slides) (Chương 3456)



EQUILIBRIUM OF A PARTICLE, THE FREE-BODY
DIAGRAM & COPLANAR FORCE SYSTEMS
Today’s Objectives:
Students will be able to :
a) Draw a free-body diagram (FBD),
and,
b) Apply equations of equilibrium to
solve a 2-D problem.

Mechanics for Engineers: Statics, 13th SI Edition
R. C. Hibbeler and Kai Beng Yap



In-Class Activities:
‡ Reading Quiz
‡ Applications
‡ What, Why, and How
of a FBD
‡ Equations of
Equilibrium
‡ Analysis of Spring and
Pulleys
‡ Concept Quiz
‡ Group Problem
Solving
‡ Attention Quiz
© Pearson Education South Asia Pte Ltd
2013. All rights reserved.

READING QUIZ
1) When a particle is in equilibrium, the sum of forces acting
on it equals ___ . (Choose the most appropriate answer)

A) A constant
D) A negative number

B) A positive number
E) An integer

C) Zero

2) For a frictionless pulley and cable, tensions in the
cable (T1 and T2) are related as _____ .
A) T1 > T2
B) T1 = T2
C) T1 < T2

T1

D) T1 = T2 sin T

T2

Mechanics for Engineers: Statics, 13th SI Edition
R. C. Hibbeler and Kai Beng Yap



© Pearson Education South Asia Pte Ltd
2013. All rights reserved.

APPLICATIONS
The crane is lifting a load. To
decide if the straps holding the
load to the crane hook will fail,
you need to know the force in
the straps. How could you find
the forces?

Straps

Mechanics for Engineers: Statics, 13th SI Edition
R. C. Hibbeler and Kai Beng Yap



© Pearson Education South Asia Pte Ltd
2013. All rights reserved.


APPLICATIONS
(continued)
For a spool of given
weight, how would you
find the forces in cables
AB and AC? If designing
a spreader bar like this

one, you need to know
the forces to make sure
the rigging doesn’t fail.

Mechanics for Engineers: Statics, 13th SI Edition
R. C. Hibbeler and Kai Beng Yap



© Pearson Education South Asia Pte Ltd
2013. All rights reserved.

APPLICATIONS
(continued)

For a given force exerted on the boat’s towing pendant, what are
the forces in the bridle cables? What size of cable must you use?

Mechanics for Engineers: Statics, 13th SI Edition
R. C. Hibbeler and Kai Beng Yap



© Pearson Education South Asia Pte Ltd
2013. All rights reserved.


COPLANAR FORCE SYSTEMS
(Section 3.3)
This is an example of a 2-D or

coplanar force system.
If the whole assembly is in
equilibrium, then particle A is
also in equilibrium.

To determine the tensions in
the cables for a given weight
of cylinder, you need to
learn how to draw a freebody diagram and apply the
equations of equilibrium.

Mechanics for Engineers: Statics, 13th SI Edition
R. C. Hibbeler and Kai Beng Yap



© Pearson Education South Asia Pte Ltd
2013. All rights reserved.

THE WHAT, WHY, AND HOW OF A
FREE-BODY DIAGRAM (FBD)
Free-body diagrams are one of the most important things for
you to know how to draw and use for statics and other subjects!
What? - It is a drawing that shows all external forces
acting on the particle.

Why? - It is key to being able to write the equations of
equilibrium—which are used to solve for the unknowns
(usually forces or angles).

Mechanics for Engineers: Statics, 13th SI Edition
R. C. Hibbeler and Kai Beng Yap



© Pearson Education South Asia Pte Ltd
2013. All rights reserved.


How?
1. Imagine the particle to be isolated or cut free from its
surroundings.
2. Show all the forces that act on the particle.
Active forces: They want to move the particle.
Reactive forces: They tend to resist the motion.
3. Identify each force and show all known magnitudes and
directions. Show all unknown magnitudes and / or
directions as variables.
y

FBD at A

FB
A

FD A



x

FC = 392.4 N (What is this?)

Note : Cylinder mass = 40 Kg
Mechanics for Engineers: Statics, 13th SI Edition
R. C. Hibbeler and Kai Beng Yap

30˚

© Pearson Education South Asia Pte Ltd
2013. All rights reserved.

EQUATIONS OF 2-D EQUILIBRIUM
FBD at A

FD A

y

A

FB

Since particle A is in equilibrium, the
net force at A is zero.

x

So FB + FC + FD = 0

30˚

or

A

6F=0

FC = 392.4 N

FBDat
atA A
FBD
In general, for a particle in equilibrium,

6 F = 0 or
6 Fx i + 6 Fy j

=

0

= 0i + 0j

(a vector equation)

Or, written in a scalar form,
6 Fx = 0 and 6 Fy = 0
These are two scalar equations of equilibrium (E-of-E).
They can be used to solve for up to two unknowns.

Mechanics for Engineers: Statics, 13th SI Edition
R. C. Hibbeler and Kai Beng Yap



© Pearson Education South Asia Pte Ltd
2013. All rights reserved.


EQUATIONS OF 2-D EQUILIBRIUM (continued)
y

FBD at A

A

FDA

30˚

FB
x

FC = 392.4 N
Note : Cylinder mass = 40 Kg

Write the scalar E-of-E:
+ o 6 Fx = FB cos 30º – FD =

0

+ n 6 Fy = FB sin 30º – 392.4 N = 0

Solving the second equation gives: FB = 785 N ĺ
From the first equation, we get: FD = 680 N ĸ
Mechanics for Engineers: Statics, 13th SI Edition
R. C. Hibbeler and Kai Beng Yap



© Pearson Education South Asia Pte Ltd
2013. All rights reserved.

SIMPLE SPRINGS

Spring Force = spring constant * deformation of spring
or F = k * s

Mechanics for Engineers: Statics, 13th SI Edition
R. C. Hibbeler and Kai Beng Yap



© Pearson Education South Asia Pte Ltd
2013. All rights reserved.


CABLES AND PULLEYS
With a frictionless pulley and cable
T1 = T2.

T1
T2

Mechanics for Engineers: Statics, 13th SI Edition
R. C. Hibbeler and Kai Beng Yap



© Pearson Education South Asia Pte Ltd
2013. All rights reserved.

EXAMPLE
Given: The box weighs 550 N and
geometry is as shown.
Find:

The forces in the ropes AB
and AC.

Plan:
1. Draw a FBD for point A.

2. Apply the E-of-E to solve for the forces in ropes AB
and AC.

Mechanics for Engineers: Statics, 13th SI Edition
R. C. Hibbeler and Kai Beng Yap



© Pearson Education South Asia Pte Ltd
2013. All rights reserved.


EXAMPLE (continued)
y
FB

FC
5

30˚

FBD at point A

3

4

A

x

FD = 550 N

Applying the scalar E-of-E at A, we get;
+ o ¦ F x = FB cos 30° – FC (4/5) = 0
+ o ¦ F y = FB sin 30° + FC (3/5) - 550 N = 0
Solving the above equations, we get;

FB = 478 N
and FC = 518 N

Mechanics for Engineers: Statics, 13th SI Edition
R. C. Hibbeler and Kai Beng Yap



© Pearson Education South Asia Pte Ltd
2013. All rights reserved.

CONCEPT QUESTIONS

1000 N

1000 N
(A)

(B)

1000 N
(C)

1) Assuming you know the geometry of the ropes, you cannot
determine the forces in the cables in which system above?
2) Why?
A) The weight is too heavy.
B) The cables are too thin.
C) There are more unknowns than equations.
D) There are too few cables for a 1000 N

weight.
Mechanics for Engineers: Statics, 13th SI Edition
R. C. Hibbeler and Kai Beng Yap



© Pearson Education South Asia Pte Ltd
2013. All rights reserved.


GROUP PROBLEM SOLVING
Given: The mass of lamp is 20
kg and geometry is as
shown.
Find:

The force in each cable.

Plan:

1. Draw a FBD for Point D.
2. Apply E-of-E at Point D to solve for the unknowns (FCD & FDE).
3. Knowing FCD, repeat this process at point C.

Mechanics for Engineers: Statics, 13th SI Edition
R. C. Hibbeler and Kai Beng Yap



© Pearson Education South Asia Pte Ltd

2013. All rights reserved.

GROUP PROBLEM SOLVING (continued)
FBD at point D
y

FDE

FCD

30˚
D

x

W = 20 (9.81) N

Applying the scalar E-of-E at D, we get;
+n ¦ Fy = FDE sin 30° – 20 (9.81) = 0
+o ¦ Fx = FDE cos 30° – FCD = 0
Solving the above equations, we get:
FDE = 392 N

and

FCD = 340 N

Mechanics for Engineers: Statics, 13th SI Edition
R. C. Hibbeler and Kai Beng Yap



© Pearson Education South Asia Pte Ltd
2013. All rights reserved.


GROUP PROBLEM SOLVING (continued)
FBD at point C
y

FAC
5

4

FCD =340 N

3

C

FBC

x

45˚

Applying the scalar E-of-E at C, we get;
+o ¦ Fx = 340 – FBC sin 45° – FAC (3/5) = 0
+ n ¦ Fy = FAC (4/5) – FBC cos 45° = 0

Solving the above equations, we get;
FBC = 275 N

and

FAC = 243 N

Mechanics for Engineers: Statics, 13th SI Edition
R. C. Hibbeler and Kai Beng Yap



© Pearson Education South Asia Pte Ltd
2013. All rights reserved.

ATTENTION QUIZ
1. Select the correct FBD of particle A.
30q

A

40q

100 N
F1

A)

A

B)

100 N

F2

30q

40ƒ
A

C)

F
30ƒ

D)
A

F2

F1

30ƒ

40ƒ
A

100 N

100 N

Mechanics for Engineers: Statics, 13th SI Edition
R. C. Hibbeler and Kai Beng Yap



© Pearson Education South Asia Pte Ltd
2013. All rights reserved.


ATTENTION QUIZ
2. Using this FBD of Point C, the sum of

F2

forces in the x-direction (6 FX) is ___ .
Use a sign convention of + o .

50ƒ
C

A) F2 sin 50ƒ – 20 = 0

F1

B) F2 cos 50ƒ – 20 = 0

C) F2 sin 50ƒ – F1

20 N

=0

D) F2 cos 50ƒ + 20 = 0

Mechanics for Engineers: Statics, 13th SI Edition
R. C. Hibbeler and Kai Beng Yap

© Pearson Education South Asia Pte Ltd
2013. All rights reserved.

Mechanics for Engineers: Statics, 13th SI Edition
R. C. Hibbeler and Kai Beng Yap

© Pearson Education South Asia Pte Ltd
2013. All rights reserved.








THREE-DIMENSIONAL FORCE SYSTEMS
Today’s Objectives:
Students will be able to solve 3-D particle equilibrium problems by
a) Drawing a 3-D free-body diagram, and
b) Applying the three scalar equations (based on one vector

equation) of equilibrium.

In-class Activities:
‡ Check Homework
‡ Reading Quiz
‡ Applications
‡ Equations of Equilibrium
‡ Concept Questions
‡ Group Problem Solving
‡ Attention Quiz

Mechanics for Engineers: Statics, 13th SI Edition
R. C. Hibbeler and Kai Beng Yap



© Pearson Education South Asia Pte Ltd
2013. All rights reserved.


READING QUIZ
1. Particle P is in equilibrium with five (5) forces acting on it in
3-D space. How many scalar equations of equilibrium can be
written for point P?
A) 2

B) 3

D) 5

E) 6

C) 4

2. In 3-D, when a particle is in equilibrium, which of the
following equations apply?
A) (6 Fx) i + (6 Fy) j + (6 Fz) k = 0
B) 6 F = 0
C) 6 Fx = 6 Fy = 6 Fz = 0
D) All of the above.
E) None of the above.

Mechanics for Engineers: Statics, 13th SI Edition
R. C. Hibbeler and Kai Beng Yap



© Pearson Education South Asia Pte Ltd
2013. All rights reserved.

APPLICATIONS
You know the weight of
the electromagnet and its
load. But, you need to
know the forces in the
chains to see if it is a safe
assembly. How would you
do this?

Mechanics for Engineers: Statics, 13th SI Edition

R. C. Hibbeler and Kai Beng Yap



© Pearson Education South Asia Pte Ltd
2013. All rights reserved.


APPLICATIONS
(continued)
This shear leg derrick
Offset distance is to be designed to
lift a maximum of 200
kg of fish.
How would you find
the effect of different
offset distances on
the forces in the cable
and derrick legs?

Mechanics for Engineers: Statics, 13th SI Edition
R. C. Hibbeler and Kai Beng Yap



© Pearson Education South Asia Pte Ltd
2013. All rights reserved.

THE EQUATIONS OF 3-D EQUILIBRIUM
When a particle is in equilibrium, the vector

sum of all the forces acting on it must be
zero (6 F = 0 ) .
This equation can be written in terms of its
x, y, and z components. This form is written
as follows.
(6 Fx) i + (6 Fy) j + (6 Fz) k = 0
This vector equation will be satisfied only when
6 Fx = 0
6 Fy = 0
6 Fz = 0
These equations are the three scalar equations of equilibrium.
They are valid for any point in equilibrium and allow you to
solve for up to three unknowns.
Mechanics for Engineers: Statics, 13th SI Edition
R. C. Hibbeler and Kai Beng Yap



© Pearson Education South Asia Pte Ltd
2013. All rights reserved.


EXAMPLE I
Given: The four forces and
geometry shown.
Find: The force F5 required to
keep particle O in
equilibrium.
Plan:
1) Draw a FBD of particle O.

2) Write the unknown force as
F5 = {Fx i + Fy j + Fz k} N
3) Write F1, F2 , F3 , F4 , and F5 in Cartesian vector form.
4) Apply the three equilibrium equations to solve for the three
unknowns Fx, Fy, and Fz.
Mechanics for Engineers: Statics, 13th SI Edition
R. C. Hibbeler and Kai Beng Yap



© Pearson Education South Asia Pte Ltd
2013. All rights reserved.

EXAMPLE I (continued)
Solution:
F1 = {300(4/5) j + 300 (3/5) k} N
F1 = {240 j + 180 k} N
F2 = {– 600 i} N
F3 = {– 900 k} N

F4 = F4 (rB/ rB)
=

200 N [(3i – 4 j + 6 k)/(32 + 42 + 62)½]

= {76.8 i

F5 = { Fx i

– 102.4 j + 153.6 k} N

– Fy j + Fz k} N

Mechanics for Engineers: Statics, 13th SI Edition
R. C. Hibbeler and Kai Beng Yap



© Pearson Education South Asia Pte Ltd
2013. All rights reserved.


EXAMPLE I (continued)
Equating the respective i, j, k components to zero, we have
6 Fx = 76.8 – 600 + Fx
6 Fy =

= 0 ; solving gives Fx = 523.2 N

240 – 102.4 + Fy = 0 ; solving gives Fy = – 137.6 N

6 Fz = 180 – 900 + 153.6 + Fz = 0 ; solving gives Fz = 566.4 N

Thus, F5 = {523 i – 138 j + 566 k} N
Using this force vector, you can determine the force’s
magnitude and coordinate direction angles as needed.

Mechanics for Engineers: Statics, 13th SI Edition
R. C. Hibbeler and Kai Beng Yap



© Pearson Education South Asia Pte Ltd
2013. All rights reserved.

EXAMPLE II
Given: A 600-N load is
supported by three cords
with the geometry as
shown.
Find:

The tension in cords AB,
AC and AD.

Plan:
1) Draw a free-body diagram of Point A. Let the unknown
force magnitudes be FB, FC, FD .
2) Represent each force in its Cartesian vector form.
3) Apply equilibrium equations to solve for the three
unknowns.
Mechanics for Engineers: Statics, 13th SI Edition
R. C. Hibbeler and Kai Beng Yap



© Pearson Education South Asia Pte Ltd
2013. All rights reserved.

EXAMPLE II (continued)
FBD at A
FD

z

FC

2m

1m

2m

30˚

A

y
FB

x
600 N

FB

= FB (sin 30q i + cos 30q j) N
= {0.5 FB i + 0.866 FB j} N

FC

= – FC i N

FD

= FD (rAD /rAD)
= FD { (1 i – 2 j + 2 k) / (12 + 22 + 22)½ } N
= { 0.333 FD i – 0.667 FD j + 0.667 FD k } N
Mechanics for Engineers: Statics, 13th SI Edition
R. C. Hibbeler and Kai Beng Yap



© Pearson Education South Asia Pte Ltd
2013. All rights reserved.

EXAMPLE II (continued)
FBD at A

Now equate the respective i , j , k
components to zero.
¦ Fx = 0.5 FB – FC + 0.333 FD = 0
¦ Fy = 0.866 FB – 0.667 FD = 0
¦ Fz = 0.667 FD – 600 = 0

z
FD

FC

2m

y
1m

2m

A

30˚

FB

x
600 N

Solving the three simultaneous equations yields
FC = 646 N (since it is positive, it is as assumed, e.g., in tension)
FD = 900 N
FB = 693 N

Mechanics for Engineers: Statics, 13th SI Edition
R. C. Hibbeler and Kai Beng Yap



© Pearson Education South Asia Pte Ltd
2013. All rights reserved.

CONCEPT QUIZ
1. In 3-D, when you know the direction of a force but not its
magnitude, how many unknowns corresponding to that force
remain?
A) One

B) Two

C) Three

D) Four

2. If a particle has 3-D forces acting on it and is in static
equilibrium, the components of the resultant force (6 Fx, 6
Fy, and 6 Fz ) ___ .
A) have to sum to zero, e.g., -5 i + 3 j + 2 k
B) have to equal zero, e.g., 0 i + 0 j + 0 k
C) have to be positive, e.g., 5 i + 5 j + 5 k
D) have to be negative, e.g., -5 i - 5 j - 5 k

Mechanics for Engineers: Statics, 13th SI Edition
R. C. Hibbeler and Kai Beng Yap



© Pearson Education South Asia Pte Ltd
2013. All rights reserved.

GROUP PROBLEM SOLVING
Given: A 17500-N (≈ 1750-kg)

motor and plate, as
shown, are in equilibrium
and supported by three
cables and
d = 1.2 m
Find:

Magnitude of the tension
in each of the cables.

Plan:
1) Draw a free-body diagram of Point A. Let the unknown force
magnitudes be FB, FC, F D.

2) Represent each force in the Cartesian vector form.
3) Apply equilibrium equations to solve for the three unknowns.
Mechanics for Engineers: Statics, 13th SI Edition
R. C. Hibbeler and Kai Beng Yap



© Pearson Education South Asia Pte Ltd
2013. All rights reserved.


GROUP PROBLEM SOLVING (continued)
FBD of Point A

z
W

y
x

FD

FB

FC

W = load or weight of unit = 17500 k N
FB

= FB (rAB/rAB) = FB {(1.2 i – 0.9 j – 3 k) / (3.354)} N

FC

= FC (rAC/rAC) = FC { (0.9 j – 3 k) / (3.132) } N

FD

= FD (rAD/rAD) = FD { (– 1.2 i + 0.3 j – 3 k) / (3.245) } N
Mechanics for Engineers: Statics, 13th SI Edition
R. C. Hibbeler and Kai Beng Yap



© Pearson Education South Asia Pte Ltd
2013. All rights reserved.

GROUP PROBLEM SOLVING (continued)
The particle A is in equilibrium, hence
FB + FC + FD + W = 0
Now equate the respective i, j, k components to zero
(i.e., apply the three scalar equations of equilibrium).
¦ Fx = (1.2/ 3.354)FB – (1.2/ 3.245)FD = 0
¦ Fy = (– 0.9/ 3.354)FB + (0.9/ 3.132)FC + (0.3/ 3.245)FD = 0
¦ Fz = (– 3/ 3.354)FB – (3/ 3.132)FC – (3/ 3.245)FD + 17500 = 0

Solving the three simultaneous equations gives the tension forces
FB = 7337 N
FC = 4568 N
FD = 7098 N
Mechanics for Engineers: Statics, 13th SI Edition
R. C. Hibbeler and Kai Beng Yap



© Pearson Education South Asia Pte Ltd
2013. All rights reserved.


ATTENTION QUIZ
z

1. Four forces act at point A and
point A is in equilibrium. Select the
correct force vector P.
A) {-20 i + 10 j – 10 k} N

P
F1 = 20 N

B) {-10 i – 20 j – 10 k} N
C) {+ 20 i – 10 j – 10 k} N

F3 = 10 N
F2 = 10 N
A

y

x

D) None of the above.

2. In 3-D, when you don’t know the direction or the
magnitude of a force, how many unknowns do you have
corresponding to that force?
A) One

B) Two

C) Three

D) Four

Mechanics for Engineers: Statics, 13th SI Edition
R. C. Hibbeler and Kai Beng Yap

© Pearson Education South Asia Pte Ltd
2013. All rights reserved.

Mechanics for Engineers: Statics, 13th SI Edition
R. C. Hibbeler and Kai Beng Yap

© Pearson Education South Asia Pte Ltd
2013. All rights reserved.








MOMENT OF A FORCE (SCALAR FORMULATION),
CROSS PRODUCT, MOMENT OF A FORCE (VECTOR
FORMULATION), & PRINCIPLE OF MOMENTS
Today’s Objectives :
Students will be able to:

a) understand and define moment, and In-Class Activities :
b) determine moments of a force in 2-D • Check Homework
• Reading Quiz
and 3-D cases.
• Applications
• Moment in 2-D
• Moment in 3-D
• Concept Quiz

• Group Problem
Solving
• Attention Quiz

Mechanics for Engineers: Statics, 13th SI Edition
R. C. Hibbeler and Kai Beng Yap



© Pearson Education South Asia Pte Ltd
2013. All rights reserved.


READING QUIZ
F = 12 N

1. What is the moment of the 12 N force
about point A (MA)?
A) 3 N·m

B) 36 N·m

D) (12/3) N·m

E) 7 N·m

C) 12 N·m
• A

d=3m

2. The moment of force F about point O
is defined as MO = ___________ .
A) r x F

B) F x r

C) r • F

D) r * F

Mechanics for Engineers: Statics, 13th SI Edition
R. C. Hibbeler and Kai Beng Yap



© Pearson Education South Asia Pte Ltd
2013. All rights reserved.

APPLICATIONS

Beams are often used to bridge gaps in walls.
We have to know what the effect of the force
on the beam will have on the supports of the
beam.
What do you think is happening at points A and B?
Mechanics for Engineers: Statics, 13th SI Edition
R. C. Hibbeler and Kai Beng Yap



© Pearson Education South Asia Pte Ltd
2013. All rights reserved.


APPLICATIONS (continued)

Carpenters often use a hammer in this way to pull a
stubborn nail. Through what sort of action does the force
FH at the handle pull the nail? How can you mathematically
model the effect of force FH at point O?
Mechanics for Engineers: Statics, 13th SI Edition
R. C. Hibbeler and Kai Beng Yap



© Pearson Education South Asia Pte Ltd
2013. All rights reserved.

MOMENT OF A FORCE – SCALAR FORMULATION
(Section 4.1)

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

Mechanics for Engineers: Statics, 13th SI Edition
R. C. Hibbeler and Kai Beng Yap



© Pearson Education South Asia Pte Ltd
2013. All rights reserved.


MOMENT OF A FORCE - SCALAR FORMULATION
(continued)
In a 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 (CW) or
counter-clockwise (CCW), depending on the tendency for
rotation.
Mechanics for Engineers: Statics, 13th SI Edition
R. C. Hibbeler and Kai Beng Yap

© Pearson Education South Asia Pte Ltd
2013. All rights reserved.



MOMENT OF A FORCE - SCALAR FORMULATION
(continued)
a
b
O

F

For example, MO = F d and

the direction is counterclockwise.

d
Often it is easier to determine
MO by using the components
of F as shown.

F

a

b

y

F
F

x

O

Then 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.
Mechanics for Engineers: Statics, 13th SI Edition
R. C. Hibbeler and Kai Beng Yap



© Pearson Education South Asia Pte Ltd
2013. All rights reserved.


VECTOR CROSS PRODUCT (Section 4.2)
While finding the moment of a force in 2-D is straightforward
when you know the perpendicular distance d, finding the
perpendicular distances can be hard—especially when you
are working with forces in three dimensions.
So a more general approach to finding the moment of a
force exists. This more general approach is usually used
when dealing with three dimensional forces but can be used
in the two dimensional case as well.

This more general method of finding the moment of a force
uses a vector operation called the cross product of two
vectors.

Mechanics for Engineers: Statics, 13th SI Edition
R. C. Hibbeler and Kai Beng Yap



© Pearson Education South Asia Pte Ltd
2013. All rights reserved.

CROSS PRODUCT (Section 4.2)

In general, the cross product of two vectors A and B results
in another vector, C , i.e., C = A u B. The magnitude and
direction of the resulting vector can be written as
C = A u B = A B sin T uC
As shown, uC is the unit vector perpendicular to both A and
B vectors (or to the plane containing the A and B vectors).

Mechanics for Engineers: Statics, 13th SI Edition
R. C. Hibbeler and Kai Beng Yap



© Pearson Education South Asia Pte Ltd
2013. All rights reserved.