E-Portfolio Mohd Fardzlee bin Abd Patah GB190077 MBV

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EXERCISE:

REFERENCE :
1. Hibbeler, R.C & S.C Fan 1997. Engineering Mechanics, Statics. SI Ed. New Jersey:
Prentice Hall. ISBN 0135995981
2. Yusof Ahmad. 1999. Mekanik Statik. Cetakan Ketiga. Malaysia: Penerbit UTM Skudai Johor
Page 34 of 34

DPP C2(b) -2

Kolej Kolej Kemahiran Tinggi MARA
Masjid Tanah, Melaka.

INFORMATION SHEET

PROGRAMME : DIPLOMA IN AUTOMOTIVE ENGINEERING TECHNOLOGY
SESSION :
CODE/COURSE : OCTOBER – DECEMBER 2021 SEMESTER : 3
LECTURER : SHEET NO : 6
DKV 21273 STATICS & WEEK : 6
DYNAMICS
MOHD FARDZLEE BIN ABD
PATAH

TOPIC : FORCE SYSTEM RESULTANTS

SUB-TOPIC : 3.1 Moment of a Force-Scalar Formulation
3.2 Cross Product
3.3 Moment of Force-Vector Formulation

LEARNING After completing the course, students should be able to:
OUTCOME : 1. Express moment of a force using scalar formulation 2D and
vector formulation 3D. C2
2. Apply cross product to solve force resultant. C3
3. Apply the principle of moment. C2
4. Solve the problems moment of a force. C3
5. Solve the problem moment of a couple. C3

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CONTENT:
MOMENT OF A FORCE (SCALAR FORMULATION)

The moment of a force about a point provides a measure of the tendency for rotation
sometimes called a torque, moment of a force, or simply moment).
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
2D, the direction of MO is either clockwise or counter-clockwise, depending on the
tendency for rotation.

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For example, MO = F d and the direction is counter-clockwise.
Often it is easier to determine MO by using the components of F as shown.

Then MO = (FYa) – (FXb). Note the different signs on the terms!
The typical sign convention for a moment in 2D 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.
Example

Given:
A 100 N force is applied to the frame.
Find:
The moment of the force at point O

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Solution
1) Resolve the 100 N force along the x and y axes.
2) Determine MO using a scalar analysis for the two force components and add those two
moments together.

CROSS PRODUCT
Scalar method works well for 2D, where F and d are easy to visualize. Now we come up
with a more rigorous definition using the cross product of 2 vectors.
Remember M0 is a vector!

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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 θ 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).
The right-hand rule is a useful tool for determining the direction of the vector resulting from
a cross product. The direction of C is given by the RHR, rotating A into B
Some identities: i × j = k and j × i = -k
And a vector crossed into itself is zero, e.g., i × i = 0

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MOMENT OF A FORCE (VECTOR FORMULATION)

Moments in 3D can be calculated using scalar (2D) 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.
Here r is the position vector from point O to any point on the line of action of F
So, using the cross product, a moment can be expressed as:

Always write this
By expanding the above equation using 2  2 determinants, we get (sample units are N - m
or lb - ft)

The physical meaning of the above equation becomes evident by considering the force
components separately and using a 2D formulation.

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Example

EXERCISE:
1. Given a 20 lb force is applied to the hammer. Find the moment of the force at A.

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2. Given a 40 N force is applied to the wrench. Find the moment of the force at O.

REFERENCE :
1. Hibbeler, R.C & S.C Fan 1997. Engineering Mechanics, Statics. SI Ed. New Jersey:
Prentice Hall. ISBN 0135995981
2. Yusof Ahmad. 1999. Mekanik Statik. Cetakan Ketiga. Malaysia: Penerbit UTM Skudai Johor

Page 9 of 9

DPP C2 (b)-2

Kolej Kolej Kemahiran Tinggi MARA
Masjid Tanah, Melaka.

INFORMATION SHEET

PROGRAMME : DIPLOMA IN AUTOMOTIVE ENGINEERING TECHNOLOGY
SESSION :
CODE/COURSE : OCTOBER – DECEMBER 2021 SEMESTER : 3
LECTURER : SHEET NO : 7
DKV 21273 STATICS & WEEK : 7
DYNAMICS
MOHD FARDZLEE BIN ABD
PATAH

TOPIC : FORCE SYSTEM RESULTANTS

SUB-TOPIC : 3.4 Principle of Moment
3.5 Moment of A Force About A Specified Axis
3.6 Moment of A Couple

LEARNING After completing the course, students should be able to:
OUTCOME : 1. Express moment of a force using scalar formulation 2D and
vector formulation 3D. C2
2. Apply cross product to solve force resultant. C3
3. Apply the principle of moment. C2
4. Solve the problems moment of a force. C3
5. Solve the problem moment of a couple. C3

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CONTENT:

Principle of Moments

The principle of moments states that when in equilibrium the total sum of the anti clockwise
moment is equal to the total sum of the clockwise moment.
When a system is stable or balance it is said to be in equilibrium as all the forces acting on
the system cancel each other out.

In equilibrium

Total Anticlockwise Moment = Total Clockwise Moment

This principle can be explained by considering two people on a seesaw.

Moments Acting On A Seesaw
Both people exert a downward force on the seesaw due to their weights.
Person A’s weight is trying to turn the seesaw anticlockwise whilst person B’s weight is
trying to turn the seesaw clockwise.

Person A’s Moment = Force x perpendicular distance from fulcrum
1000 x 1 = 1000 Nm

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Person B’s Moment = Force x perpendicular distance from fulcrum
500 x 2 = 1000 Nm

Persons A’s moment = Persons B’s Moment
Anticlockwise moment = Clockwise moment

Therefore seesaw is in equilibrium.
Moment Of A Force About A Specified Axis
The moment of a force about an axis p is defined as the projection OB on p of the
moment

Moment of the force F about the axis p.
Denoting the unit vector of p, the moment Mp of force about an axis p can be
expressed as scalar product (dot product)

Mp = .
or as the mixed triple product of the unit vector , the position vector , and the force :

Mp = . ( × )
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Using the determinant form for the mixed triple product, we have the magnitude of the
moment

(3.24)

Mp =

where , are direction cosines of axis p

,
x, y,

are components of
z
Fx,
Fy, are components of
Fz

Example
A force F = 40 N is applied at a point M(xM;yM;zM) M(3;2;4) [m]. The line oF of action of

the force F is described by direction angles ( ; ; ) (80o;60o; acute angle). An

axis a which passes through the origin O has direction angles ( ; ; ) (60o;100o;
acute angle). Determine:

• The moment of the force F about O.
• The moments Mx, My, and Mz of the force F about axes x, y, and z, respectively.

• The moment of the force F about a axis.

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Moment of the force F about the axis a.
Notice: An acute angle is such an angle for which the condition 0 < < 90o is valid.

Solution
We determine the angle first. Since the expression

cos + cos + cos =1
is valid for any set of direction cosines, we have

cos = + = = 0.848
According to the definition of the moment of the force F, we may write

= ×= = =

= ==

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The magnitude of is

MO = = = 88.06 Nm
The direction cosines of are

cos = = = - 0.138 ... 98o

cos = = = - 0.84 ... 147o

cos = = = 0.524 ... 58o30'

As the components MOx, MOy, and MOz of are equal to the moments Mx, My, and Mz of

F about x, y, and z axes respectively, the following is valid:

Mx = MOx, My = MOy, Mz = MOz

According to the definition of the moment Ma of the force F about an axis a, we have

Ma = T

where the unit vector of a is

= [cos , cos , cos ]T
As

cos = + = 0.8484

we conclude that

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Ma = [0.5, 0.1736, 0.8484] = 45.92 Nm

Moment of a Couple

A couple is a pair of forces, equal in magnitude, oppositely directed, and displaced by
perpendicular distance, d.

Since the forces are equal and oppositely directed, the resultant force is zero. But the
displacement of the force couple (d) does create a couple moments. The moment, M,
about some arbitrary point O can be calculated.

If point O is placed on the line of action of one of the forces, say FB, then that force
causes no rotation (or tendency toward rotation) and the calculation of the moment is
simplified.

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This is a significant result: The couple moment, M, depends only on the position vector r
between forces FA and FB. The couple moment does not have to be determined relative to
the location of a point or an axis.
Example: Moment from a Large Hand Wheel

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EXERCISE:

1. Given a 20 lb force is applied to the hammer. Find the moment of the force at A.

2. Given a 40 N force is applied to the wrench. Find the moment of the force at O.

3. The moment resulting from applying the same forces to the smaller hand wheel can
be determined using any of the three solution procedures outlined above.
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REFERENCE :
1. Hibbeler, R.C & S.C Fan 1997. Engineering Mechanics, Statics. SI Ed. New Jersey:
Prentice Hall. ISBN 0135995981
2. Yusof Ahmad. 1999. Mekanik Statik. Cetakan Ketiga. Malaysia: Penerbit UTM Skudai Johor

Page 11 of 11

DPP C2(b) -2

Kolej Kolej Kemahiran Tinggi MARA
Masjid Tanah, Melaka.

INFORMATION SHEET

PROGRAMME : DIPLOMA IN AUTOMOTIVE ENGINEERING TECHNOLOGY
SESSION :
CODE/COURSE : OCTOBER – DECEMBER 2021 SEMESTER : 3
LECTURER : SHEET NO : 8
DKV 21273 STATICS & WEEK : 8
DYNAMICS
MOHD FARDZLEE BIN ABD
PATAH

TOPIC : FORCE SYSTEM RESULTANTS

SUB-TOPIC : 3.5 Moment of A Force About A Specified Axis
3.6 Moment of A Couple

LEARNING After completing the course, students should be able to:
OUTCOME : 1. Express moment of a force using scalar formulation 2D and
vector formulation 3D. C2
2. Apply cross product to solve force resultant. C3
3. Apply the principle of moment. C2
4. Solve the problems moment of a force. C3
5. Solve the problem moment of a couple. C3

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CONTENT:
Moment Of A Force About A Specified Axis
The moment of a force about an axis p is defined as the projection OB on p of the
moment

Moment of the force F about the axis p.
Denoting the unit vector of p, the moment Mp of force about an axis p can be
expressed as scalar product (dot product)

Mp = .
or as the mixed triple product of the unit vector , the position vector , and the force :

Mp = . ( × )
Using the determinant form for the mixed triple product, we have the magnitude of the

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moment DPP C2(b) -2
(3.24)

Mp =

where , are direction cosines of axis p

,
x, y,

are components of
z
Fx,
Fy, are components of
Fz

Example
A force F = 40 N is applied at a point M(xM;yM;zM) M(3;2;4) [m]. The line oF of action of

the force F is described by direction angles ( ; ; ) (80o;60o; acute angle). An

axis a which passes through the origin O has direction angles ( ; ; ) (60o;100o;
acute angle). Determine:

• The moment of the force F about O.
• The moments Mx, My, and Mz of the force F about axes x, y, and z, respectively.

• The moment of the force F about a axis.

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Moment of the force F about the axis a.
Notice: An acute angle is such an angle for which the condition 0 < < 90o is valid.

Solution
We determine the angle first. Since the expression

cos + cos + cos =1
is valid for any set of direction cosines, we have

cos = + = = 0.848
According to the definition of the moment of the force F, we may write

= ×= = =

= ==

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The magnitude of is

MO = = = 88.06 Nm
The direction cosines of are

cos = = = - 0.138 ... 98o

cos = = = - 0.84 ... 147o

cos = = = 0.524 ... 58o30'

As the components MOx, MOy, and MOz of are equal to the moments Mx, My, and Mz of

F about x, y, and z axes respectively, the following is valid:

Mx = MOx, My = MOy, Mz = MOz

According to the definition of the moment Ma of the force F about an axis a, we have

Ma = T

where the unit vector of a is

= [cos , cos , cos ]T
As

cos = + = 0.8484

we conclude that

Ma = [0.5, 0.1736, 0.8484] = 45.92 Nm
Moment of a Couple

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A couple is a pair of forces, equal in magnitude, oppositely directed, and displaced by
perpendicular distance, d.

Since the forces are equal and oppositely directed, the resultant force is zero. But the
displacement of the force couple (d) does create a couple moments. The moment, M,
about some arbitrary point O can be calculated.

If point O is placed on the line of action of one of the forces, say FB, then that force
causes no rotation (or tendency toward rotation) and the calculation of the moment is
simplified.

This is a significant result: The couple moment, M, depends only on the position vector r
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between forces FA and FB. The couple moment does not have to be determined relative to
the location of a point or an axis.
Example: Moment from a Large Hand Wheel

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EXERCISE:
1. Given a 20 lb force is applied to the hammer. Find the moment of the force at A.

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2. Given a 40 N force is applied to the wrench. Find the moment of the force at O.

3. The moment resulting from applying the same forces to the smaller hand wheel can
be determined using any of the three solution procedures outlined above.

REFERENCE :
1. Hibbeler, R.C & S.C Fan 1997. Engineering Mechanics, Statics. SI Ed. New Jersey:
Prentice Hall. ISBN 0135995981
2. Yusof Ahmad. 1999. Mekanik Statik. Cetakan Ketiga. Malaysia: Penerbit UTM Skudai Johor

Page 9 of 9

DPP C2(b) -2

Kolej Kolej Kemahiran Tinggi MARA
Masjid Tanah, Melaka.

INFORMATION SHEET

PROGRAMME : DIPLOMA IN AUTOMOTIVE ENGINEERING TECHNOLOGY
SESSION :
CODE/COURSE : OCTOBER – DECEMBER 2021 SEMESTER : 3
LECTURER : SHEET NO : 8
DKV 21273 STATICS & WEEK : 9
DYNAMICS
MOHD FARDZLEE BIN ABD
PATAH

TOPIC : EQUILIBRIUM OF A RIGID BODY

SUB-TOPIC : 4.1 Condition For Rigid Body Equilibrium

LEARNING After completing the unit, students should be able to:
OUTCOME : 1.Generate the equations of equilibrium for a rigid body.
2.Explain the concept of the free body diagram for a rigid body.
3.Calculate and solve rigid body equilibrium problem using the equations

of equilibrium.

CONTENT:

Conditions for Rigid-Body Equilibrium

Static equilibrium is defined as a state where an object is not accelerating in any way.
The two conditions for the equilibrium of a rigid body (such as a meter stick) are
1. the vector sum of forces on the body must be zero and
2. The vector sum of torques on the body must be zero.

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For a simple linear body fixed at some pivot, torque depends on two things; the force
vector and lever arm. The torque is
Torqueses that act to rotate the body in the counter-clockwise direction are positive and
those that rotate the body clockwise are negative. The pivot point may be taken to be any
point on the rigid body.
•The equilibrium of a body is expressed as

•Consider summing moments about some other point, such as point A, we require

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Free Body Diagrams
Support Reactions
• If a support prevents the translation of a body in a given direction, then a force is
developed on the body in that direction.
• If rotation is prevented, a couple moments is exerted on the body.

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Internal Forces
• External and internal forces can act on a rigid body
• For FBD, internal forces act between particles which are contained within the boundary
of the FBD, are not represented
• Particles outside this boundary exert external forces on the system

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Weight and Center of Gravity
• Each particle has a specified weight
• System can be represented by a single resultant force, known as weight W of the body
• Location of the force application is known as the center of gravity

Procedure for Drawing a FBD
1. Draw Outlined Shape
• Imagine body to be isolated or cut free from its constraints
• Draw outline shape
2. Show All Forces and Couple Moments
• Identify all external forces and couple moments that act on the body
3. Identify Each Loading and Give Dimensions
• Indicate dimensions for calculation of forces
• Known forces and couple moments should be properly labeled with their magnitudes and
directions

EXERCISE:
1. Draw the free body diagram of the uniform beam. The beam has a mass of 100kg.

REFERENCE :
1. Hibbeler, R.C & S.C Fan 1997. Engineering Mechanics, Statics. SI Ed. New Jersey:
Prentice Hall. ISBN 0135995981
2. Yusof Ahmad. 1999. Mekanik Statik. Cetakan Ketiga. Malaysia: Penerbit UTM Skudai Johor

Page 6 of 6

DPP C2(b) -2

Kolej Kolej Kemahiran Tinggi MARA
Masjid Tanah, Melaka.

INFORMATION SHEET

PROGRAMME : DIPLOMA IN AUTOMOTIVE ENGINEERING TECHNOLOGY
SESSION :
CODE/COURSE : OCTOBER – DECEMBER 2021 SEMESTER : 3
LECTURER : SHEET NO : 9
DKV 21273 STATICS & WEEK : 10
DYNAMICS
MOHD FARDZLEE BIN ABD
PATAH

TOPIC : EQUILIBRIUM OF A RIGID BODY

SUB-TOPIC : 4.2 Equilibrium in Two Dimensions
4.2.1 Free Body Diagrams
4.2.2 Equation of Equilibrium
4.2.3 Two and Three Force Members

LEARNING After completing the unit, students should be able to:
OUTCOME : 1.Generate the equations of equilibrium for a rigid body.
2.Explain the concept of the free body diagram for a rigid body.
3.Calculate and solve rigid body equilibrium problem using the equations

of equilibrium.

CONTENT:

EQUILIBRIUM IN TWO DIMENSIONS

If the resultant of all external forces acting on a rigid body is zero, then the body is said to
be in equilibrium. Therefore, in order for the rigid body to be in equilibrium, both the
resultant force and the resultant couple must be zero.

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Forces and moments acting on a rigid body could be external forces/moments or internal
forces/moments. Forces acting from one body to another by direct physical contact or from
the Earth are examples of external forces. Fluid pressure acting to the wall of a water tank
or a force exerted by the tire of a truck to the road is all external forces. The weight of a
body is also an external force. Internal forces, on the other hand, keep the particles which
constitute the body intact. Since internal forces occur in pairs that are equal in magnitude
opposite in direction, they are not considered in the equilibrium of rigid bodies.
When examining the equilibrium of rigid bodies, it is extremely important to consider all
the forces acting on the body and keep out all the other forces that are not directly exerted
on the body. Therefore, the first step in the analysis of the equilibrium of rigid bodies must
be to draw the “free body diagram” of the body in question. Since the rigid body is
considered by its real shape and dimensions, the correct placement of the forces on the
diagram is of great importance.
In rigid bodies subjected to two dimensional force systems, the forces exerted from
supports and connection elements are shown in the free body diagram as follows:
It should be kept in mind that reaction will occur along the direction in which the motion of
the body is restricted.

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If all the forces acting on the rigid body are planar and all the couples are perpendicular to
the plane of the body, equations of equilibrium become two dimensional.

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or in scalar form,

At most three unknowns can be determined.

In two dimensional problems, in alternative to the above set of equations, two more sets of
equations can be employed in the solution of problems.

Points A, B and C in the latter set cannot lie along the same line, if they do, trivial
equations will be obtained.

Two and Three Force Member

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EXERCISE:

1. Draw the free body diagram of the uniform beam. The beam has a mass of 100kg.

REFERENCE :
1. Hibbeler, R.C & S.C Fan 1997. Engineering Mechanics, Statics. SI Ed. New Jersey:
Prentice Hall. ISBN 0135995981
2. Yusof Ahmad. 1999. Mekanik Statik. Cetakan Ketiga. Malaysia: Penerbit UTM Skudai Johor

Page 9 of 9

DPP C2 (b)-2

Kolej Kolej Kemahiran Tinggi MARA
Masjid Tanah, Melaka.

INFORMATION SHEET

PROGRAMME : DIPLOMA IN AUTOMOTIVE ENGINEERING TECHNOLOGY
SESSION :
CODE/COURSE : OCTOBER – DECEMBER 2021 SEMESTER : 3
LECTURER : SHEET NO : 10
DKV 21273 STATICS & WEEK : 12
DYNAMICS
MOHD FARDZLEE BIN ABD
PATAH

TOPIC : CENTER OF GRAVITY AND CENTROID

SUB-TOPIC : 5.1 Center of Gravity and Center of Mass For a System
of Particle

LEARNING After completing the unit, students should be able to :
OUTCOME : 1. Understand the concept of the center of gravity, center of

mass, and the centroid.
2. Know how to determine the location of the center of gravity

and centroid for a system of discrete particles and a body
of arbitrary shape.
3. Use the theorems of Pappus and Guldinus for finding the
area and volume for a surface of revolution.
4. Present a method for finding the resultant of a general
distributed loading and show how it applies to finding the
resultant of a fluid.

Content
CENTER OF GRAVITY AND CENTER OF MASS FOR A SYSTEM OF PARTICLES

Center of Gravity
The center of gravity G is a point which locates the resultant weight of a system of
particles. If we consider a system of n particles fixed within a region of space as shown in
figure 5.1, it is evident that the weights of the particles comprise a system of parallel forces

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which can be replaced by a single equivalent resultant weight having the defined point G of
application.

Center of gravity G of a system of particles in space.
As established previously, it is possible to obtain an equivalent force of a system of forces
by adding the vectors representing each one of the individual forces. Also, it was
determined that the point of application of that equivalent force is found by doing
summation of moments about a reference point. Similarly, in a system of n particles with
weight W i(i= 1, 2...n). The total weight is found by adding up the individual weight of
the particles and the location at which this weight acts is determined by doing summation
of moments about a point of reference. Thus, the resultant of the weight is equal to the
total weight of all n particles:

The summation of the moments of the weights of all the particles about the x, y, and z axes
is equal to the moment of the resultant weight about these axe
s. Then, to determine the x coordinate of G, it is necessary to do summation of moments

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about the y axis to obtain

In a similar manner, to determine the y coordinate of G, calculation of summation of
moments about the x axis is effectuated,

The weights do not produce moment about z axis; therefore, the z coordinate of G is
calculated by rotating the coordinate system, with the particles attached to it, by 90o about
either the x or y axis, and then proceed in a similar manner as before

Thus, following this procedure is possible to determine the coordinates of the center of
gravity G of a system composed by n particles as

Center of Mass. The determination of the point known as center of mass is necessary to
carry out the study of problems concerning to the motion of matter under the influence of a
force. Provided that the acceleration due to the gravity
g for every particle of the system is constant, then W = mg. In this way, can be written as

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EXERCISE: Where is the Center of Mass?
1.

2. A thin rod of length 3 is bent at right angles at a distance L from one end.
Locate the CM with respect to the corner. Take L=1.2 m.

REFERENCE :
1. Hibbeler, R.C & S.C Fan 1997. Engineering Mechanics, Statics. SI Ed. New Jersey:
Prentice Hall. ISBN 0135995981
2. Yusof Ahmad. 1999. Mekanik Statik. Cetakan Ketiga. Malaysia: Penerbit UTM Skudai Johor

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