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Ishik University / Sulaimani

Architecture Department

Structure

ARCH 214

Chapter -2-

Force Vectors

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Contents :

1. Scalars and Vectors

2. Vector Operations

3. Vector Addition of Forces

4. Addition of a System of Coplanar Force

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Objectives

To show how to add forces and resolve them into

components using the Parallelogram Law and

Trigonometry Analysis.

To express force and position in Cartesian vector

form and explain how to determine the vector’s

magnitude and direction.

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Scalar :

– A quantity characterized by a positive or negative number.

– Indicated by letters in italic such as A.

Example: mass, volume and length

Vector :

– A quantity that has both magnitude and direction.

Example: position, force and moment

– Represent by a letter with an arrow over it such as or A

– Magnitude is designated as or simply A.

– In this subject, vector is presented as A and its magnitude (positive

quantity) as A.

1. Scalars and Vectors

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A

A

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Vector

– Represented graphically as an arrow.

– Length of arrow = Magnitude of Vector.

– Angle between the reference axis and arrow’s line of action

= Direction of Vector.

– Arrowhead = Sense of Vector

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A. Vector Addition

- Addition of two vectors A and B gives a resultant vector R by the

parallelogram law.

- Result R can be found by triangle construction.

Example: R = A + B = B + A

2. Vector Operations

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B. Vector Subtraction

The resultant of the difference between two vectors 𝐀 and 𝐁 of the same type

may be expressed as:

- Special case of addition

Example: R’ = A – B = A + ( - B )

- Rules of Vector Addition Applies

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C. Multiplication and division of vector by a scalar:

If a vector is multiplied by a positive scalar, its magnitude is

increased by that amount.

When multiplied by a negative scalar it will also change the

directional sense of the vector

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3. Vector Addition of Forces

Experimental evidence has shown that a force is a vector

quantity since it has a specified magnitude, direction, and sense

and it adds according to the parallelogram law.

When two or more forces are added, successive applications of

the parallelogram law is carried out to find the resultant

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Finding a resultant force:

The two component forces 𝐅𝟏 and 𝐅𝟐 acting on the pin in figure,

can be added together to form the resultant force.

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Finding the components of a force:

o Sometimes it is necessary to resolve a force into two components in

order to study its pulling and pushing effect in two specific directions.

o For example, in figure below, F is to be resolved into two components

along two members, defined by u and v.

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Addition of several forces:

o If more than two forces are to be added successive applications of the

parallelogram law can be carried out in order to obtain the resultant

force.

o For example if the three forces 𝐅𝟏, 𝐅𝟐, 𝐅𝟑 act at a point o, the resultant

of any two of the forces is found (𝐅𝟏 + 𝐅𝟐) and then this resultant is

added to the third force yielding the resultant of all three forces.

(𝐅𝐑 = (𝐅𝟏 + 𝐅𝟐) + 𝐅𝟑)

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a) Parallelogram Law

b) Trigonometry 1. sine law 2. cosine law

Procedure of Analysis / Review

Two "component" forces F1 and F2 in Fig.(a) add according to the parallelogram

law yielding, a resultant force FR that forms the diagonal of the parallelogram.

If a force F is to be resolved into components along two axes u and v, Fig.(b).

Then start at the head of force F, and construct lines parallel to the axes, thereby

forming the parallelogram.

The sides of the parallelogram represent the components Fu and Fv.

a) Parallelogram Law

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• Redraw a half portion of the parallelogram to illustrate the triangular

head to tail addition of the components.

• From this triangle, the magnitude of the resultant force can be

determined using the law of cosines, and its direction is determined from

the law of sines.

• The magnitudes of two force components are determined from the law

of sines.

b) Trigonometry Analysis

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Example -1-

The screw eye is subjected to two forces F1 and F2. Determine the

magnitude and the direction of a Resultant force.

Ans:

𝐹𝑅 = 213 N

∅ = 54.7°

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Solution 1: Parallelogram

Unknown: magnitude of FR and angle θ

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N

N

NNNNFR

213

6.212

4226.0300002250010000

115cos150100215010022

Law of Cosines:

Solution 2: Trigonometry

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8.39sin

9063.06.212

150sin

115sin

6.212

sin

150

N

N

NN

Law of Sines

Direction Φ of FR measured from the

horizontal is,

8.54

158.39

Note: The results seem reasonable since it shows FR to have a magnitude larger

than its components and a direction that is between them.

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Example -2-

Determine the magnitude of the component force 𝐅 in figure below

and the magnitude of the resultant force 𝐅𝐑 if 𝐅𝐑 is directed a long

the positive y axis.

Ans:

𝐹 = 245 lb

𝐹𝑅 = 273 lb

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Solution 2: Trigonometry Solution 1: Parallelogram

The magnitudes of FR and F are the two unknowns. They can be

determined by applying the law of sines.

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Example -3-

Determine the magnitude of the resultant force acting on the

screw eye and its direction measured clockwise from the x

axis.

Ans:

𝐹𝑅 = 6.80 kN

θ = 103°

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Example -4-

Resolve the horizontal 600lb force in figure below, into

components action along the u and v axes and determine the

magnitudes of these components.

Ans:

𝐹𝑢 = 1039 lb

𝐹𝑣 = 600 lb

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4. Addition of a System of Coplanar Force

For resultant of two or more forces:

• Find the components of the forces in the specified axes

• Add them algebraically

• Form the resultant

In this subject, we resolve each force into rectangular forces along the x and

y axes.

yx FFF

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Scalar Notation

- x and y axes are designated positive and negative

- Components of forces expressed as algebraic scalars

Example: Sense of direction along positive x and y axes

Instead of using the angle 𝜽, the direction of 𝐅 can also be defined using a

small "slope" triangle, shown in figure.

It is also possible to represent the x and y components of a force in terms of

Cartesian unit vectors i and j

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Cartesian Vector Notation

F = Fxi + Fyj

• We can express 𝐅 as a Cartesian vector.

Coplanar Force Resultants

• In coplanar force resultant case, each force is resolved into its x and

y components.

• Then the respective components are added using scalar algebra since

they are collinear.

• For example,

consider the three concurrent forces,

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• Each force is represented as a Cartesian vector.

Cartesian vector notation;

F1 = F1xi + F1yj

F2 = - F2xi + F2yj

F3 = F3xi – F3yj

• Vector resultant is therefore,

FR = F1 + F2 + F3

= F1xi + F1yj - F2xi + F2yj + F3xi – F3yj

= (F1x - F2x + F3x)i + (F1y + F2y – F3y)j

= (FRx)i + (FRy)j

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• We can represent the components of the resultant force of any number of

coplanar forces symbolically by the algebraic sum the x and y

components of all the forces.

• In all cases, FRx = ∑Fx

FRy = ∑Fy

• Take note of sign conventions

• Magnitude of FR can be found by

Pythagorean Theorem, Ry

2Rx

2

RFFF

• Direction angle θ (orientation of the force)

can be found by trigonometry, Rx

Ry

F

Ftan

1

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Example -5-

The link is subjected to two forces F1 and F2. Determine the magnitude and

orientation of the resultant force.

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Solution :

N

NNF

FF

N

NNF

FF

Ry

yRy

Rx

xRx

8.582

45cos40030sin600

:

8.236

45sin40030cos600

:

Solution 1: Scalar Notation;

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Resultant Force

From vector addition, Direction angle θ is

N

NNFR

629

8.5828.23622

9.67

8.236

8.582tan 1

N

N

Solution 2: Cartesian Vector Notation

F1 = { 600cos30°i + 600sin30°j } N F2 = { -400sin45°i + 400cos45°j } N Thus, FR = F1 + F2 = (600cos30°N - 400sin45°N)i + (600sin30°N + 400cos45°N)j = {236.8i + 582.8j}N

*The magnitude and direction of FR are determined in the S.1nle manner as before.

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Example -6-

Determine the x and y components of 𝐅𝟏 and 𝐅𝟐 acting on the boom shown in

figure, express each force as a Cartesian vector.

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Solution;

NNNF

NNNF

y

x

17317330cos200

10010030sin200

1

1

Scalar Notation;

N10013

5260

N24013

12260

2

2

y

x

F

FBy similar triangles we have;

NNF

NF

y

x

100100

240

2

2

Scalar Notation;

Cartesian Vector Notation;

NjiF

NjiF

100240

173100

2

1

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Example -7-

The link is subjected to two forces F1 and F2. Determine the magnitude and

orientation of the resultant force.

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N

NNF

FF

N

NNF

FF

Ry

yRy

Rx

xRx

8.582

45cos40030sin600

:

8.236

45sin40030cos600

:

Scalar Notation:

N629N8.582N8.236F22

R

Resultant Force;

From vector addition, direction angle θ is;

9.67N8.236

N8.582tan 1

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Resultant Force; N629N8.582N8.236F22

R

From vector addition, direction angle θ is; 9.67

N8.236

N8.582tan 1

Cartesian Vector Notation;

F1 = { 600cos30°i + 600sin30°j } N

F2 = { -400sin45°i + 400cos45°j } N

Thus,

FR = F1 + F2

= (600cos30ºN - 400sin45ºN)i

+ (600sin30ºN + 400cos45ºN)j

= {236.8i + 582.8j}N

The magnitude and direction of FR are determined in the same manner as

before.

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Example -8-

Resolve each force acting on the post into its x and y components.

Ans:

F1𝑥 = O N

F1y = 300 N

F2𝑥 = −318 N

F2y = 318 N

𝐹3𝑥 = 360 N

𝐹3𝑦 = 480 N

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Example -9-

Determine the magnitude and direction of the resultant force.

Ans:

𝐹𝑅 = 567 N

𝜃 = 38.1°

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Example -10-

Determine the magnitude of the resultant force acting on the corbel and its

direction θ measured counterclockwise from the x axis.

Ans:

𝐹𝑅 = 1254 lb

𝛷 = 78.68°

𝜃 = 180 + 𝛷 = 259°

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Example -11-

If the resultant force acting on the bracket is to be 750 N directed along the

positive x axis, determine the magnitude of 𝐅 and its direction θ.

Ans:

𝜃 = 31.76°

𝐹 = 236 N

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Example -12-

Determine the magnitude of the resultant force and its direction θ measured

counterclockwise from the positive x axis.

Ans:

𝜃 = 39.8°

𝐹 = 31.2 kN

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References :

1. Engineering Mechanics-Statics by R.C.-Hibbeler, 12th Edition.

2. Lecture Notes and Exercises on Statics, by Dr. Abdulwahab Amrani.