Chapter II a

Force Systems 2D

fig by J.L. Meriam, L.G. Kraige, Engineering Mechanics I

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 Lecture Overview
1.) - Classification of Vectors

- Vector addition, parallelogram law

- Decomposition of vectors, force components

2.) - Moment of a vector

- Force couples

ADDIS ABABA UNIVERSITY Engineering Mechanics Department of

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 Statics concerning rigid bodies

- focus on sliding vectors

- principle of transmissibility
(force can be applied on any point along its line of action without
changing its external effect on a rigid body)

fig by J.L. Meriam, L.G. Kraige, Engineering Mechanics I

ADDIS ABABA UNIVERSITY Engineering Mechanics Department of

Faculty of Technology material by Karsten Schlesier Civil Engineering
Vector Addition

1. parallelogram law

2. triangle rule

3. analytic method

ADDIS ABABA UNIVERSITY Engineering Mechanics Department of

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Vector Addition – graphical method

the parallelogram law – resultant force

ADDIS ABABA UNIVERSITY Engineering Mechanics Department of

Faculty of Technology material by Karsten Schlesier Civil Engineering
Vector Addition – graphical method

the parallelogram law – resultant force

2 forces replaced by a single resultant force

scale

ADDIS ABABA UNIVERSITY Engineering Mechanics Department of

Faculty of Technology material by Karsten Schlesier Civil Engineering
Vector Addition – graphical method

the parallelogram law – resultant force

2 forces replaced by a single resultant force

scale

ADDIS ABABA UNIVERSITY Engineering Mechanics Department of

Faculty of Technology material by Karsten Schlesier Civil Engineering
Vector Addition – graphical method

the parallelogram law – resultant force

2 forces replaced by a single resultant force

A
point of intersection

ADDIS ABABA UNIVERSITY Engineering Mechanics Department of

Faculty of Technology material by Karsten Schlesier Civil Engineering
Vector Addition – graphical method

the parallelogram law – resultant force

2 forces replaced by a single resultant force

ADDIS ABABA UNIVERSITY Engineering Mechanics Department of

Faculty of Technology material by Karsten Schlesier Civil Engineering
Vector Addition – graphical method

the parallelogram law – resultant force

Line of action of R through point of intersection

of A and B

scale

A R

ADDIS ABABA UNIVERSITY Engineering Mechanics Department of

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Vector Addition – graphical method

the triangle rule (from parallelogram law)

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Vector Addition – graphical method

the triangle rule (from parallelogram law)

line of action of one force to be moved!

ADDIS ABABA UNIVERSITY Engineering Mechanics Department of

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Vector Addition – graphical method

the triangle rule (from parallelogram law)

for two forces

r r r r r
R = A+B = B+A

B
R
A
R A
B

ADDIS ABABA UNIVERSITY Engineering Mechanics Department of

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Vector Addition – graphical method

the triangle rule (from parallelogram law)

for three forces

r r r r r r r
R = A + B + C = B + A + C = ...

B
C A C
A R
R
B

ADDIS ABABA UNIVERSITY Engineering Mechanics Department of

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Vector Addition – graphical method

the triangle rule (from parallelogram law)

location of line of action of resultant force R?

Initial situation: Result of addition:

A B
A C
B
R
C

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Vector Addition – graphical method

the triangle rule (from parallelogram law)

location of line of action of resultant force R?

Initial situation:

A A

B B

C C

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Vector Addition – graphical method

lines of action – point of intersection

Initial situation:

A A

B B

C C

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Vector Addition – graphical method

Initial situation:

A
A

B B

C C

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Vector Addition – graphical method

Initial situation:

B
A
A

C C

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Vector Addition – graphical method

result for 2 forces A and B : A + B = RAB

location of resultant force RAB

Initial situation:

B
A
A
RAB
B

C C

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Vector Addition – graphical method

location of RAB and C

Initial situation:

RAB
B

C C

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Vector Addition – graphical method

lines of action

Initial situation:

RAB
B

C C

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Vector Addition – graphical method

point of intersection

Initial situation:

A
RAB

C C

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Vector Addition – graphical method

Initial situation:

C A
RAB

ADDIS ABABA UNIVERSITY Engineering Mechanics Department of

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Vector Addition – graphical method

resultant force of the 3 forces A, B and C

Initial situation:

C A
RAB

RABC B

ADDIS ABABA UNIVERSITY Engineering Mechanics Department of

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Vector Addition – graphical method

location of line of action of resultant force

Initial situation:

C A
RAB

RABC B

ADDIS ABABA UNIVERSITY Engineering Mechanics Department of

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Vector Addition – graphical method

location of line of action of resultant force

Initial situation:

A B
A C

B RABC
R

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Vector Addition – graphical method

EXCURSION: equilibrium at particle

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Vector Addition – graphical method

e.g.: equilibrium at particle:

addition of all vectors = 0

situation at
particle:

A resultant
B
reaction force A C

B R
R

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Recalling equilibrium at particle

ADDIS ABABA UNIVERSITY Engineering Mechanics Department of

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Recalling equilibrium at particle

ADDIS ABABA UNIVERSITY Engineering Mechanics Department of

Faculty of Technology material by Karsten Schlesier Civil Engineering
Recalling equilibrium at particle

ADDIS ABABA UNIVERSITY Engineering Mechanics Department of

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Vector Addition – analytic method

ADDIS ABABA UNIVERSITY Engineering Mechanics Department of

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Vector Addition – analytic method

trigonometric rules

B R
B
β θ
A
2 2 2
R = A + B − 2 A B ⋅ cosθ magnitude

sinβ sinθ
= Inclination or R from A
B R
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Vector Decomposition

1. components along arbitrary axis

a) graphical method
b) analytical method

2. rectangular components
vector addition by rectangular components

ADDIS ABABA UNIVERSITY Engineering Mechanics Department of

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Vector Decomposition – graphical method

ADDIS ABABA UNIVERSITY Engineering Mechanics Department of

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Vector Decomposition – graphical method
vector components along 2 axis of arbitrary incline
b

F
φ
θ
a

given: Force F, 2 axis of arbitrary incline a and b

ADDIS ABABA UNIVERSITY Engineering Mechanics Department of

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Vector Decomposition – graphical method
vector components along 2 axis of arbitrary incline
b

F
φ
θ
a

applying parallelogram law

ADDIS ABABA UNIVERSITY Engineering Mechanics Department of

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Vector Decomposition – graphical method
vector components along 2 axis of arbitrary incline
b

Fb F
φ
θ
Fa a

ADDIS ABABA UNIVERSITY Engineering Mechanics Department of

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Vector Decomposition – analytical method

ADDIS ABABA UNIVERSITY Engineering Mechanics Department of

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Vector Decomposition – analytical method
vector components along 2 axis of arbitrary incline
b

Fb F α = 180 o − (φ + α)
φ B
θ α
Fa a
sinφ sinθ sinα
by trigonometric rules: = =
Fa Fb F
components in sinφ sinθ
a and b directions: Fa = F and Fb = F
sinα sinα
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Rectangular Components

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Rectangular Components
y

F
θ
x

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Rectangular Components

F
θ
x

horizontal component of F: Fx = F ⋅ cosθ

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Rectangular Components

F
Fy
θ
Fx x

horizontal component of F: Fx = F ⋅ cosθ

vertical component of F: Fy = F ⋅ sinθ
magnitude of F: F = Fx2 + Fy2
ADDIS ABABA UNIVERSITY Engineering Mechanics Department of
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Vector addition by rectangular components

ADDIS ABABA UNIVERSITY Engineering Mechanics Department of

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Vector addition by rectangular components

fig by J.L. Meriam, L.G. Kraige, Engineering Mechanics I

R =R= (F1x + F2 x ) + (F1 y + F2 y )

2 2

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2 examples

ADDIS ABABA UNIVERSITY Engineering Mechanics Department of

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example 2.1

given: detail of a symmetric cable support

θ = 30˚, P = 10 kN

determine: tensile force (horizontal) and shear force

(vertical) exerted to the connection.

fig by J.L. Meriam, L.G. Kraige, Engineering Mechanics I

ADDIS ABABA UNIVERSITY Engineering Mechanics Department of

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example 2.1

θ = 30˚, P = 10 kN

equilibrium condition: reaction force support –

Py position of bolt
R= -P

Px

Px and Py are the rectangular

components of P!
fig by J.L. Meriam, L.G. Kraige, Engineering Mechanics I

ADDIS ABABA UNIVERSITY Engineering Mechanics Department of

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example 2.1
Py
θ = 30˚, P = 10 kN R= -P

Px

fig by J.L. Meriam, L.G. Kraige, Engineering Mechanics I

║Px ║= cosθ·║P║ = cos30˚·10kN = 8.66kN

║Py ║= sinθ·║P║ = sin30˚·10kN = 5.00kN

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example 2.2 (from meriam textbook 2/13)

fig by J.L. Meriam, L.G. Kraige, Engineering Mechanics I

ADDIS ABABA UNIVERSITY Engineering Mechanics Department of

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example 2.2

T R=?
R

T = 400 N

fig by J.L. Meriam, L.G. Kraige, Engineering Mechanics I

ADDIS ABABA UNIVERSITY Engineering Mechanics Department of

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example 2.2

y
T R=?
R
x

T = 400 N fig by J.L. Meriam, L.G. Kraige, Engineering Mechanics I

R x = T + T ⋅ cos60 = 400N(1 + 0.5) = 600.0N

o

R y = T ⋅ sin60 = 400N ⋅ 3 2 = 346.4N

0

R = R 2x + R 2y = 692.8N
ADDIS ABABA UNIVERSITY Engineering Mechanics Department of
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example 2.2

y
T R=?
R
x 60˚

T fig by J.L. Meriam, L.G. Kraige, Engineering Mechanics I

R x = T + T ⋅ cos60 = 400N(1 + 0.5) = 600.0N

o

R y = T ⋅ sin60 = 400N ⋅ 3 2 = 346.4N

0

R = R 2x + R 2y = 692.8N
ADDIS ABABA UNIVERSITY Engineering Mechanics Department of
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example 2.2

y
T R=?
R
x 60˚

T fig by J.L. Meriam, L.G. Kraige, Engineering Mechanics I

R x = T + T ⋅ cos60 = 400N(1 + 0.5) = 600.0N

o

R y = T ⋅ sin60 = 400N ⋅ 3 2 = 346.4N

0

R = R 2x + R 2y = 692.8N
ADDIS ABABA UNIVERSITY Engineering Mechanics Department of
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example 2.2

y
T R=?
R
x 60˚

T
fig by J.L. Meriam, L.G. Kraige, Engineering Mechanics I

R x = T + T ⋅ cos60 = 400N(1 + 0.5) = 600.0N

o

R y = T ⋅ sin60 = 400N ⋅ 3 2 = 346.4N

0

R = R 2x + R 2y = 692.8N
ADDIS ABABA UNIVERSITY Engineering Mechanics Department of
Faculty of Technology material by Karsten Schlesier Civil Engineering
example 2.2

y
T R=?
R
x 60˚

T
fig by J.L. Meriam, L.G. Kraige, Engineering Mechanics I

R x = T + T ⋅ cos60 = 400N(1 + 0.5) = 600.0N

o

R y = T ⋅ sin60 = 400N ⋅ 3 2 = 346.4N

0

R = R 2x + R 2y = 692.8N
ADDIS ABABA UNIVERSITY Engineering Mechanics Department of
Faculty of Technology material by Karsten Schlesier Civil Engineering
example 2.2

y
T R=?
R
x 60˚

T
fig by J.L. Meriam, L.G. Kraige, Engineering Mechanics I

R x = T + T ⋅ cos60 = 400N(1 + 0.5) = 600.0N

o

R y = T ⋅ sin60 = 400N ⋅ 3 2 = 346.4N

0

R = R + R = 692.8N
2
x
2
y
ADDIS ABABA UNIVERSITY Engineering Mechanics Department of
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 Lecture Overview
1.) - Classification of Vectors

- Vector addition, parallelogram law

- Decomposition of vectors, force components

2.) - Moment of a vector

- Force couples

ADDIS ABABA UNIVERSITY Engineering Mechanics Department of

Faculty of Technology material by Karsten Schlesier Civil Engineering
Moment

1. moment of a force about a point

- scalar product
- cross product

2. force couples

ADDIS ABABA UNIVERSITY Engineering Mechanics Department of

Faculty of Technology material by Karsten Schlesier Civil Engineering
Moment

1. moment of a force about a point

fig by J.L. Meriam, L.G. Kraige, Engineering Mechanics I

ADDIS ABABA UNIVERSITY Engineering Mechanics Department of

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moment of a force

force acting in a plane

ADDIS ABABA UNIVERSITY Engineering Mechanics Department of

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moment of a force

line of action and support

ADDIS ABABA UNIVERSITY Engineering Mechanics Department of

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moment of a force

line of action of reaction force

F distance

hinge support: no equilibrium - rotation!

ADDIS ABABA UNIVERSITY Engineering Mechanics Department of

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moment of a force

axis of rotation

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moment of a force

Moment of Force F M = F·d [kNm]

d = moment arm
perpendicular to line of action!

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moment of a force

sign convention – right hand rule (+)

M (+)

ADDIS ABABA UNIVERSITY Engineering Mechanics Department of

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moment of a force, scalar product

Moment of Force F M = F·d [kNm]

d = moment arm

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examples - moment of a force

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ADDIS ABABA UNIVERSITY Engineering Mechanics Department of
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ADDIS ABABA UNIVERSITY Engineering Mechanics Department of
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ADDIS ABABA UNIVERSITY Engineering Mechanics Department of
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 M = F·d

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 Equilibrium!
left side right side

M = +F·d M = -F·d

F F
+ M M -

d d

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 Equilibrium!

d d

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 d d

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 2F

d d

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left side right side

M = +F·d M = -2F·d
2F

d d

ADDIS ABABA UNIVERSITY Engineering Mechanics Department of

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 Rotation!

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ADDIS ABABA UNIVERSITY Engineering Mechanics Department of
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ADDIS ABABA UNIVERSITY Engineering Mechanics Department of
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 d d/2

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 Equilibrium!
left side right side

M = +F·d M = -2F·d/2
2F

d d/2

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ADDIS ABABA UNIVERSITY Engineering Mechanics Department of
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 5m

1m

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 F = ??
5m

500 N

1m

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ADDIS ABABA UNIVERSITY Engineering Mechanics Department of
Faculty of Technology material by Karsten Schlesier Civil Engineering
 600 N
???

1m 3m

ADDIS ABABA UNIVERSITY Engineering Mechanics Department of

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Moment

1. moment of a force about a point

- scalar product
- cross product

2. force couples

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Moment - Cross Product (vector product)

Moment of a vector V about any point 0

r is a position vector from reference point 0 to any point on

the line of action of vector V.
r r r r r 0
M0 = r × V = F ⋅ r ⋅ sinθ ║r║·sinθ
r r r r
i j k θ
r F
M 0 = rx ry rz
Vx Vy Vz

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 Definition of Cross Product (vector product)

Determination of resulting vector by three by three matrix with

unit vectors i, j and k.
r r r
i j k
ax ay az
bx by bz
r r r r r
A × B = (a y b z - a z b y )i + (a z b x - a x b z ) j + (a x b y - a y b x )k
for 2D system
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Moment – Varignon’s Theorem

Moment of a force F about any point equals the sum of

the moments of the components of the force about the
same point.

by cross product:

fig by J.L. Meriam, L.G. Kraige, Engineering Mechanics I

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Moment – Varignon’s Theorem

Moment of a force F about any point equals the sum of

the moments of the components of the force about the
same point.

by rectangular moment arms:

fig by J.L. Meriam, L.G. Kraige, Engineering Mechanics I

ADDIS ABABA UNIVERSITY Engineering Mechanics Department of

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example 2.3 (from meriam textbook 2/5)

Calculate the magnitude of the moment about the base point 0 in

different ways.

fig by J.L. Meriam, L.G. Kraige, Engineering Mechanics I

ADDIS ABABA UNIVERSITY Engineering Mechanics Department of

Faculty of Technology material by Karsten Schlesier Civil Engineering
example 2.3

Calculate the magnitude of the moment about the base point 0 in

different ways.
1.) product of force by moment arm

fig by J.L. Meriam, L.G. Kraige, Engineering Mechanics I

ADDIS ABABA UNIVERSITY Engineering Mechanics Department of

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example 2.3

Calculate the magnitude of the moment about the base point 0 in

different ways.

fig by J.L. Meriam, L.G. Kraige, Engineering Mechanics I

ADDIS ABABA UNIVERSITY Engineering Mechanics Department of

Faculty of Technology material by Karsten Schlesier Civil Engineering
example 2.3

Calculate the magnitude of the moment about the base point 0 in

different ways.

fig by J.L. Meriam, L.G. Kraige, Engineering Mechanics I

ADDIS ABABA UNIVERSITY Engineering Mechanics Department of

Faculty of Technology material by Karsten Schlesier Civil Engineering
example 2.3

Calculate the magnitude of the moment about the base point 0 in

different ways.
2.) replace force by rectangular
components

x
fig by J.L. Meriam, L.G. Kraige, Engineering Mechanics I

ADDIS ABABA UNIVERSITY Engineering Mechanics Department of

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example 2.3

Calculate the magnitude of the moment about the base point 0 in

different ways.

fig by J.L. Meriam, L.G. Kraige, Engineering Mechanics I

ADDIS ABABA UNIVERSITY Engineering Mechanics Department of

Faculty of Technology material by Karsten Schlesier Civil Engineering
example 2.3

Calculate the magnitude of the moment about the base point 0 in

different ways.

fig by J.L. Meriam, L.G. Kraige, Engineering Mechanics I

ADDIS ABABA UNIVERSITY Engineering Mechanics Department of

Faculty of Technology material by Karsten Schlesier Civil Engineering
example 2.3

Calculate the magnitude of the moment about the base point 0 in

different ways.

fig by J.L. Meriam, L.G. Kraige, Engineering Mechanics I

ADDIS ABABA UNIVERSITY Engineering Mechanics Department of

Faculty of Technology material by Karsten Schlesier Civil Engineering
example 2.3

Calculate the magnitude of the moment about the base point 0 in

different ways.
3.) by cross product

x
fig by J.L. Meriam, L.G. Kraige, Engineering Mechanics I

ADDIS ABABA UNIVERSITY Engineering Mechanics Department of

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example 2.3

Calculate the magnitude of the moment about the base point 0 in

different ways.

x
fig by J.L. Meriam, L.G. Kraige, Engineering Mechanics I

ADDIS ABABA UNIVERSITY Engineering Mechanics Department of

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example 2.3

Calculate the magnitude of the moment about the base point 0 in

different ways.

r r r x
fig by J.L. Meriam, L.G. Kraige, Engineering Mechanics I
i j k
2 4 0
600 ⋅ cos40o - 600 ⋅ sin40 o 0
r r r r r
A × B = (a y b z - a z b y )i + (a z b x - a x b z ) j + (a x b y - a y b x )k
=0 =0

ADDIS ABABA UNIVERSITY Engineering Mechanics Department of

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Moment

1. moment of a force about a point

- scalar product
- cross product

2. force couples

ADDIS ABABA UNIVERSITY Engineering Mechanics Department of

Faculty of Technology material by Karsten Schlesier Civil Engineering
Moment

2. force couples

fig by J.L. Meriam, L.G. Kraige, Engineering Mechanics I

ADDIS ABABA UNIVERSITY Engineering Mechanics Department of

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moment of a couple

moment produced by 2 equal and opposite

forces is known as a couple.

d F

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moment of a couple

moment of couple about any point of reference 0.

0
F
a

d F

ADDIS ABABA UNIVERSITY Engineering Mechanics Department of

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moment of a couple

moment of couple about any point of reference 0.

0
F
a
fig by J.L. Meriam, L.G. Kraige, Engineering Mechanics I

d F R=F–F=0

M = F(a+d) - Fa
M = Fd
ADDIS ABABA UNIVERSITY Engineering Mechanics Department of
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moment of a couple

moment of couple about any point of reference 0.

0
F
a
r fig by J.L. Meriam, L.G. Kraige, Engineering Mechanics I

d F R=F–F=0

M = F(a+d) - Fa
cross product: M = r x F M = Fd
ADDIS ABABA UNIVERSITY Engineering Mechanics Department of
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moment of a couple

fig by J.L. Meriam, L.G. Kraige, Engineering Mechanics I

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Force-couple systems

resolution of a force into a force and a couple –

equal external effect.

fig by J.L. Meriam, L.G. Kraige, Engineering Mechanics I

ADDIS ABABA UNIVERSITY Engineering Mechanics Department of

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example 2.4 (from meriam textbook 2/55)

fig by J.L. Meriam, L.G. Kraige, Engineering Mechanics I

ADDIS ABABA UNIVERSITY Engineering Mechanics Department of

Faculty of Technology material by Karsten Schlesier Civil Engineering
example 2.4

M = -F·y
y = -M/F = - 0.375 kNm / 5 kN
y = -75 mm

fig by J.L. Meriam, L.G. Kraige, Engineering Mechanics I

ADDIS ABABA UNIVERSITY Engineering Mechanics Department of

Faculty of Technology material by Karsten Schlesier Civil Engineering
example 2.4

M = -F·y
y = -M/F = - 0.375 kNm / 5 kN
y = -75 mm 75 mm

=
initial
fig by J.L. Meriam, L.G. Kraige, Engineering Mechanics I

ADDIS ABABA UNIVERSITY Engineering Mechanics Department of

Faculty of Technology material by Karsten Schlesier Civil Engineering
ADDIS ABABA UNIVERSITY Engineering Mechanics Department of
Faculty of Technology material by Karsten Schlesier Civil Engineering