6161103 2.4 addition of a system

2.4 Addition of a System of Coplanar Forces

2.4 Addition of a System of Coplanar Forces

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 +=

2.4 Addition of a System of Coplanar Forces

2.4 Addition of a System of Coplanar Forces

� Scalar Notation- x and y axes are designated positive and negative

- Components of forces expressed as algebraic - Components of forces expressed as algebraic scalars

Eg:

Sense of direction

along positive x and

y axes

yx FFF +=

2.4 Addition of a System of Coplanar Forces

2.4 Addition of a System of Coplanar Forces

�Scalar NotationEg:

Sense of direction

yx FFF ''' +=

Sense of direction

along positive x and

negative y axes

2.4 Addition of a System of Coplanar Forces

2.4 Addition of a System of Coplanar Forces

�Scalar Notation- Head of a vector arrow = sense of the vector graphically (algebraic signs not vector graphically (algebraic signs not used)

- Vectors are designated using boldface notations

- Magnitudes (always a positive quantity) are designated using italic symbols

2.4 Addition of a System of Coplanar Forces

2.4 Addition of a System of Coplanar Forces

� Cartesian Vector Notation- Cartesian unit vectors i and j are used to designate the x and y directions

- Unit vectors i and j have dimensionless - Unit vectors i and j have dimensionless magnitude of unity ( = 1 )

- Their sense are indicated by a positive or negative sign (pointing in the positive or negative x or y axis)

- Magnitude is always a positive quantity, represented by scalars Fx and Fy

2.4 Addition of a System of Coplanar Forces

2.4 Addition of a System of Coplanar Forces

� Cartesian Vector Notation F = Fxi + Fyj F’ = F’xi + F’y(-j)

F’ = F’xi – F’yj

2.4 Addition of a System of Coplanar Forces

2.4 Addition of a System of Coplanar Forces

� Coplanar Force ResultantsTo determine resultant of several coplanar forces:

- Resolve force into x and y - Resolve force into x and y components

- Addition of the respective components using scalar algebra

- Resultant force is found using the parallelogram law

2.4 Addition of a System of Coplanar Forces

2.4 Addition of a System of Coplanar Forces

� Coplanar Force ResultantsExample: Consider three coplanar forces

Cartesian vector notation

F1 = F1xi + F1yj

F2 = - F2xi + F2yj

F3 = F3xi – F3yj

� Coplanar Force ResultantsVector resultant is therefore

FR = F1 + F2 + F3

= F i + F j - F i + F j + F i – F j

2.4 Addition of a System of Coplanar Forces

2.4 Addition of a System of Coplanar Forces

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

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

= (FRx)i + (FRy)j

2.4 Addition of a System of Coplanar Forces

2.4 Addition of a System of Coplanar Forces

� Coplanar Force ResultantsIf scalar notation are usedFRx = (F1x - F2x + F3x)

FRy = (F1y + F2y – F3y)FRy = (F1y + F2y – F3y)

In all cases,

FRx = ∑Fx

FRy = ∑Fy

* Take note of sign conventions

2.4 Addition of a System of Coplanar Forces

2.4 Addition of a System of Coplanar Forces

� Coplanar Force Resultants- Positive scalars = sense of direction along the positive coordinate axes

- Negative scalars = sense of direction - Negative scalars = sense of direction along the negative coordinate axes

- Magnitude of FR can be found by Pythagorean Theorem

RyRxR FFF 22 +=

2.4 Addition of a System of Coplanar Forces

2.4 Addition of a System of Coplanar Forces

� Coplanar Force Resultants- Direction angle θ (orientation of the force) can be found by trigonometry

Rx

Ry

F

F1tan−=θ

2.4 Addition of a System of Coplanar Forces

2.4 Addition of a System of Coplanar Forces

Example 2.5

Determine x and y components of F1 and F2

acting on the boom. Express each force as a

Cartesian vector Cartesian vector

2.4 Addition of a System of Coplanar Forces

2.4 Addition of a System of Coplanar Forces

Solution

Scalar Notation

←=−=−= NNNF x 10010030sin2001o

Hence, from the slope triangle

↑=== NNNF y

x

17317330cos2001

1

o

= −

12

5tan1θ

2.4 Addition of a System of Coplanar Forces

2.4 Addition of a System of Coplanar Forces

Solution

Alt, by similar triangles

N

F x

13

12

2602 =

Similarly,

NNF

N

x 24013

12260

13260

2 =

=

=

NNF y 10013

52602 =

=

2.4 Addition of a System of Coplanar Forces

2.4 Addition of a System of Coplanar Forces

Solution

Scalar Notation

→== NNF x 2402402

Cartesian Vector Notation

F1 = {-100i +173j }N

F2 = {240i -100j }N

↓=−=

→==

NNF

NNF

y

x

100100

240240

2

2

2.4 Addition of a System of Coplanar Forces

2.4 Addition of a System of Coplanar Forces

Example 2.6

The link is subjected to two forces F1 and

F2. Determine the magnitude and F2. Determine the magnitude and orientation of the resultant force.

2.4 Addition of a System of Coplanar Forces

2.4 Addition of a System of Coplanar Forces

Solution

Scalar Notation

Σ= FF xRx :

↑=

+=

Σ=

→=

−=

N

NNF

FF

N

NNF

Ry

yRy

Rx

xRx

8.582

45cos40030sin600

:

8.236

45sin40030cos600

oo

oo

2.4 Addition of a System of Coplanar Forces

2.4 Addition of a System of Coplanar Forces

Solution

Resultant Force

( ) ( )N

NNFR

629

8.5828.236 22

=

+=

From vector addition,

Direction angle θ is

N629=

o9.67

8.236

8.582tan 1

=

= −

N

Nθ

2.4 Addition of a System of Coplanar Forces

2.4 Addition of a System of Coplanar Forces

Solution

Cartesian Vector Notation

F1 = { 600cos30°i + 600sin30°j } N

F2 = { -400sin45°i + 400cos45°j } NF2 = { -400sin45°i + 400cos45°j } N

Thus,

FR = F1 + F2

= (600cos30°N - 400sin45°N)i + (600sin30°N + 400cos45°N)j

= {236.8i + 582.8j}N

2.4 Addition of a System of Coplanar Forces

2.4 Addition of a System of Coplanar Forces

Example 2.7

The end of the boom O is subjected to three

concurrent and coplanar forces. Determine

the magnitude and orientation of the the magnitude and orientation of the

resultant force.

2.4 Addition of a System of Coplanar Forces

2.4 Addition of a System of Coplanar Forces

Solution

Scalar Notation

Σ= FF xRx

4

:

↑=

+=

Σ=

←=−=

−+−=

N

NNF

FF

NN

NNNF

Ry

yRy

Rx

8.296

5

320045cos250

:

2.3832.383

5

420045sin250400

o

o

Solution

Resultant Force

2.4 Addition of a System of Coplanar Forces

2.4 Addition of a System of Coplanar Forces

( ) ( )NNFR 8.2962.383 22 +−=

From vector addition,

Direction angle θ is

N485=

o8.37

2.383

8.296tan 1

=

= −

N

Nθ