1

Chapter 2

: (Concurrent Forces)Space Statics

Resultant and Equilibrium of a Particle

Prof. Imam Morgan

Head of MCTR Department

(2)

TBH TBG

FA

FD

P, TBG , TBH …. three Nonconcurrent

Forces (known)

P, Q1 ,Q2 …. three concurrent Known

Forces

F

Forces in Space (in 3D)

P, TBG , TBH and FA , FD five Nonconcurrent

[Resultant ? ??]

[Find F and T for Equilibrium]

or: Add F =? To get [Equilibrium] for

the four forces

[Resultant ]

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Head of MCTR Department

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Forces in Space ? …The forces are called in space as long as they

are not contained in one plane.

►The tensions in the

cables supporting the

platform are in space.

The platform is

considered as a particle.

►The tensions in the

cables supporting the

container are also

space forces. The

tensions intersect at

one point.

We study:

■ Resultant of

concurrent forces ■ Equilibrium of a particle.

Concurrent Forces

Lines of action of

forces intersect at

a point

Statics in Space [3D]? 1

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Head of MCTR Department

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►The plane forces exerted by the

four tugboats on the sub-marine

can be replaced by a single

equivalent force exerted by one

tugboat. (Resultant)

►Three or more space forces direct

the shipping boat along certain

direction. They could be replaced by

two boats for the same action.

(Resultant)

Non-concurrent

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Head of MCTR Department

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►Three space forces act

on a bracket. It is

required to replace them

by an equivalent effect

at O.

►The sign plate is kept in

Equilibrium in the shown

position by means of two

cables and a support at A

Non-concurrent We need to study:

■ Equivalent Systems

■ Equilibrium of

Rigid Body.

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Head of MCTR Department

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►► Resolve into horizontal and

vertical components:

F

Finally, has been resolved into three

rectangular components Fx , Fy , and Fz F

2 Force ( ) in Vector Form F

► Consider a force acting at the origin O of the orthogonal rectangular frame. The

angle defines the plane containing

while θy defines the position of in that plane.

F

F F

►►►Then, resolve Fh into rectangular components:

Fx = Fh cos = (F sin θy) cos

and Fz = (F sin θy))sin

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Head of MCTR Department

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θ θx

θz

kFjFiFF zyx

Therefore: Force in vector form

F

F

Where: are called Rectangular Components zyx F,F,F

θx , θy , θz are called the direction angles of

cos θx , cos θy , cos θz are the direction cosines

k,j,i are the unit vectors along x, y, and z, respectively

of

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Head of MCTR Department

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► In general, the space can be divided

into eight regions as shown in the figure.

The regions are generated using three

intersecting planes which are mutually

orthogonal. Their point of intersection is

the origin O of the shown chosen frame.

The location of any force, given in vector

form, can be easily specified in space.

►Three forces are shown in

figure. Although they have the same

magnitude (see later), however they are

completely different in their effects due to

the different signs of the components.

321 ,, FFF

kjiF ˆ3ˆ4ˆ61

kjiF ˆ3ˆ4ˆ62

kjiF ˆ3ˆ4ˆ63

4N

6N

4N

3N

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Head of MCTR Department

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θ θx

θz

Given: The force in vector form

kFjFiFF zyx

Required: ● its magnitude,

● its direction

►Magnitude: 222

zyx FFFF

►Direction:

F

F

F

F

F

F

zz

y

y

xx

cos

cos

cos

NF

kjiF

70206030

206030

222

For example: a force is given by:

◄

and the direction angles are:

7020

7060

7030

z

y

x

cos

cos

cos

θx = 64.6o

θy= 31o

θz = 73.4o

A Magnitude and Direction of ( ) F

Basic Relations

x

y

z

A

B

C

D

E

O Fx

Fy

Fz

F

x

y

z

A

B

C

D

E

O Fx

Fy

Fz

F

x

x

y

z

A

B

C

D

E

O Fx

Fy

Fz

F

y

z

The Three direction

angles can be

obtained as:

F

F

F

F

F

F

zz

yy

xx

cos

cos

cos

More Details

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Head of MCTR Department

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x

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Head of MCTR Department

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θθx

θz

θθx

θz

Unit Vector of a Force ( ) F

Given : The force in vector form

kFjFiFF zyx

Required : ● a unit vector ( ) directed

along its line of action.

F

In general the unit vector is obtained by

dividing the vector by its magnitude.

F

FF

For example, find the unit vector for the

vector given in the previous example

kji

kjiF

7

2

7

6

7

3

70

206030

Note that the components of the

unit vector are the direction cosines of the same vector

unit vector F

B

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Head of MCTR Department

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θ θx

θz

Given : the magnitude F, of a force, as

well as its direction (θx , θy , θz)

Required : F in vector form.

We have:

zz

yy

xx

cosFF

cosFF

cosFF

Direction Angles are dependent

The direction angles of the force vector

are dependent. They are related by:

1222 zyx coscoscos

Therefore, only two angles are

sufficient for determination of

the direction of any force in

space (fact).

D

Determination of ( ) on the bases

of its magnitude and direction

F

C

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Head of MCTR Department

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Example (1)

The magnitude of the shown force is

300 N. Write down this force in vector form

and hence find its rectangular components.

Solution:

From the figure θx= 60o and θy= 45o

1222 zyx coscoscos

50

2502

.cos

.cos

z

z

θz = 60o or 120o Consider θz= 120o

NcosFF

N.cosFF

NcosFF

zz

yy

xx

150

13212

150

kj.iF 15013212150

z

y

x

F

θz

60o

45o

refused

First Method

F1x F1h

F1y

F1z

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Head of MCTR Department

(14)

Example (2)

x

y

z

Express each force in vector form and

hence determine its direction with

respect to the coordinate frame.

k.j.i.F 363588636351

1F

Solution:

► Force

Second Method

The angles 60 and 45 defining the direction of

are not direction angles. Therefore, we must

use the geometry to resolve this force.

As shown: F1 is resolved into F1y and F1h. Then,

F1h is resolved into F1x and F1z:

F1y= 100 sin 60 = +86.6 lb

and F1h= 100 cos 60= 50 lb

F1x = 50 sin45 = 35.36 N

F1z = + 50 cos45= + 35.36 N

1F

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Head of MCTR Department

(15)

x

y

z

ozz

oyy

oxxDirection

3.69100

36.35cos

30100

8.86cos

111100

36.35cos

x

y

z

ozz

oyy

oxx

z

x

hy

direction

kjiF

IbF

IbF

IbFandIbF

FForce

3.69300

07.106cos

135300

13.212cos

2.52300

71.183cos:

ˆ07.106ˆ13.212ˆ71.183

07.10630sin13.212

71.18330cos13.212

13.21245cos30013.21245sin300

:

2

2

2

22

2

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Head of MCTR Department

(16)

In many applications, the forces are applied along certain defined directions.

● For example the tension in the chain AB is directed through the shown

direction where the two points A and B are well defined.

● Also, the force in the cable AB is directed as shown.

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Head of MCTR Department

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Position Vector

►The position vector ( ) is the vector that specifies the position

of point B with respect to point A. In other words, the position of

B as seen from point A. It is drawn from A to B.

ABr /

kzzjyyixxr

rrr

kzjyixrrandkzjyixrr

ABABABAB

ABAB

BBBOBBAAAOAA

ˆ)(ˆ)(ˆ)(

ˆˆˆˆˆˆ

/

/

//

Note that if a force is applied along AB and directed from A to B,

then, both will have the same unit vector. F

ABrandF /

z

y

x

ixx ABˆ

jyy ABˆ

kzz ABˆ

ABr /

O

z

y

x

ABr /

O

Prof. Imam Morgan

Head of MCTR Department

(18)

F

Very Important Force Defined by its Magnitude and

Two Points on its Line of Action

In this case, the force is known in

magnitude (F) and its line of action

passes through two given points, for

example M and N .

■ Find the unit vector from: F

212

2

12

2

12

12212

zzyyxx

kzzjyyixxF

■ Then, find the force in vector form as:

FFF

Third Method

The two steps

are carried out

in one step.

Prof. Imam Morgan

Head of MCTR Department

(19)

Example (3)

A tower guy wire is anchored by means of a

bolt at A. The tension in the wire is 2500 N.

Determine: (a) the rectangular components

the force acting on the bolt,

(b) the angles defining the

direction of the force.

Solution:

(a) Determination of F

kji

kjiF

79521201060

308040

3000804002500

222

Rectangular components

F

(40, 0, -30)

(0, 80, 0)

A = (40,0,-30)

B = (0,80,0)

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Head of MCTR Department

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(b) Direction angles at A:

o

zz

o

yy

o

xx

.cos

cos

.cos

5712500

795

322500

2120

11152500

1060

F

NF

kjiF

2500

ˆ795ˆ2120ˆ1060

F

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Head of MCTR Department

(21)

Example (4)

The tension in the shown cable is 700 N. Find the

cartesian components of the tension at A.

Determine its direction.

z

y

x

(2,3,-2)

(0,-3,1)

ozz

oyy

oxxDirection

NTCheck

kji

kjiT

4.115700

300cos

0.31700

600cos

4.73700

200cos:

700300600200

ˆ300ˆ600ˆ200

7

ˆ3ˆ6ˆ2700

222

θz=115.4o

θy=31.0o

θx=73.4o

z’

y’

x’

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Head of MCTR Department

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►When the particle is acted upon by several

forces, then, these forces can be replaced by

one force called the equivalent or the “Resultant”.

R

As shown, this resultant must pass through the point of intersection (Cocurrent Forces)

kFjFiFF

kFjFiFF

kFjFiFF

nznynxn

zyx

zyx

2222

1111

kRjRiRR zyx

zz

yy

xx

FR

FR

FR

Resultant of Concurrent Forces 3

. . . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . . . .

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Head of MCTR Department

(23)

Example (5 )

A wall section of precast concrete is

temporarily held by the cables shown.

Knowing that the tension in cables AB and

AC are 840 lb and 1200 lb, respectively,

determine the magnitude and direction of the

resultant of the forces at A.

Solution

Choose the coordinate frame as shown and

then write each force in vector form. 2T

(16,0,-11)

(0,8,0)

(0,8,-27)

kji

kji

kjiT

ˆ440ˆ320ˆ640

21

ˆ11ˆ8ˆ16840

11816

ˆ11ˆ8ˆ16840

2221

1T

1T

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Head of MCTR Department

(24)

kji

kji

kjiT

ˆ800ˆ400ˆ800

24

ˆ11ˆ8ˆ161200

16816

ˆ16ˆ8ˆ161200

2222

2T

So, the resultant which must pass through A is:

kji

TTR

ˆ360ˆ720ˆ1440

21

● R = 1650 lb

● θx = 150.8o ● θy = 64.1o ● θz = 102.6o Draw a sketch to show the direction of

the resultant and try to find its point of

intersection, D, with the wall (Option)

Answer: D ≡ (0, 8, - 15) ft

2T

(16,0,-11)

(0,8,-27)

and we have kjiT ˆ440ˆ320ˆ6401

1T

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Head of MCTR Department

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Consider a particle A under the

action of several space forces: F1 , F2 , …… , Fn.

■ Find the resultant R, then:

►the particle will move along

a space curve.

►the particle will move along a

curvilinear pass contained in xy plane

►the particle will perform a

rectilinear motion along x direction. ■ However, if

0R

The particle

is in case of

equilibrium 0

0

0

zz

yy

xx

FR

FR

FR

Conditions of

Equilibrium

Fn F3

F2 F1

A

kRjRiRRisRIf zyxˆˆˆ

0ˆˆ jRiRRIf yx

00ˆ iRRIf x

Equilibrium of Concurrent Forces 4

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Head of MCTR Department

(26)

Example (6)

A 200 kg cylinder is hung by means of

two cables AB and AC. A horizontal force

P holds the cylinder in the shown

position. Determine the magnitude of P

and the tension in each cable.

2

1

(1.2,2,0)

(0,12,8)

(0,12,-10)

Solution:

Write down each force in vector form:

kTjTiT

kjiTT

kTjTiT

kjiTT

jmgW

iPP

ˆ)705.0(ˆ)705.0(ˆ)085.0(

13.14

ˆ10ˆ10ˆ2.1

ˆ)622.0(ˆ)778.0(ˆ)093.0(

86.12

ˆ8ˆ10ˆ2.1

0ˆ0

00ˆ

222

22

111

11

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Head of MCTR Department

(27)

Then, apply the conditions of equilibrium:

0085.0093.00 21 TTPFx

0705.0778.019620 21 TTFy

0705.0622.00 21 TTFz

kTjTiT

kjiTT

kTjTiT

kjiTT

jmgW

iPP

ˆ)705.0(ˆ)705.0(ˆ)085.0(

13.14

ˆ10ˆ10ˆ2.1

ˆ)622.0(ˆ)778.0(ˆ)093.0(

86.12

ˆ8ˆ10ˆ2.1

0ˆ0

00ˆ

222

22

111

11

2

1

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Head of MCTR Department

(28)

0085.0093.00 21 TTPFx

0705.0778.019620 21 TTFy

0705.0622.00 21 TTFz

T1 = 1402 N

T2 = 1238 N

P = 235 N

Think over!!!

● Determine the location of point C such that the

two tensions will have the same value and find

the tension in this case (distance 1.2 is kept constant).

● If the distance 1.2 is required to be doubled,

find P, T1, and T2 in this case (same cables lengths).

Comment on theobtained results.

● If the two points B and C coincide at one point,

determine the location of this point such that

the particle keeps its equilibrium in the shown

position (consider variable lengths of cables).

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Head of MCTR Department

(29)

Example (7)

The 100 kg cylinder is suspended from

the ceiling by cables attached at points

B, C, and D. What are the tensions in

the cables AB, AC, and AD?

Prof. Imam Morgan

Head of MCTR Department

(30)

Solution:

TAD

TAC TAB

W

NT

NT

NT

getweequationstheseSolving

TTT

TTT

TTT

mEquilibriuofConditionsFrom

jjW

kjiTkji

TT

kjiTkji

TT

kjoiTkji

TT

AB

AB

AB

ADACAB

ADACAB

ADACAB

ADABAD

ACACAC

ABABAB

7.168

1.636

1.519

:,

0514.0408.0333.0

981868.0816.0667.0

0514.0408.0667.0

0ˆ9810ˆ81.9100

ˆ514.0ˆ686.0ˆ514.034

34ˆ3

ˆ408.0ˆ816.0ˆ408.024

2ˆ4ˆ2

ˆ333.0ˆ667.ˆ667.036

2ˆ4ˆ4