# 2.0 Statics of Particles

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29 Νοε 2013 (πριν από 4 χρόνια και 7 μήνες)

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IZMIR INSTITUTE OF TECHNOLOGY

Department of Architecture

AR23
1

Fall12/13

29.11.2013

Dr. Engin Aktaş

1

2.0 Statics of Particles

IZMIR INSTITUTE OF TECHNOLOGY

Department of Architecture

AR23
1

Fall12/13

29.11.2013

Dr. Engin Aktaş

2

Forces on a Particle

30
o

Line of Action

Point of Application

A

direction

magnitude

Since the vector has a well defined point of application, it is a
fixed vector, therefore can not be moved without modifications

IZMIR INSTITUTE OF TECHNOLOGY

Department of Architecture

AR23
1

Fall12/13

29.11.2013

Dr. Engin Aktaş

3

Q

P

P+Q=Q+P

Commutative

A

P

Q

Q

P

A

A

Associative

A

P

Q

S

A

Q

P+Q+S=(P+Q)+S=P+(Q+S)

P

IZMIR INSTITUTE OF TECHNOLOGY

Department of Architecture

AR23
1

Fall12/13

29.11.2013

Dr. Engin Aktaş

4

Resultant of several concurrent
forces

Q

P

S

Using polygon rule

A

A

P

Q

S

IZMIR INSTITUTE OF TECHNOLOGY

Department of Architecture

AR23
1

Fall12/13

29.11.2013

Dr. Engin Aktaş

5

Resolution of a force into
components

A

F

P

Q

A

F

P

Q

A

F

Q

P

IZMIR INSTITUTE OF TECHNOLOGY

Department of Architecture

AR23
1

Fall12/13

29.11.2013

Dr. Engin Aktaş

6

A

P

Q

R

a

Example
(Beer & Johnston)

R
B
Q
A
sin
sin

25
o

20
o

Q
=60 N

P
=40 N

A

The two forces P and Q act on a bolt A. Determine
their resultant.

20
o

25
o

180
-
25=155
o

R
2
=
P
2
+
Q
2
-
2
PQ
cos
B

B

R
2
=(40N)
2
+(60N)
2
-
2(40N)(60N)cos155
0

R=97.73 N

Law of cosines

Law of sines

N
N
R
B
Q
A
73
.
97
155
sin
60
sin
sin

IZMIR INSTITUTE OF TECHNOLOGY

Department of Architecture

AR23
1

Fall12/13

29.11.2013

Dr. Engin Aktaş

7

Example (cntd.)

R
B
Q
A
sin
sin

Law of sines

N
N
R
B
Q
A
73
.
97
155
sin
60
sin
sin

A=15.04
o

a
=20
o
+A=35.04
o

R=97.7 kN

a
=35.0
o

IZMIR INSTITUTE OF TECHNOLOGY

Department of Architecture

AR23
1

Fall12/13

29.11.2013

Dr. Engin Aktaş

8

Example
(Beer and Johnston)

105
sin
25
30
sin
45
sin
2
1
kN
T
T
B

A

C

1

2

30
o

45
o

A barge is pulled by two tugboats. If
the resultant of the forces exerted
by t
ugboats

is a 25 kN force
directed along the axis of barge,
determine the tension in each of the
ropes.

R=25 kN

45
o

30
o

T
2

T
1

105
o

Using law of sines

kN
kN
T
30
.
18
105
sin
45
sin
25
1

kN
kN
T
94
.
12
105
sin
30
sin
25
2

IZMIR INSTITUTE OF TECHNOLOGY

Department of Architecture

AR23
1

Fall12/13

29.11.2013

Dr. Engin Aktaş

9

Rectangular Components

F

x

y

q

F
x

F
y

IZMIR INSTITUTE OF TECHNOLOGY

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Fall12/13

29.11.2013

Dr. Engin Aktaş

10

F

y

q

IZMIR INSTITUTE OF TECHNOLOGY

Department of Architecture

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Fall12/13

29.11.2013

Dr. Engin Aktaş

11

Unit Vectors

x

y

i

j

Magnitude=1

F

F
x
=
F
x
i

F
y
=
F
y
j

IZMIR INSTITUTE OF TECHNOLOGY

Department of Architecture

AR23
1

Fall12/13

29.11.2013

Dr. Engin Aktaş

12

Example
(Beer and Johnston)

x
y
x
y
F
F
F
F
1
tan
tan

q
q

0
.
65
5
.
3
5
.
7
tan
1
kN
kN
q
F
=(3.5 kN)
i
+(7.5kN)
j

A

q

Determine the magnitude of the force and angle
q
.

A

x

y

F
x
=3.5 kN

F
y
=7.5 kN

F

q

kN
kN
F
F
y
28
.
8
0
.
65
sin
5
.
7
sin

q
IZMIR INSTITUTE OF TECHNOLOGY

Department of Architecture

AR23
1

Fall12/13

29.11.2013

Dr. Engin Aktaş

13

Addition of Forces by Summing x and y
components

A

x

y

F
1
=150 N

30
o

F
2
=80 N

20
o

F
3
=110 N

F
4
=100 N

15
o

Determine the resultant of the
forces on the bolt

F
2

cos 20
o

-
(F
2

sin

2
0
o
)

-
F
3

F
1

sin 30
o

F
1

cos 30
o

F
4

cos 15
o

-
(
F
4

sin

15
o
)

Forces

Magnitude, N

x Component, N

y Component, N

F
1

150

+129.9

+75.0

F
2

80

-
27.4

+75.2

F
3

110

0

-
110.0

F
4

100

+96.6

-
25.9

R
x
=199.1

R
y
=+14.3

IZMIR INSTITUTE OF TECHNOLOGY

Department of Architecture

AR23
1

Fall12/13

29.11.2013

Dr. Engin Aktaş

14

Example

(cntd)

R
=
R
x
i
+
R
y
j

R
=(199.1N)
i
+(14.30N)
j

R

a

R
x
=199.1 N

R
y
=14.30 N

10
.
4
1
.
199
30
.
14
tan
tan
1
1
N
N
R
R
x
y
a
IZMIR INSTITUTE OF TECHNOLOGY

Department of Architecture

AR23
1

Fall12/13

29.11.2013

Dr. Engin Aktaş

15

Equilibrium of a Particle

When the resultant of all the forces acting on a particle is
zero, the particle is in equilibrium.

x

y

A

F
1
=300 N

F
2
=173.2 N

F
3
=200 N

F
4
=
4
00 N

F
1
=300 N

F
2
=173.2 N

F
3
=200 N

F
4
=
4
00 N

R=
S
F=0

(
S
F
x
)
i
+ (
S
F
y
)
j
=0

S
F
x
=0

S
F
y
=0

S
F
x
=300 N
-
(200 N) sin30
o
-
(400 N) sin 30
o
=300 N
-
100 N
-
200 N = 0

S
F
y
=
-
173.2 N
-
(200 N) cos30
o
+ (400 N) cos 30
o
=
-
173.2 N
-
173.2 N
+
346.4 N = 0

30
o

30
o

IZMIR INSTITUTE OF TECHNOLOGY

Department of Architecture

AR23
1

Fall12/13

29.11.2013

Dr. Engin Aktaş

16

Free
-
Body Diagram

A sketch showing the physical conditions of the problem is known as
space
diagram
.

Choose a significant particle and draw a separate diagram showing that
particle and forces on it. This is the
Free
-
Body diagram
.

A

B

C

50
o

30
o

Space Diagram

T
AC

30
o

50
o

T
AB

736 N

60
o

80
o

40
o

736 N

Free
-
Body Diagram

T
AC

T
AB

Using law of sines

80
sin
736
40
sin
60
sin
AC
AB
T
T
T
AB
=647 N

T
AC
=
4
80 N

IZMIR INSTITUTE OF TECHNOLOGY

Department of Architecture

AR23
1

Fall12/13

29.11.2013

Dr. Engin Aktaş

17

Analytical Solution

Two unknowns
T
AB

and
T
AC

Equilibrium Equations

S
F
x
=0 and
S
F
y
=0

The system of equations then

0
30
cos
50
cos
0

AC
AB
x
T
T
F
0
736
30
sin
50
sin
0

N
T
T
F
AC
AB
y
Solving

the above system
(2 unknowns 2 equations)

T
AB
=647 N

T
AC
=480 N

IZMIR INSTITUTE OF TECHNOLOGY

Department of Architecture

AR23
1

Fall12/13

29.11.2013

Dr. Engin Aktaş

18

Su
mary

of Solution techniques

equilibrium under three forces

may use
force triangle

rule

equilibrium under more than three forces

may use
force polygon rule

If analytical solution is desired

may use
equations of equilibrium

IZMIR INSTITUTE OF TECHNOLOGY

Department of Architecture

AR23
1

Fall12/13

29.11.2013

Dr. Engin Aktaş

19

Example

(Beer and Johnston)

E

B

C

A

60
o

20
o

Flow

As

a

part

of

the

design

of

a

new

sailboat,

it

is

desired

to

determine

the

drag

force

which

may

be

expected

at

a

given

speed
.

To

do

so,

a

model

of

the

proposed

hull

is

placed

in

a

test

channel

and

three

cables

are

used

to

keep

its

bow

on

the

centerline

of

the

channel
.

Dynamometer

indicate

that

for

a

given

speed,

the

tension

is

200

N

in

cable

AB

and

300

N

in

cable

AE
.

Determine

the

drag

force

exerted

on

the

hull

and

the

tension

in

cable

AC
.

Unknowns;
F
D
,
T
AC

Free
-
Body Diagram

T
AE
=300 N

T
AB
=200 N

F
D

T
AC

60
o

20
o

Equilibrium condition

Resultant of the all forces should be zero

R
=
T
AB
+
T
AC
+
T
AE
+
F
D
=
0

Let’s write all the forces in
x

and
y

components

T
AB
=
-
(200 N) sin 60
o
i
+(200 N) cos 60
o

j

T
AB
=
-
173.2
i
+100
j

T
AC
= (
T
AC
) sin 20
o
i
+(
T
AC
) cos 20
o

j

T
AC
= 0.342(
T
AC
)
i
+0.9397(
T
AC
)
j

T
AE
=
-
(300 N)
j

F
D
=
F
D

i

IZMIR INSTITUTE OF TECHNOLOGY

Department of Architecture

AR23
1

Fall12/13

29.11.2013

Dr. Engin Aktaş

20

Example

(cntd.)

R
=
T
AB
+
T
AC
+
T
AE
+
F
D
=
0

R
=(
-
173.2 N)
i
+ 100 N
j
+ 0.342(
T
AC
)
i
+ 0.9397 (
T
AC
)
j

+ (
-
300 N)
j

+
F
D

i
=
0

R
=
{
-
173.2 + 0.342(
T
AC
) +
F
D
}

i
+
{
100 + 0.9397 (
T
AC
) + (
-
300 N)
}

j
=

0

-
173.2 N + 0.342(
T
AC
) +
F
D

=

0

S
F
x
=0

S
F
y
=0

100 + 0.9397 (
T
AC
) + (
-
300 N)

=

0

T
AC
=213 N

F
D
=100 N

IZMIR INSTITUTE OF TECHNOLOGY

Department of Architecture

AR23
1

Fall12/13

29.11.2013

Dr. Engin Aktaş

21

Forces
in
Space

x

y

z

V

V
z
k

k

j

i

V
x
i

V
y
j

q
x

q
y

q
z

V
=
V
x
i
+
V
y
j
+
V
z
k

V
x
=
V
cos
q
x

V
y
=
V
cos

q
y

V
z
=
V

cos

q
z

IZMIR INSTITUTE OF TECHNOLOGY

Department of Architecture

AR23
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Fall12/13

29.11.2013

Dr. Engin Aktaş

22

Direction cosines

l
= cos
q
x

m

= cos
q
y

n

= cos
q
z

V
x
=
l V

V
y
=
mV

V
z
=
nV

V
2
=
V
x
2
+
V
y
2
+
V
z
2

l
2
+
m
2
+
n
2
=1

IZMIR INSTITUTE OF TECHNOLOGY

Department of Architecture

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Fall12/13

29.11.2013

Dr. Engin Aktaş

23

x

y

z

B
(
x
B
, y
B
, z
B
)

A
(
x
A
, y
A
, z
A
)

V

V=

(
x
B
-
x
A
)

i

+

(
y
B
-
y
A
)

j

+

(
z
B
-
z
A
)

k

n

V
V
n

Unit vector along AB

IZMIR INSTITUTE OF TECHNOLOGY

Department of Architecture

AR23
1

Fall12/13

29.11.2013

Dr. Engin Aktaş

24

Rectangular Coordinates in Space

q
y

f

B

A

C

O

F

F
y

F
h

y

x

z

f

B

C

O

F
x

F
y

y

x

z

F
h

F
z

F
y
=
F
cos
q
y

F
h
=
F
sin
q
y

F
x

= F
h
cos
f
=
F
sin
q
y
cos
f

F
z

= F
h
sin
f
=
F
sin
q
y
sin
f

E

D

F
2
=
F
y
2
+
F
h
2

F
h
2
=
F
x
2
+
F
z
2

2
2
2
z
y
x
F
F
F
F

IZMIR INSTITUTE OF TECHNOLOGY

Department of Architecture

AR23
1

Fall12/13

29.11.2013

Dr. Engin Aktaş

25

Problems

IZMIR INSTITUTE OF TECHNOLOGY

Department of Architecture

AR23
1

Fall12/13

29.11.2013

Dr. Engin Aktaş

26

Problem 1
-
(Meriam and Kraige)

A

4

3

F
=1800 N

x

y

z

The
1800 N

force
F

is applied at the end of the
I
-
beam. Express
F

as a vector using the unit
vectors
i

and
j
.

First let’s find the
x

and
y

components of the force
F
.

F
x
=
-
1800 N 3/5 =
-
1080 N

F
y
=
-
1800 N 4/5 =
-
1440 N

F

=
F
x
i+
F
y
j

=
-
1080
i
-
1440
j

N

IZMIR INSTITUTE OF TECHNOLOGY

Department of Architecture

AR23
1

Fall12/13

29.11.2013

Dr. Engin Aktaş

27

Problem 2
-
(Meriam and Kraige)

The

ratio

of

the

lift

force

L

to

the

drag

force

D

for

the

simple

airfoil

is

L
/
D

=

10
.

If

the

lift

force

on

a

short

section

of

the

airfoil

is

50

N
,

compute

the

magnitude

of

the

resultant

force

R

and

the

angle

q

which

it

makes

with

the

horizontal
.

C

D
=
5 N

L
=
50 N

Air flow

D

L

C

q

tan
q

= 50/5

q
= tan
-
1
10 =84.3
o

F
= (50
2
+5
2
)
0.50
= 50.2 N

IZMIR INSTITUTE OF TECHNOLOGY

Department of Architecture

AR23
1

Fall12/13

29.11.2013

Dr. Engin Aktaş

28

Problem 3
-
(Meriam and Kraige)

The

gusset

plate

is

subjected

to

the

two

forces

shown
.

Replace

them

by

two

equivalent

force,

F
x

in

the

x
-
direction

and

F
a

in

the

a

direction
.

Determine

the

magnitudes

of

F
x

and

F
a
.

y

x

900 N

800 N

10
o

45
o

25
o

A

a

A

800 N

900 N

R

10
o

25
o

10
o

65
o

From the law of cosines

R
2
=800
2
+900
2
-
2(800)(900) cos75

R
=1040 N

a

From the law of sines

7
.
56
1040
75
sin
900
sin
75
sin
900
sin
sin
900
75
sin
1
1
R
R
a
a
66.7
o

45
o

45
o

113.3
o

A

F
x

F
a

21.7
o

1040 N

N
F
F
x
x
544
7
.
21
sin
45
sin
1040

N
F
F
a
a
1351
3
.
113
sin
45
sin
1040

IZMIR INSTITUTE OF TECHNOLOGY

Department of Architecture

AR23
1

Fall12/13

29.11.2013

Dr. Engin Aktaş

29

Problem 4
-
(Meriam and Kraige)

B

A

C

50

m

40

m

40

m

The

guy

cables

AB

and

AC

are

attached

to

the

top

of

the

transmission

tower
.

The

tension

in

cable

AC

is

8

kN
.

Determine

the

required

tension

T

in

cable

AB

such

that

the

net

effect

of

the

two

cable

tensions

is

a

downward

force

at

point

A
.

Determine

the

magnitude

R

of

this

downward

force

R

8 kN

This

time

let’s

use

another

approach
.

Since

the

resultant

force

is

downward

the

sum

of

horizontal

components

of

the

two

forces

should

up

to

zero
.

N
T
T
AB
AB
68
.
5
0
1
.
72
40
8
64
50

The magnitude of
R

N
R
21
.
10
1
.
72
60
8
64
40
68
.
5

IZMIR INSTITUTE OF TECHNOLOGY

Department of Architecture

AR23
1

Fall12/13

29.11.2013

Dr. Engin Aktaş

30

Problem 5
-
(Beer and Johnston)

Two

cables

are

tied

together

at

C

and

as

shown
.

Determine

the

tension

in

AC

and

BC
.

1600 kg

500 mm

1375 mm

1200 mm

A

C

B

Let’s draw the Free Body Diagram (FBD) at C

C

W=1600*9.81=15700N

y

x

T
BC

T
AC

Writing Equilibrium Equations

S
F
x
=0

S
F
y
=0

0
1825
1375
1300
500

CB
AC
T
T
0
15700
1825
1200
1300
1200

CB
AC
T
T
T
AC
=12470 N

T
CB
=6370 N

IZMIR INSTITUTE OF TECHNOLOGY

Department of Architecture

AR23
1

Fall12/13

29.11.2013

Dr. Engin Aktaş

31

Problem 6
-
(Beer and Johnston)

A

precast
-
concrete

wall

section

is

temporarily

held

by

the

cables

shown
.

Knowing

that

tension

is

4200

N

in

cable

AB

and

6000

N

in

cable

AC
,

determine

the

magnitude

and

direction

of

the

resultant

of

the

forces

exerted

by

cables

AB

and

AC

on

stake

A
.

D

A

B

C

5.5 m

8 m

4 m

x

y

z

Coordinates of points A, B and C

A

(
8
,

-
4
,

-
5
.
5
)

B

(
0
,

0
,

0
)

C

(
0
,

0
,

-
13
.
5
)

AB
=(
0
-
8
)

i

+

(
0
-
(
-
4
))

j

+

(
0
-
(
-
5
.
5
))

k

=

-
8

i

+

4

j

+

5
.
5

k

50
.
10
5
.
5
4
8
2
2
2

AB
k
j
i
k
j
i
n
AB
524
.
0
381
.
0
762
.
0
50
.
10
5
.
5
4
8

N
k
j
i
F
AB
524
.
0
381
.
0
762
.
0
4200

AC
=(
0
-
8
)

i

+

(
0
-
(
-
4
))

j

+

(
-
13
.
5
-
(
-
5
.
5
))

k

=

-
8

i

+

4

j

-

8

k

00
.
12
8
4
8
2
2
2

AC
k
j
i
k
j
i
n
AC
667
.
0
333
.
0
667
.
0
00
.
12
8
4
8

N
k
j
i
F
AC
667
.
0
333
.
0
667
.
0
6000

IZMIR INSTITUTE OF TECHNOLOGY

Department of Architecture

AR23
1

Fall12/13

29.11.2013

Dr. Engin Aktaş

32

Problem 6
-
(Beer and Johnston)

A

precast
-
concrete

wall

section

is

temporarily

held

by

the

cables

shown
.

Knowing

that

tension

is

4200

N

in

cable

AB

and

6000

N

in

cable

AC
,

determine

the

magnitude

and

direction

of

the

resultant

of

the

forces

exerted

by

cables

AB

and

AC

on

stake

A
.

D

A

B

C

5.5 m

8 m

4 m

x

y

z

N
k
j
i
F
AB
524
.
0
381
.
0
762
.
0
4200

N
k
j
i
F
AC
667
.
0
333
.
0
667
.
0
6000

R

=
F
AB

+
F
AC

R

=(
-
3200
-
4002
)

i

+(
1600
+
1998
)

j

+

(
2201
-
4002
)

k

=

-
7200

i

+

3600

j

-

1800

k

x

y

z

R
= 8250 N

Direction cosines

l

=

0
.
873

m

=

0
.
436

n

=

0
.
218

IZMIR INSTITUTE OF TECHNOLOGY

Department of Architecture

AR23
1

Fall12/13

29.11.2013

Dr. Engin Aktaş

33

Problem 7
-
(Beer and Johnston)

A

crate

is

supported

by

three

cables

as

shown
.

Determine

the

weight

W

of

the

crate

knowing

that

the

tension

in

cable

AB

is

4620

N
.

y

x

z

B

A

O

C

D

700 mm

600 mm

1125
mm

650 mm

450 mm

Coordinates of points A, B, C and D

A

(
0
,

-
1125
,

0
)

Let’s draw FBD first

W

F
AB

F
AC

F

Unknowns :

F
,
F
AC

and
W

B

(
700
,

0
,

0
)

C

(
0
,

0
,

-
600
)

D

(
-
650
,

0
,
450
)

AB
=((
700
-
0
)
i
+(
0
-
(
-
1125
))
j
+
0
k
=
700
i
+
1125
j
+
0
k

AB
=(
700
2
+
1125
2
)
0
.
5
=
1325

n
AB
=(
700
i
+
1125
j
+
0
k
)/
1325
=
0
.
5283
i
+
0
.
8491
j

F
AB
=
4620
(
0
.
5283
i
+
0
.
8491
j
)

N

IZMIR INSTITUTE OF TECHNOLOGY

Department of Architecture

AR23
1

Fall12/13

29.11.2013

Dr. Engin Aktaş

34

Problem 7
-
(Beer and Johnston)

y

x

z

B

A

O

C

D

700 mm

600 mm

1125
mm

650 mm

450 mm

Coordinates of points A, B, C and D

A

(
0
,

-
1125
,

0
)

B

(
700
,

0
,

0
)

C

(
0
,

0
,

-
600
)

D

(
-
650
,

0
,
450
)

AC
=(
0
i
+(
0
-
(
-
1125
))
j
+(
-
600
-
0
)
k
=
0
i
+
1125
j
-
600
k

AC
=(
600
2
+
1125
2
)
0
.
5
=
1275

n
AC
=(
0
i
+
1125
j
-
600
k
)/
1275
=
0
.
8824
j
-
0
.
4706
k

F
AC
=
F
AC
(
0
.
8824
j
-
0
.
4706
k
)

N

=((
-
650
-
0
)
i
+(
0
-
(
-
1125
))
j
+(
450
-
0
)
k
=
-
650
i
+
1125
j
+
450
k

=(
650
2
+
1125
2

+
450
2
)
0
.
5
=
1375

n
=(
-
650
i
+
1125
j
+
450
k
)/
1375
=

-
0
.
4727
i
+
0
.
8182
j
+
0
.
3273
k

F
=
F
(
-
0
.
4727
i
+
0
.
8182
j
+
0
.
3273
k
)

N

W
=
-
W

j

IZMIR INSTITUTE OF TECHNOLOGY

Department of Architecture

AR23
1

Fall12/13

29.11.2013

Dr. Engin Aktaş

35

Problem 7
-
(Beer and Johnston)

y

x

z

B

A

O

C

D

700 mm

600 mm

1125
mm

650 mm

450 mm

F
AB
+
F
AC
+
F
+
W
=0

4620
(
0
.
5283
i
+
0
.
8491
j
)

+
F
AC
(
0
.
8824
j
-
0
.
4706
k
)+
F
(
-
0
.
4727
i
+
0
.
8182
j
+
0
.
3273
k
)
-
W

j
=
0

2441
-
0.4727
F
=0

F
=

5160

N

-
0.4706
F
AC
+0.3273
*5160
=0

F
AC
=

3590

N

3922+3170+4225
-
W
= 0

W
=

11320

N

IZMIR INSTITUTE OF TECHNOLOGY

Department of Architecture

AR23
1

Fall12/13

29.11.2013

Dr. Engin Aktaş

36

Problem 8
-
(Beer and Johnston)

A

16

kg

triangular

plate

is

supported

by

three

wires

as

shown
.

Determine

the

tension

in

each

wire

.

y

x

z

600 mm

200 mm

400 mm

200 mm

200 mm

D

B

C

A