IZMIR INSTITUTE OF TECHNOLOGY
Department of Architecture
AR231 Fall12/13
2.0 Statics of Particles
15.3.2016
Dr. Engin Aktaş
1
IZMIR INSTITUTE OF TECHNOLOGY
Department of Architecture
AR231 Fall12/13
Forces on a Particle
Line of Action
magnitude
Point of Application
A
30o
direction
Since the vector has a well defined point of application, it is a
fixed vector, therefore can not be moved without modifications
15.3.2016
Dr. Engin Aktaş
2
IZMIR INSTITUTE OF TECHNOLOGY
Department of Architecture
AR231 Fall12/13
Properties of Vector Addition
P
Commutative
P+Q=Q+P
Q
P
A
A
Q
P
P
A
A
Associative
Q
P
S
Q
Q
A
P+Q+S=(P+Q)+S=P+(Q+S)
15.3.2016
Dr. Engin Aktaş
3
IZMIR INSTITUTE OF TECHNOLOGY
Department of Architecture
AR231 Fall12/13
Resultant of several concurrent
forces
P
Using polygon rule
P
A
Q
S
Q
A
S
15.3.2016
Dr. Engin Aktaş
4
IZMIR INSTITUTE OF TECHNOLOGY
Department of Architecture
AR231 Fall12/13
Resolution of a force into
components
Q
Q
F
F
A
A
P
P
F
Q
A
P
15.3.2016
Dr. Engin Aktaş
5
IZMIR INSTITUTE OF TECHNOLOGY
Q=60 N
o
25
A
Department of Architecture
AR231 Fall12/13
Example (Beer & Johnston)
The two forces P and Q act on a bolt A. Determine
P=40 N their resultant.
20o
180-25=155o
Q
25o
R
a
o
A
Law of cosines
Law of sines
15.3.2016
20
P
B
R2=P2+Q2-2PQcosB
R2=(40N)2+(60N)2-2(40N)(60N)cos1550
sin A sin B

Q
R
R=97.73 N
Q sin B 60N sin 155
sin A 

R
97.73N
Dr. Engin Aktaş
6
IZMIR INSTITUTE OF TECHNOLOGY
Department of Architecture
AR231 Fall12/13
Example (cntd.)
Law of sines
A=15.04o
The answer is
15.3.2016
Q sin B 60N sin 155
sin A 

R
97.73N
sin A sin B

Q
R
a=20o+A=35.04o
R=97.7 kN
a=35.0o
Dr. Engin Aktaş
7
IZMIR INSTITUTE OF TECHNOLOGY
Department of Architecture
AR231 Fall12/13
Example (Beer and Johnston)
A
A barge is pulled by two tugboats. If
the resultant of the forces exerted
by tugboats is a 25 kN force
directed along the axis of barge,
determine the tension in each of the
ropes.
1
B
30o
45o
2
C
Using law of sines
R=25 kN
o
30
45
105o
T1
o
T2
T1
T2
25kN


sin 45 sin 30 sin 105
25kN sin 45
T1 
 18.30 kN
sin 105
25kN sin 30
T2 
 12.94 kN
sin 105
15.3.2016
Dr. Engin Aktaş
8
IZMIR INSTITUTE OF TECHNOLOGY
Department of Architecture
AR231 Fall12/13
Rectangular Components
y
F
Fy
q
Fx
15.3.2016
Dr. Engin Aktaş
x
9
IZMIR INSTITUTE OF TECHNOLOGY
Department of Architecture
AR231 Fall12/13
y
F
q
15.3.2016
Dr. Engin Aktaş
10
IZMIR INSTITUTE OF TECHNOLOGY
Department of Architecture
AR231 Fall12/13
Unit Vectors
y
Magnitude=1
Fy=Fyj
F
j
i
15.3.2016
Fx=Fxi
Dr. Engin Aktaş
x
11
IZMIR INSTITUTE OF TECHNOLOGY
Department of Architecture
AR231 Fall12/13
Example (Beer and Johnston)
F=(3.5 kN)i+(7.5kN)j
Determine the magnitude of the force and angle q.
q
A
tan q 
Fy=7.5 kN
y
A
15.3.2016
Fx
 q  tan
1
Fy
Fx
7.5kN
q  tan
 65.0
3.5kN
1
F
q
Fx=3.5 kN
Fy
x
Fy
7.5kN
F

 8.28kN
sin q sin 65.0
Dr. Engin Aktaş
12
IZMIR INSTITUTE OF TECHNOLOGY
Department of Architecture
AR231 Fall12/13
Additiony of Forces by Summing x and y
components
20
F2=80 N
A
o
Determine the resultant of the
forces on the bolt
F1=150 N
30o
F2 cos 20o
F1 sin 30o
x
15o
F4=100 N
-(F2 sin 20o)
F3=110 N
-(F4 sin 15o)
Forces Magnitude, N x Component, N
F1
150
+129.9
-27.4
F2
80
0
110
F3
+96.6
F4
100
Rx=199.1
15.3.2016
F1 cos 30o
F4 cos 15o
y Component, N
+75.0
+75.2
-110.0
-25.9
Ry=+14.3
Dr. Engin Aktaş
-F3
13
IZMIR INSTITUTE OF TECHNOLOGY
Department of Architecture
Example
AR231 Fall12/13
(cntd)
R=Rxi+Ryj
R=(199.1N)i+(14.30N)j
Ry=14.30 N
R
a
Rx=199.1 N
a  tan
15.3.2016
1
Ry
14.30 N
 tan
 4.10
Rx
199.1N
1
Dr. Engin Aktaş
14
IZMIR INSTITUTE OF TECHNOLOGY
Department of Architecture
AR231 Fall12/13
Equilibrium of a Particle
When the resultant of all the forces acting on a particle is
zero, the particle is in equilibrium.
y
F1=300 N
F4=400 N
30o
F2=173.2 N
A
F3=200 N
F1=300 N
F2=173.2 N
30o
x
F4=400 N
F3=200 N
R=SF=0
(SFx)i+ (SFy)j=0
SFx=0
SFy=0
SFx=300 N -(200 N) sin30o-(400 N) sin 30o=300 N -100 N -200 N = 0
SFy=-173.2 N -(200 N) cos30o+ (400 N) cos 30o=-173.2 N -173.2 N +346.4 N = 0
15.3.2016
Dr. Engin Aktaş
15
IZMIR INSTITUTE OF TECHNOLOGY
Department of Architecture
AR231 Fall12/13
Free-Body
Diagram
A sketch showing the physical conditions of the problem is known as space
diagram.
B
50o
A
C
30o
Space Diagram
Choose a significant particle and draw a separate diagram showing that
particle and forces on it. This is the Free-Body diagram.
TAB
50o
Using law of sines
TAC
o
30
736 N
736 N
40o
60
TAB=647 N
TAC
Dr. Engin Aktaş
TAC
TAB
736


sin 60 sin 40 sin 80
80o
o
Free-Body Diagram
15.3.2016
TAB
TAC=480 N
16
IZMIR INSTITUTE OF TECHNOLOGY
Department of Architecture
AR231 Fall12/13
Analytical Solution
• Two unknowns TAB and TAC
• Equilibrium Equations
SFx=0 and SFy=0
The system of equations then
 Fx  0  TAB cos 50  TAC cos 30  0
 Fy  0  TAB sin 50  TAC sin 30  736 N  0
Solving the above system (2 unknowns 2 equations)
TAB=647 N
TAC=480 N
15.3.2016
Dr. Engin Aktaş
17
IZMIR INSTITUTE OF TECHNOLOGY
Department of Architecture
AR231 Fall12/13
Sumary 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
15.3.2016
Dr. Engin Aktaş
18
IZMIR INSTITUTE OF TECHNOLOGY
Department of Architecture
Example
AR231 Fall12/13
(Beer and Johnston)
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
20o
60o
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
A
Flow
the centerline of the channel. Dynamometer reading
indicate that for a given speed, the tension is 200 N in cable
AB and 300 N in cable AE. Determine the drag force
E
exerted on the hull and the tension in cable AC.
Equilibrium condition
Start with drawing Free-Body Diagram
Resultant of the all forces should be zero
R=TAB+TAC+TAE+FD=0
20o
TAC
Let’s write all the forces in x and y components
TAB=200 N 60o
TAB=-(200 N) sin 60o i+(200 N) cos 60o j
FD
TAB=-173.2 i+100 j
TAC= (TAC) sin 20o i+(TAC) cos 20o j
Unknowns; FD, TAC
TAC= 0.342(TAC) i+0.9397(TAC) j
B
C
TAE=-(300 N) j
TAE=300 N
15.3.2016
FD=FD i
Dr. Engin Aktaş
19
IZMIR INSTITUTE OF TECHNOLOGY
Department of Architecture
Example
AR231 Fall12/13
(cntd.)
R=TAB+TAC+TAE+FD=0
R=(-173.2 N)i + 100 N j + 0.342(TAC) i + 0.9397 (TAC) j + (-300 N) j + FD i =0
R={-173.2 + 0.342(TAC) + FD} i + {100 + 0.9397 (TAC) + (-300 N)} j = 0
SFx=0
-173.2 N + 0.342(TAC) + FD = 0
SFy=0
100 + 0.9397 (TAC) + (-300 N) = 0
TAC=213 N
FD=100 N
15.3.2016
Dr. Engin Aktaş
20
IZMIR INSTITUTE OF TECHNOLOGY
Department of Architecture
AR231 Fall12/13
y
z
j
V
k
Forces
in
Space
Vzk
Vyj qy
qz
qx
x
Vxi
V=Vxi+Vyj+Vzk
Vx=V cos qx
15.3.2016
Vy=V cos qy
Dr. Engin Aktaş
i
Vz=V cos qz
21
IZMIR INSTITUTE OF TECHNOLOGY
Department of Architecture
AR231 Fall12/13
Direction cosines
l = cos qx
m = cos qy
n = cos qz
Vx=l V
Vy=mV
Vz=nV
V 2=Vx2+Vy2+Vz2
l 2 + m 2 + n 2=1
15.3.2016
Dr. Engin Aktaş
22
IZMIR INSTITUTE OF TECHNOLOGY
z
Department of Architecture
y
AR231 Fall12/13
B(xB, yB, zB)
V
A(xA, yA, zA)
n
V= (xB-xA) i + (yB-yA) j + (zB-zA) k
x
Unit vector along AB
15.3.2016
V
n
V
Dr. Engin Aktaş
23
IZMIR INSTITUTE OF TECHNOLOGY
Department of Architecture
AR231 Fall12/13
Rectangular Coordinates in Space
y
y
B
Fy
B
A
qy
Fy
F
x
O
Fh
O
f
f
E
Fy=F cosqy
z
Fh
Fz
C
Fx
D
x
C
Fx = Fh cosf = F sinqy cosf
z
Fh=F sinqy
Fz = Fh sinf = F sinqy sinf
F 2 = F y2 + F h 2
Fh2 = Fx2 + Fz2
F F F F
2
x
15.3.2016
2
y
Dr. Engin Aktaş
2
z
24
IZMIR INSTITUTE OF TECHNOLOGY
Department of Architecture
AR231 Fall12/13
Problems
15.3.2016
Dr. Engin Aktaş
25
IZMIR INSTITUTE OF TECHNOLOGY
y
A
Department of Architecture
AR231 Fall12/13
Problem 1-(Meriam and Kraige)
3
F=1800 N
4
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.
x
First let’s find the x and y components of the force F.
z
Fx= -1800 N 3/5 = -1080 N
Fy= -1800 N 4/5 = -1440 N
F = Fx i+ Fy j = -1080 i -1440 j N
15.3.2016
Dr. Engin Aktaş
26
IZMIR INSTITUTE OF TECHNOLOGY
Department of Architecture
AR231 Fall12/13
Problem 2-(Meriam and Kraige)
L
C
D
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.
L= 50 N
Air flow
C
tan q = 50/5
q
D=5 N
q = tan-110 =84.3o
F = (502+52)0.50 = 50.2 N
15.3.2016
Dr. Engin Aktaş
27
IZMIR INSTITUTE OF TECHNOLOGY
Department of Architecture
AR231 Fall12/13
Problem 3-(Meriam and Kraige)
y
A
10o
900 N
45o
25o
a
x
800 N
The gusset plate is subjected to the two forces
shown. Replace them by two equivalent force, Fx in
the x-direction and Fa in the a direction. Determine
the magnitudes of Fx and Fa.
From the law of cosines
R2=8002+9002-2(800)(900) cos75
R=1040 N
A
800 N
10o
a
R
10
o
65o
25o
900 N
From the law of sines
R
900
 900 sin 75 
1  900 sin 75 

 a  sin 1 
  sin 
  56.7
sin 75 sin a
R


 1040 
Fx
A
Fx
1040
45
66.7

 Fx  544 N
113.3
sin 45 sin 21.7
45
1040 N
21.7
Fa
1040
Fa

 Fa  1351 N
sin 45 sin 113.3
o
o
o
o
o
15.3.2016
Dr. Engin Aktaş
28
IZMIR INSTITUTE OF TECHNOLOGY
Department of Architecture
AR231 Fall12/13
Problem 4-(Meriam and Kraige)
40 m
A
B
50 m
40 m
8 kN
R
The magnitude of R
15.3.2016
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
C force at point A. Determine the
magnitude R of this downward
force
This time let’s use another approach. Since the
resultant force is downward the sum of
horizontal components of the two forces should
add up to zero.
50
40
 TAB
8
 0  TAB  5.68 N
64
72.1
40
60
R  5.68  8
 10.21 N
64
72.1
Dr. Engin Aktaş
29
IZMIR INSTITUTE OF TECHNOLOGY
Department of Architecture
AR231 Fall12/13
Problem 5-(Beer and Johnston)
500 mm
1375 mm
1200 mm
A
B
C
1600 kg
Two cables are tied together at C and loaded as
shown. Determine the tension in AC and BC.
Let’s draw the Free Body Diagram (FBD) at C
y
TBC
TAC
x
C
W=1600*9.81=15700N
Writing Equilibrium Equations
SFx=0
SFy=0
 TAC
TAC
500
1375
 TCB
0
1300
1825
1200
1200
 TCB
 15700  0
1300
1825
TAC=12470 N
TCB=6370 N
15.3.2016
Dr. Engin Aktaş
30
IZMIR INSTITUTE OF TECHNOLOGY
y
Department of Architecture
Problem 6-(Beer and Johnston)
C
4m
D
AR231 Fall12/13
B
A
z
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
x exerted by cables AB and AC on stake A.
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

5.5 m
8m
n AB

AB   8  42  5.52  10.50
 8i  4 j  5.5k

 0.762i  0.381j  0.524k
10.50
FAB  4200 0.762i  0.381j  0.524k  N
2
AC=(0-8) i + (0-(-4)) j + (-13.5-(-5.5)) k = -8 i + 4 j - 8 k
n AC
 8  4

  8  12.00
FAC  6000 0.667i  0.333j  0.667k  N
 8i  4 j  8k

 0.667i  0.333j  0.667k
12.00
AC 
15.3.2016
2
2
2
Dr. Engin Aktaş
31
IZMIR INSTITUTE OF TECHNOLOGY
y
AR231 Fall12/13
Problem 6-(Beer and Johnston)
C
4m
D
Department of Architecture
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
x exerted by cables AB and AC on stake A.
FAB  4200 0.762i  0.381j  0.524k  N
B
FAC  6000 0.667i  0.333j  0.667k  N
A
z
5.5 m
R = FAB + FAC
8m
R =(-3200-4002) i +(1600+1998) j + (2201-4002) k = -7200 i + 3600 j - 1800 k
R = 8250 N
y
Direction cosines
x
l = 0.873
m = 0.436
n = 0.218
z
15.3.2016
Dr. Engin Aktaş
32
IZMIR INSTITUTE OF TECHNOLOGY
y
700 mm
O
D
AR231 Fall12/13
Problem 7-(Beer and Johnston)
C
450 mm
Department of Architecture
FAD
Let’s draw FBD first
FAC
FAB
600 mm
B
650 mm
z
A
x
1125
mm
W
Unknowns : FAD, FAC and W
Coordinates of points A, B, C and D
A crate is supported by A (0, -1125, 0) B (700, 0, 0) C (0, 0, -600) D (-650, 0,450)
three cables as shown.
AB=((700-0)i+(0-(-1125))j+0k=700i+1125j+0k
Determine the weight W
AB=(7002+11252)0.5=1325
of the crate knowing that
nAB=(700i+1125j+0k)/1325=0.5283i+0.8491j
the tension in cable AB is
4620 N.
FAB=4620(0.5283i+0.8491j) N
15.3.2016
Dr. Engin Aktaş
33
IZMIR INSTITUTE OF TECHNOLOGY
y
Department of Architecture
AR231 Fall12/13
Problem 7-(Beer and Johnston)
Coordinates of points A, B, C and D
C
450 mm
A (0, -1125, 0) B (700, 0, 0) C (0, 0, -600) D (-650, 0,450)
700 mm
O
D
AC=(0i+(0-(-1125))j+(-600-0)k=0i+1125j-600k
600 mm
B
650 mm
z
A
x
1125
mm
AC=(6002+11252)0.5=1275
nAC=(0i+1125j-600k)/1275=0.8824j-0.4706k
FAC=FAC(0.8824j-0.4706k) N
AD=((-650-0)i+(0-(-1125))j+(450-0)k=-650i+1125j+450k
AD=(6502+11252 +4502)0.5=1375
nAD=(-650i+1125j+450k)/1375= -0.4727i+0.8182j+0.3273k
FAD=FAD(-0.4727i+0.8182j+0.3273k) N
W=-W j
15.3.2016
Dr. Engin Aktaş
34
IZMIR INSTITUTE OF TECHNOLOGY
y
AR231 Fall12/13
Problem 7-(Beer and Johnston)
C
450 mm
Department of Architecture
700 mm
O
D
600 mm
B
650 mm
z
A
x
FAB+FAC+FAD+W=0
1125
mm
4620(0.5283i+0.8491j) +FAC(0.8824j-0.4706k)+FAD(-0.4727i+0.8182j+0.3273k)-W j=0
2441-0.4727FAD=0
FAD= 5160 N
-0.4706 FAC+0.3273*5160=0
FAC= 3590 N
W= 11320 N
3922+3170+4225-W = 0
15.3.2016
Dr. Engin Aktaş
35
IZMIR INSTITUTE OF TECHNOLOGY
y
Department of Architecture
AR231 Fall12/13
Problem 8-(Beer and Johnston)
D
600 mm
C
200 mm
A
200 mm
B
200 mm
x
400 mm
z
A 16 kg triangular plate is
supported by three wires
as shown. Determine the
tension in each wire .
15.3.2016
Dr. Engin Aktaş
36