Download - statics of particles
IZMIR INSTITUTE OF TECHNOLOGYIZMIR INSTITUTE OF TECHNOLOGY Department of ArchitectureDepartment of Architecture AR23AR2311 Fall12/13Fall12/13
21.04.23 Dr. Engin Aktaş 1
2.0 Statics of Particles
IZMIR INSTITUTE OF TECHNOLOGYIZMIR INSTITUTE OF TECHNOLOGY Department of ArchitectureDepartment of Architecture AR23AR2311 Fall12/13Fall12/13
21.04.23 Dr. Engin Aktaş 2
Forces on a Particle
50 N30o
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 TECHNOLOGYIZMIR INSTITUTE OF TECHNOLOGY Department of ArchitectureDepartment of Architecture AR23AR2311 Fall12/13Fall12/13
21.04.23 Dr. Engin Aktaş 3
Properties of Vector Addition
Q
P P+Q=Q+PCommutative
AP
Q
P+Q
Q+P
QP
A
AAssociative
A
PQ
S
P+Q
P+Q+SA
Q
Q+S
P+Q+S
P+Q+S=(P+Q)+S=P+(Q+S)
P S
IZMIR INSTITUTE OF TECHNOLOGYIZMIR INSTITUTE OF TECHNOLOGY Department of ArchitectureDepartment of Architecture AR23AR2311 Fall12/13Fall12/13
21.04.23 Dr. Engin Aktaş 4
Resultant of several concurrent forces
Q
P
S
Using polygon rule
A
A
P
Q
SP+Q+S
IZMIR INSTITUTE OF TECHNOLOGYIZMIR INSTITUTE OF TECHNOLOGY Department of ArchitectureDepartment of Architecture AR23AR2311 Fall12/13Fall12/13
21.04.23 Dr. Engin Aktaş 5
Resolution of a force into components
A
F
P
Q
A
F
P
Q
AFQ
P
IZMIR INSTITUTE OF TECHNOLOGYIZMIR INSTITUTE OF TECHNOLOGY Department of ArchitectureDepartment of Architecture AR23AR2311 Fall12/13Fall12/13
21.04.23 Dr. Engin Aktaş 6
A
P
Q
R
Example (Beer & Johnston)
R
B
Q
A sinsin
25o
20o
Q=60 N
P=40 NA
The two forces P and Q act on a bolt A. Determine their resultant.
20o
25o
180-25=155o
R2=P2+Q2-2PQcosB
B
R2=(40N)2+(60N)2-2(40N)(60N)cos1550 R=97.73 NLaw of cosines
Law of sines N
N
R
BQA
73.97
155sin60sinsin
IZMIR INSTITUTE OF TECHNOLOGYIZMIR INSTITUTE OF TECHNOLOGY Department of ArchitectureDepartment of Architecture AR23AR2311 Fall12/13Fall12/13
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Example (cntd.)
R
B
Q
A sinsinLaw of sines
N
N
R
BQA
73.97
155sin60sinsin
A=15.04o =20o+A=35.04o
The answer is R=97.7 kN =35.0o
IZMIR INSTITUTE OF TECHNOLOGYIZMIR INSTITUTE OF TECHNOLOGY Department of ArchitectureDepartment of Architecture AR23AR2311 Fall12/13Fall12/13
21.04.23 Dr. Engin Aktaş 8
Example (Beer and Johnston)
105sin
25
30sin45sin21 kNTT
B
A
C
1
2
30o
45o
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.
R=25 kN
45o 30o
T2 T1
105o
Using law of sines
kNkN
T 30.18105sin
45sin251
kNkN
T 94.12105sin
30sin252
IZMIR INSTITUTE OF TECHNOLOGYIZMIR INSTITUTE OF TECHNOLOGY Department of ArchitectureDepartment of Architecture AR23AR2311 Fall12/13Fall12/13
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Rectangular Components
F
x
y
Fx
Fy
IZMIR INSTITUTE OF TECHNOLOGYIZMIR INSTITUTE OF TECHNOLOGY Department of ArchitectureDepartment of Architecture AR23AR2311 Fall12/13Fall12/13
21.04.23 Dr. Engin Aktaş 10
F
x
y
F
x
F y
IZMIR INSTITUTE OF TECHNOLOGYIZMIR INSTITUTE OF TECHNOLOGY Department of ArchitectureDepartment of Architecture AR23AR2311 Fall12/13Fall12/13
21.04.23 Dr. Engin Aktaş 11
Unit Vectors
x
y
i
j
Magnitude=1 F
Fx=Fxi
Fy=Fyj
IZMIR INSTITUTE OF TECHNOLOGYIZMIR INSTITUTE OF TECHNOLOGY Department of ArchitectureDepartment of Architecture AR23AR2311 Fall12/13Fall12/13
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Example (Beer and Johnston)
x
y
x
y
F
F
F
F 1tantan
0.655.3
5.7tan 1
kN
kN
F=(3.5 kN)i+(7.5kN)j
A Determine the magnitude of the force and angle .
Ax
y
Fx=3.5 kN
F y=
7.5
kN
F
kN
kNFF y 28.8
0.65sin
5.7
sin
IZMIR INSTITUTE OF TECHNOLOGYIZMIR INSTITUTE OF TECHNOLOGY Department of ArchitectureDepartment of Architecture AR23AR2311 Fall12/13Fall12/13
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Addition of Forces by Summing x and y components
A x
y
F1=150 N
30o
F2=80 N 20o
F3=110 N
F4=100 N
15o
Determine the resultant of the forces on the bolt
F2 cos 20o
-(F2 sin 20o)
-F3
F1 sin 30o
F1 cos 30o
F4 cos 15o
-(F4 sin 15o)
Forces Magnitude, N x Component, N y Component, NF1 150 +129.9 +75.0
F2 80 -27.4 +75.2
F3 110 0 -110.0
F4 100 +96.6 -25.9
Rx=199.1 Ry=+14.3
IZMIR INSTITUTE OF TECHNOLOGYIZMIR INSTITUTE OF TECHNOLOGY Department of ArchitectureDepartment of Architecture AR23AR2311 Fall12/13Fall12/13
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Example (cntd)
R=Rxi+Ryj
R=(199.1N)i+(14.30N)j
R
Rx=199.1 N
Ry=14.30 N
10.41.199
30.14tantan 11
N
N
R
R
x
y
IZMIR INSTITUTE OF TECHNOLOGYIZMIR INSTITUTE OF TECHNOLOGY Department of ArchitectureDepartment of Architecture AR23AR2311 Fall12/13Fall12/13
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Equilibrium of a ParticleWhen the resultant of all the forces acting on a particle is
zero, the particle is in equilibrium.
x
y
A F1=300 N
F2=173.2 NF3=200 N
F4=400 NF1=300 N
F2=173.2 N
F3=200 N
F4=400 N
R=F=0
(Fx)i+ (Fy)j=0
Fx=0 Fy=0
Fx=300 N -(200 N) sin30o-(400 N) sin 30o=300 N -100 N -200 N = 0
Fy=-173.2 N -(200 N) cos30o+ (400 N) cos 30o=-173.2 N -173.2 N +346.4 N = 0
30o
30o
IZMIR INSTITUTE OF TECHNOLOGYIZMIR INSTITUTE OF TECHNOLOGY Department of ArchitectureDepartment of Architecture AR23AR2311 Fall12/13Fall12/13
21.04.23 Dr. Engin Aktaş 16
Free-Body DiagramA 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
BC
50o 30o
Space Diagram
TAC
30o50o
TAB
736 N 60o80o
40o
736 N
Free-Body Diagram TAC
TAB
Using law of sines
80sin
736
40sin60sinACAB TT
TAB=647 N
TAC=480 N
IZMIR INSTITUTE OF TECHNOLOGYIZMIR INSTITUTE OF TECHNOLOGY Department of ArchitectureDepartment of Architecture AR23AR2311 Fall12/13Fall12/13
21.04.23 Dr. Engin Aktaş 17
Analytical Solution• Two unknowns TAB and TAC
• Equilibrium Equations
Fx=0 and Fy=0
The system of equations then030cos50cos0 ACABx TTF
073630sin50sin0 NTTF ACABy
Solving the above system (2 unknowns 2 equations)
TAB=647 N
TAC=480 N
IZMIR INSTITUTE OF TECHNOLOGYIZMIR INSTITUTE OF TECHNOLOGY Department of ArchitectureDepartment of Architecture AR23AR2311 Fall12/13Fall12/13
21.04.23 Dr. Engin Aktaş 18
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
IZMIR INSTITUTE OF TECHNOLOGYIZMIR INSTITUTE OF TECHNOLOGY Department of ArchitectureDepartment of Architecture AR23AR2311 Fall12/13Fall12/13
21.04.23 Dr. Engin Aktaş 19
Example (Beer and Johnston)
E
B C
A
60o 20o
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 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 exerted on the hull and the tension in cable AC.
Unknowns; FD, TAC
Start with drawing Free-Body Diagram
TAE=300 N
TAB=200 N
FD
TAC
60o
20o
Equilibrium conditionResultant of the all forces should be zero
R=TAB+TAC+TAE+FD=0
Let’s write all the forces in x and y components
TAB=-(200 N) sin 60o i+(200 N) cos 60o j
TAB=-173.2 i+100 jTAC= (TAC) sin 20o i+(TAC) cos 20o j
TAC= 0.342(TAC) i+0.9397(TAC) j
TAE=-(300 N) j
FD=FD i
IZMIR INSTITUTE OF TECHNOLOGYIZMIR INSTITUTE OF TECHNOLOGY Department of ArchitectureDepartment of Architecture AR23AR2311 Fall12/13Fall12/13
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Example (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
-173.2 N + 0.342(TAC) + FD = 0Fx=0
Fy=0 100 + 0.9397 (TAC) + (-300 N) = 0
TAC=213 N
FD=100 N
IZMIR INSTITUTE OF TECHNOLOGYIZMIR INSTITUTE OF TECHNOLOGY Department of ArchitectureDepartment of Architecture AR23AR2311 Fall12/13Fall12/13
21.04.23 Dr. Engin Aktaş 21
Forces in
Space
x
yz
V
Vzk
k
j
iVxi
Vyj
x
y
z
V=Vxi+Vyj+Vz
kVx=V cos x Vy=V cos y Vz=V cos z
IZMIR INSTITUTE OF TECHNOLOGYIZMIR INSTITUTE OF TECHNOLOGY Department of ArchitectureDepartment of Architecture AR23AR2311 Fall12/13Fall12/13
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Direction cosines
l = cos x m = cos y n = cos z
Vx=l V Vy=mV Vz=nV
V 2=Vx
2+Vy2+Vz
2
l 2 + m 2 + n 2=1
IZMIR INSTITUTE OF TECHNOLOGYIZMIR INSTITUTE OF TECHNOLOGY Department of ArchitectureDepartment of Architecture AR23AR2311 Fall12/13Fall12/13
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x
yz
B(xB, yB, zB)
A(xA, yA, zA)V
V= (xB-xA)i
+ (yB-yA) j
+ (zB-zA)k
n
V
Vn Unit vector along AB
IZMIR INSTITUTE OF TECHNOLOGYIZMIR INSTITUTE OF TECHNOLOGY Department of ArchitectureDepartment of Architecture AR23AR2311 Fall12/13Fall12/13
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Rectangular Coordinates in Space
y
B
A
C
O
FFy
Fh
y
x
z
B
C
OFx
Fy
y
x
z
FhFz
Fy=F cosy
Fh=F siny
Fx = Fh cos= F siny cos
Fz = Fh sin= F siny sin
E
D
F 2 = Fy2 + Fh
2Fh
2 = Fx2 + Fz
2
222zyx FFFF
IZMIR INSTITUTE OF TECHNOLOGYIZMIR INSTITUTE OF TECHNOLOGY Department of ArchitectureDepartment of Architecture AR23AR2311 Fall12/13Fall12/13
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Problems
IZMIR INSTITUTE OF TECHNOLOGYIZMIR INSTITUTE OF TECHNOLOGY Department of ArchitectureDepartment of Architecture AR23AR2311 Fall12/13Fall12/13
21.04.23 Dr. Engin Aktaş 26
Problem 1-(Meriam and Kraige)
A4
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.
Fx= -1800 N 3/5 = -1080 N
Fy= -1800 N 4/5 = -1440 N
F = Fx i+ Fy j = -1080 i -1440 j N
IZMIR INSTITUTE OF TECHNOLOGYIZMIR INSTITUTE OF TECHNOLOGY Department of ArchitectureDepartment of Architecture AR23AR2311 Fall12/13Fall12/13
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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 which it makes with the horizontal.
CD=5 N
L= 5
0 N
Air flow
D
L
C
tan = 50/5
= tan-110 =84.3o
F = (502+52)0.50 = 50.2 N
IZMIR INSTITUTE OF TECHNOLOGYIZMIR INSTITUTE OF TECHNOLOGY Department of ArchitectureDepartment of Architecture AR23AR2311 Fall12/13Fall12/13
21.04.23 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, Fx in the x-direction and Fa in the a direction. Determine the magnitudes of Fx and Fa.
y
x
900 N800 N
10o
45o25o
A
a
A 800 N
900 N
R
10o
25o
10o
65o
From the law of cosines
R2=8002+9002-2(800)(900) cos75
R=1040 N
From the law of sines
7.56
1040
75sin900sin
75sin900sin
sin
900
75sin11
R
R
66.7o
45o
45o
113.3o
AFx
Fa21.7o
1040 N
NFF
xx 5447.21sin45sin
1040
NFF
aa 13513.113sin45sin
1040
IZMIR INSTITUTE OF TECHNOLOGYIZMIR INSTITUTE OF TECHNOLOGY Department of ArchitectureDepartment of Architecture AR23AR2311 Fall12/13Fall12/13
21.04.23 Dr. Engin Aktaş 29
Problem 4-(Meriam and Kraige)
B
A
C
50 m 40 m
40 m
20 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 kNThis 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.
NTT ABAB 68.501.72
408
64
50
The magnitude of R NR 21.101.72
608
64
4068.5
IZMIR INSTITUTE OF TECHNOLOGYIZMIR INSTITUTE OF TECHNOLOGY Department of ArchitectureDepartment of Architecture AR23AR2311 Fall12/13Fall12/13
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Problem 5-(Beer and Johnston)
Two cables are tied together at C and loaded 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
TBCTAC
Writing Equilibrium Equations
Fx=0
Fy=0
01825
1375
1300
500 CBAC TT
0157001825
1200
1300
1200 CBAC TT
TAC=12470 N
TCB=6370 N
IZMIR INSTITUTE OF TECHNOLOGYIZMIR INSTITUTE OF TECHNOLOGY Department of ArchitectureDepartment of Architecture AR23AR2311 Fall12/13Fall12/13
21.04.23 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
AB
C
5.5 m
8 m
4 m
13.5 m
x
y
z
Coordinates of points A, B and CA (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
50.105.548 222 AB
kjikji
nAB 524.0381.0762.050.10
5.548
NkjiFAB 524.0381.0762.04200 AC=(0-8) i + (0-(-4)) j + (-13.5-(-5.5)) k = -8 i + 4 j - 8 k
00.12848 222 AC
kjikji
nAC 667.0333.0667.000.12
848
NkjiFAC 667.0333.0667.06000
IZMIR INSTITUTE OF TECHNOLOGYIZMIR INSTITUTE OF TECHNOLOGY Department of ArchitectureDepartment of Architecture AR23AR2311 Fall12/13Fall12/13
21.04.23 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
AB
C
5.5 m
8 m
4 m
13.5 m
x
y
z
NkjiFAB 524.0381.0762.04200
NkjiFAC 667.0333.0667.06000
R = FAB + FAC
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 TECHNOLOGYIZMIR INSTITUTE OF TECHNOLOGY Department of ArchitectureDepartment of Architecture AR23AR2311 Fall12/13Fall12/13
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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
xz
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
FAB
FACFAD
Unknowns : FAD, FAC and W
B (700, 0, 0) C (0, 0, -600) D (-650, 0,450)
AB=((700-0)i+(0-(-1125))j+0k=700i+1125j+0k
AB=(7002+11252)0.5=1325
nAB=(700i+1125j+0k)/1325=0.5283i+0.8491j
FAB=4620(0.5283i+0.8491j) N
IZMIR INSTITUTE OF TECHNOLOGYIZMIR INSTITUTE OF TECHNOLOGY Department of ArchitectureDepartment of Architecture AR23AR2311 Fall12/13Fall12/13
21.04.23 Dr. Engin Aktaş 34
Problem 7-(Beer and Johnston)y
xz
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=(0i+(0-(-1125))j+(-600-0)k=0i+1125j-600k
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
IZMIR INSTITUTE OF TECHNOLOGYIZMIR INSTITUTE OF TECHNOLOGY Department of ArchitectureDepartment of Architecture AR23AR2311 Fall12/13Fall12/13
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Problem 7-(Beer and Johnston)y
xz
B
A
O
C
D
700 mm
600 mm
1125 mm
650 mm
450 mm
FAB+FAC+FAD+W=0
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
3922+3170+4225-W = 0 W= 11320 N
IZMIR INSTITUTE OF TECHNOLOGYIZMIR INSTITUTE OF TECHNOLOGY Department of ArchitectureDepartment of Architecture AR23AR2311 Fall12/13Fall12/13
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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
CA