Statics of Particles

IZMIR INSTITUTE OF TECHNOLOGYIZMIR INSTITUTE OF TECHNOLOGY Department of ArchitectureDepartment of Architecture AR23AR2311 Fall12/13Fall12/13

17.04.23 Dr. Engin Aktaş 1

2.0 Statics of Particles



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



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



Resultant of several concurrent forces

Q

P

S

Using polygon rule

A

A

P

Q

SP+Q+S



Resolution of a force into components

A

F

P

Q

A

F

P

Q

AFQ

P



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



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



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



Rectangular Components

F

x

y

Fx

Fy



F

x

y

F

x

F y



Unit Vectors

x

y

i

j

Magnitude=1 F

Fx=Fxi

Fy=Fyj




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



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



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



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



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



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



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




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



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



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



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



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



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



Problems



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




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




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




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



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



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



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




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



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



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




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

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Statics of Particles