Applied Mechanics 02

8/9/2019 Applied Mechanics 02

1/31

STATICS OF PARTICLES

Applied Mechanics


2/31

Introduction2 - 2

• The objective for the current chapter is to investigate the effects of forces

on particles:

- replacing multiple forces acting on a particle with a singleequivalent or resultant force

- relations between forces acting on a particle that is in a

state of equilibrium!

• The focus on particles does not impl" a restriction to miniscule bodies!#ather the stud" is restricted to anal"ses in which the si$e and shape ofthe bodies is not significant so that all forces ma" be assumed to beapplied at a single point!


3/31

#esultant of Two %orces2 - &

• force: action of one bod" on another'characteri$ed b" its point of applicationmagnitude line of action and sense!

• ()perimental evidence shows that thecombined effect of two forces ma" berepresented b" a single resultant force!

• The resultant is equivalent to the diagonal of

a parallelogram which contains the two forcesin adjacent legs!

• %orce is a vector quantit"!


4/31

*ectors2 - +

• Vector: parameters possessing magnitude and direction which add according to the parallelogram law! ()amples:

displacements velocities accelerations!

• *ector classifications:

- Fixed or bound vectors have well defined points ofapplication that cannot be changed without affectingan anal"sis!

- Free vectors ma" be freel" moved in space withoutchanging their effect on an anal"sis!

- Sliding vectors ma" be applied an"where along their

line of action without affecting an anal"sis!

• Equal vectors have the same magnitude and direction!

• Negative vector of a given vector has the same magnitude

and the opposite direction!

• Scalar: parameters possessing magnitude but notdirection! ()amples: mass volume temperature


5/31

Addition of *ectors2 - ,• Trape$oid rule for vector addition

• Triangle rule for vector addition

.

.

Q P R

B PQQ P R

+=

−+= cos2222

• /aw of cosines

• /aw of sines

A

C

R

B

Q

A sinsinsin==

• *ector addition is commutative

P QQ P +=+

• *ector subtraction


6/31

Addition of *ectors2 - 0• Addition of three or more vectors through

repeated application of the triangle rule

• The pol"gon rule for the addition of three or

more vectors!

• *ector addition is associative

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

• Multiplication of a vector b" a scalar


7/31

#esultant of 1everal .oncurrent %orces2 - • Concurrent forces: set of forces which all

pass through the same point!

A set of concurrent forces applied to aparticle ma" be replaced b" a single resultantforce which is the vector sum of the appliedforces!

• Vector force components: two or more force vectors which together have the same effectas a single force vector!


8/31

1ample 3roblem2 - 4

The two forces act on a bolt at A!5etermine their resultant!

16/7TI68:

• 9raphical solution - construct aparallelogram with sides in the samedirection as P and Q and lengths inproportion! 9raphicall" evaluate theresultant which is equivalent in direction

and proportional in magnitude to the thediagonal!

• Trigonometric solution - use the trianglerule for vector addition in conjunction

with the law of cosines and law of sines tofind the resultant!


9/31

1ample 3roblem2 -

• 9raphical solution - A parallelogram with sidesequal to P and Q is drawn to scale! Themagnitude and direction of the resultant or ofthe diagonal to the parallelogram are measured

°== 35 N98 α R

• 9raphical solution - A triangle is drawn with P and Q head-to-tail and to scale! The magnitudeand direction of the resultant or of the thirdside of the triangle are measured

°== 35 N98 α R


10/31

1ample 3roblem2 -;<

• Trigonometric solution - Appl" the triangle rule!

%rom the /aw of .osines

( ) ( ) ( ) ( ) °−+=−+=

155cos N60 N402 N60 N40

cos222

222 B PQQ P R

A

A

R

Q B A

R

B

Q

A

+°=°=

°=

=

=

20

04.15 N73.97

N60155sin

sinsin

sinsin

α

N73.97= R

%rom the /aw of 1ines

°= 04.35α


11/31

1ample 3roblem2 -;;

a= the tension in each of the ropesfor α > +,o

b= the value of α for which thetension in rope 2 is a minimum!

A barge is pulled b" two tugboats!If the resultant of the forcese)erted b" the tugboats is ,


12/31

1ample 3roblem2 -;2

• 9raphical solution - 3arallelogram #ule with @nown resultant direction andmagnitude @nown directions for sides!

lbf 2600lbf 3700 21 == T T

• Trigonometric solution - Triangle #ule with /aw of 1ines

°=

°=

° 105sinlbf 5000

30sin45sin

21 T T

lbf 2590lbf 3660 21 == T T


13/31

1ample 3roblem2 -;&• The angle for minimum tension in rope 2 is

determined b" appl"ing the Triangle #uleand observing the effect of variations in α!

• The minimum tension in rope 2 occurs whenT1 and T2 are perpendicular!

( ) °= 30sinlbf 50002T lbf 25002 =T

( ) °= 30coslbf 50001T lbf 43301 =T

°−°= 3090α °= 60α


14/31

#ectangular .omponents of a %orce: 7nit *ectors2 -;+

• *ector components ma" be e)pressed as products ofthe unit vectors with the scalar magnitudes of the

vector components!

F x and F y are referred to as the scalar components of j F i F F y x +=

F

• Ma" resolve a force vector into perpendicularcomponents so that the resulting parallelogram is arectangle! are referred to as rectangularvector components and

y x F F F +=

y x F F and

• 5efine perpendicular unit vectors which areparallel to the x and y a)es! ji and


15/31

Addition of %orces b" 1umming .omponents2 -;,

S Q P R ++=

• ?ish to find the resultant of & or moreconcurrent forces

( ) ( ) jS Q P iS Q P jS iS jQiQ j P i P j Ri R

y y y x x x

y x y x y x y x

+++++=

+++++=+

• #esolve each force into rectangular components

∑=++=

x x x x x

F S Q P R

• The scalar components of the resultant are equalto the sum of the corresponding scalarcomponents of the given forces!

∑= ++= y y y y y

F S Q P R

x

y y x

R

R R R R 122 tan−=+= θ

• To find the resultant magnitude and direction


16/31

1ample 3roblem2 -;0

%our forces act on bolt A as shown!5etermine the resultant of the forceon the bolt!

16/7TI68:

• #esolve each force into rectangularcomponents!

• .alculate the magnitude and directionof the resultant!

• 5etermine the components of theresultant b" adding the corresponding

force components!


17/31

1ample 3roblem2 -;

16/7TI68:

• #esolve each force into rectangular components!

9.256.96100

0.1100110

2.754.2780

0.759.129150

4

3

2

1

−+ −

+−++−−

F

F

F

F

comp ycomp xmag force

22 3.141.199 += R N6.199= R

• .alculate the magnitude and direction!

N1.199

N3.14tan =α °= 1.4α

• 5etermine the components of the resultant b"adding the corresponding force components!

1.199+= x R 3.14+= y R


18/31

(quilibrium of a 3article2 -;4

• ?hen the resultant of all forces acting on a particle is $ero the particle isin equilibrium!

• 3article acted upon b"two forces:

- equal magnitude

- same line of action

- opposite sense

• 3article acted upon b" three or more forces:- graphical solution "ields a closed pol"gon

- algebraic solution

00

0

==

==

∑∑

∑

y x F F

F R

• Newton’s First aw: If the resultant force on a particle is $ero the particle willremain at rest or will continue at constant speed in a straight line!


19/31

%ree-od" 5iagrams2 -;

Space !iagram: A s@etch showingthe ph"sical conditions of the

problem!

Free"#ody !iagram: A s@etch showingonl" the forces on the selected particle!


20/31

1ample 3roblem2 -2<

In a ship-unloading operation a

&,


21/31

1ample 3roblem2 -2;16/7TI68:

• .onstruct a free-bod" diagram for theparticle at A!

• Appl" the conditions for equilibrium!

• 1olve for the un@nown force magnitudes!

°=

°=

° 58sinlb3500

2sin120sin

AC AB T T

lb3570= ABT

lb144= AC T


22/31

1ample 3roblem2 -22

It is desired to determine the drag forceat a given speed on a protot"pe sailboathull! A model is placed in a test channeland three cables are used to align its bow on the channel centerline! %or a

given speed the tension is +< lb in cable A# and 0< lb in cable AE !

5etermine the drag force e)erted on thehull and the tension in cable AC !

16/7TI68:

• .hoosing the hull as the free bod"draw a free-bod" diagram!

• ()press the condition for equilibriumfor the hull b" writing that the sum of

all forces must be $ero!

• #esolve the vector equilibriumequation into two componentequations! 1olve for the two un@nowncable tensions!


23/31

1ample 3roblem2 -2&16/7TI68:

• .hoosing the hull as the free bod" draw afree-bod" diagram!

°=

==

25.60

75.1ft4

ft7tan

α

α

°=

==

56.20

375.0ft4

ft1.5tan

β

β

• ()press the condition for equilibriumfor the hull b" writing that the sum ofall forces must be $ero!

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


24/31

1ample 3roblem2 -2+• #esolve the vector equilibrium equation into

two component equations! 1olve for the twoun@nown cable tensions!

( ) ( )

( ) ( )

( )

( )

( ) jT

i F T

R

i F F

iT

jT iT jT iT T

ji

jiT

AC

D AC

D D

AC AC

AC AC AC

AB

609363.084.19

3512.073.34

0

lb06

9363.03512.056.20cos56.20sin

lb84.19lb73.34

26.60coslb4026.60sinlb40

−++

++−=

=

=

−=

+=°+°=

+−=

°+°−=


25/31

1ample 3roblem2 -2,

( )

( ) jT

i F T

R

AC

D AC

609363.084.19

3512.073.34

0

−++++−=

=

This equation is satisfied onl" if each componentof the resultant is equal to $ero

( )

( ) 609363.084.19003512.073.3400

−+==++−==

∑∑

AC y

D AC x

T F

F T F

lb66.19

lb9.42

+= += D AC

F

T


26/31

#ectangular .omponents in 1pace2 -20

• The vector iscontained in theplane $#AC !

F • #esolve intohori$ontal and verticalcomponents!

yh F F θ sin=

F

y y F F θ cos=

• #esolve intorectangular components

h F

φ θ

φ

φ θ

φ

sinsin

sin

cossin

cos

y

h y

y

h x

F

F F

F

F F

=

=

==


27/31


• ?ith the angles between and the a)es F

( )

k ji F

k ji F

k F j F i F F

F F F F F F

z y x

z y x

z y x

z z y y x x

θ θ θ λ λ

θ θ θ

θ θ θ

coscoscos

coscoscos

coscoscos

++==

++=

++=

===

• is a unit vector along the line of action of and are the directioncosines for

F

F

λ

z y x θ θ θ cosand,cos,cos


28/31


5irection of the force is defined b"the location of two points

( ) ( )222111 ,,and,, z y x N z y x M

( )

d

Fd F

d

Fd F

d

Fd F

k d jd id d

F F

z z d y yd x xd

k d jd id

N M d

z z

y y

x x

z y x

z y x

z y x

===

++=

=

−=−=−=

++=

=

1

and joiningvector

121212

λ

λ


29/31

1ample 3roblem2 -2

The tension in the gu" wire is 2,


30/31

1ample 3roblem2 -&<

16/7TI68:

• 5etermine the unit vector pointing from A towards #!

( ) ( ) ( )

( ) ( ) ( )

m3.94

m30m80m40

m30m80m40

222

=

++−=

++−=

AB

k ji AB

• 5etermine the components of the force!

( ) ( )( ) ( ) ( )k ji

k ji

F F

N795 N2120 N1060

318.0848.0424.0 N2500

++−=

++−=

= λ

k ji

k ji

318.0848.0424.0

3.94

30

3.94

80

3.94

40

++−=

+

+

−=λ


31/31

1ample 3roblem2 -&;

• 8oting that the components of the unit vector arethe direction cosines for the vector calculate thecorresponding angles!

k ji

k ji z y x

318.0848.0424.0

coscoscos

++−=

++= θ θ θ λ

5.71

0.32

1.115

=

=

=

z

y

x

θ

θ

θ

Applied Mechanics 02

Documents

Transcript of Applied Mechanics 02