Post on 01-Jun-2018
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STATICS OF PARTICLES
Applied Mechanics
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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$e and shape ofthe bodies is not significant so that all forces ma" be assumed to beapplied at a single point!
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#esultant of Two %orces2 - &
• force: action of one bod" on another'characteri$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"!
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*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
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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
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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
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#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!
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1ample 3roblem2 - 4
The two forces act on a bolt at A!5etermine their resultant!
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• 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!
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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
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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
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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 ,
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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
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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
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#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
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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
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1ample 3roblem2 -;0
%our forces act on bolt A as shown!5etermine the resultant of the forceon the bolt!
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• #esolve each force into rectangularcomponents!
• .alculate the magnitude and directionof the resultant!
• 5etermine the components of theresultant b" adding the corresponding
force components!
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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
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(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!
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%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!
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1ample 3roblem2 -2<
In a ship-unloading operation a
&,
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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
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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 !
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• .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!
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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
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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
−++
++−=
=
=
−=
+=°+°=
+−=
°+°−=
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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
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#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
=
=
==
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#ectangular .omponents in 1pace2 -2
• ?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
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#ectangular .omponents in 1pace2 -24
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
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1ample 3roblem2 -2
The tension in the gu" wire is 2,
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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
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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