example:%%the$massive$diving$boardhumanic/p1200_lecture19.pdf · rigid%objects%in%equilibrium...
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Rigid%Objects%in%Equilibrium
Example:%%The$Massive$Diving$Board
A"woman"whose"weight"is"530"N"is"
poised"at"the"right"end"of"a"diving"board
with"length"3.90"m.""The"board"now"has
a"weight"of"1400"N"and"is"supported"by
a"fulcrum"1.40"m"away"from"the"left
end.
Find"the"forces"that"the"bolt"and"the"
fulcrum"exert"on"the"board.
How"do"the"results"change"from"the
“Massless"Diving"Board”"example"done
earlier?
lW/2
WD
Rigid%Objects%in%Equilibrium
lW/2
WD
τ∑ = F2ℓ2 −WℓW −WDℓW2
#
$%
&
'(= 0
F2 =WℓWℓ2
+WDℓW2ℓ2
!
"#
$
%&
F2 =530 N( ) 3.90 m( )
1.40 m+ (1400 N) 3.90 m
2 1.40 m( )
!
"##
$
%&&
=1480 N+1950 N = 3430 N
Rigid%Objects%in%Equilibrium
lW/2
WD
Fy∑ = −F1 +F2 −W −WD = 0
−F1 +3430 N− 530 N−1400 N = 0
F1 =1500 N
Rigid%Objects%in%Equilibrium
Example:%%Bodybuilding
The$arm$is$horizontal$and$weighs$32.2$N.$$The$deltoid$muscle$can$supply1840$N$of$force.$$What$is$the$weight$of$the$heaviest$dumbbell$she$can$hold?
Rigid%Objects%in%Equilibrium
0=+−−=∑ Mddaa MWW τ
ℓM = 0.150 m( )sin13.0"
Rigid%Objects%in%Equilibrium
Wd =−Waℓa +MℓM
ℓd
=− 32.2 N( ) 0.280 m( )+ 1840 N( ) 0.150 m( )sin13.0!
0.620 m= 85.6 N
Chapter(10
Fluids
Mass$Density
DEFINITION(OF(MASS(DENSITY
The(mass(density(of(a(substance(is(the(mass(of(a(substance(divided(by(its(volume:
Vm
=!
SI$Unit$of$Mass$Density:$$kg/m3
Mass$Density
Mass$Density
Example:$$Blood%as%a%Fraction%of%Body%Weight
The$body$of$a$man$whose$weight$is$690$N$contains$about5.2x10=3 m3 of$blood.
(a)$Find$the$blood�s$weight$and$(b)$express$it$as$a$percentage$of$the$body$weight.
m = ρV = 1060kg m3( ) 5.2×10−3m3( ) = 5.5 kg
Mass$Density
( )( ) N 54sm80.9kg 5.5 2 === mgW(a)
(b) %8.7%100N 690
N 54Percentage =!=
Pressure
AFP =
SI(Unit(of(Pressure:((1"N/m2 ="1Pa
Pascal
Definition(of(Pressure:((The"pressure,"P,"exerted"bya"fluid"is"defined"as"the"magnitude"of"the"force,"F,acting"perpendicular"to"a"surface"divided"by"the"area,A,"over"which"the"force"acts.
The"pressure"of"the"moving"air"molecules(blue"dots)"acting"inside"of"a"tire.
Pressure
Example:--The$Force$on$a$Swimmer
Suppose'the'pressure'acting'on'the'backof'a'swimmer�s'hand'is'1.2x105 Pa.''Thesurface'area'of'the'back'of'the'hand'is'8.4x10@3m2.
(a)Determine'the'magnitude'of'the'forcethat'acts'on'it.(b)'Discuss'the'direction'of'the'force.
Pressure
AFP =
( )( )N 100.1
m104.8mN102.13
2325
!=
!!== "PAF
Since&the&water&pushes&perpendicularly&against&the&back&of&the&hand,&the&forceis&directed&downward&in&the&drawing.
Pressure
Atmospheric.Pressure.at.Sea.Level:.. 1.013x105 Pa)=)1)atmosphere
Pressure&and&Depth&in&a&Static&Fluid
012 =!!=" mgAPAPFy
mgAPAP += 12
m = ρV
Consider)a)static)fluid,)i.e.)a)fluid)atrest:
!a = 0 ⇒
!F = 0∑
Pressure&and&Depth&in&a&Static&Fluid
VgAPAP !+= 12
AhV =
AhgAPAP !+= 12
P2 = P1 + ρ gh
Pressure&and&Depth&in&a&Static&Fluid
Conceptual&Example:&&The$Hoover$Dam
Lake%Mead%is%the%largest%wholly%artificial%reservoir%in%the%United%States.%%The%waterin%the%reservoir%backs%up%behind%the%damfor%a%considerable%distance%(120%miles).
Suppose%that%all%the%water%in%Lake%Meadwere%removed%except%a%relatively%narrowvertical%column.
Would%the%Hoover%Dam%still%be%neededto%contain%the%water,%or%could%a%much%lessmassive%structure%do%the%job?
Pressure&and&Depth&in&a&Static&Fluid
Example:&&The$Swimming$Hole
Points'A'and'B'are'located'a'distance'of'5.50'm'beneath'the'surface'of'the'water.''Find'the'pressure'at'each'of'these'two'locations.
Pressure&and&Depth&in&a&Static&Fluid
( ) ( )( )( )Pa 1055.1
m 50.5sm80.9mkg1000.1Pa 1001.15
233
pressure catmospheri
52
!=
!+!=!!"!!#$
P
ghPP !+= 12
Pascal�s&Principle
PASCAL�S&PRINCIPLE
Any$change$in$the$pressure$applied$to$a$completely$enclosed$fluid$is$transmitted$undiminished$to$all$parts$of$the$fluid$and$enclosing$walls. A2
Pascal�s&Principle
A2
Situation(in(the(Figure(– The(plunger(andpiston(are(at(the(same(height.
( )m 012 gPP !+=
1
1
2
2
AF
AF
=
!!"
#$$%
&=
1
212 AAFF
Pascal�s&Principle
Example:&&A"Car"Lift
The$input$piston$has$a$radius$of$0.0120$mand$the$output$plunger$has$a$radius$of$0.150$m.
The$combined$weight$of$the$car$and$the$plunger$is$20500$N.$$Suppose$that$the$inputpiston$has$a$negligible$weight$and$the$bottomsurfaces$of$the$piston$and$plunger$are$atthe$same$level.$$
What$is$the$required$input$force$of$the$pistonIn$order$to$maintain$the$position$of$the$carand$plunger?
A2
Pascal�s&Principle
A2
F2 = F1A2A1
!
"#
$
%& ⇒ F1 = F2
A1A2
"
#$
%
&'
F1 = 20500 N( )π 0.0120 m( )2
π 0.150 m( )2 =131 N