Sta$cs
Physics1162016
Condi$onsforSta$cEquilibrium
•  Equilibrium and static
equilibrium
•  Static equilibrium
conditions
–  Net external force must
equal zero
–  Net external torque must
equal zero
•  Center of gravity
•  Solving static equilibrium
problems
Conditions for Equilibrium
•  The first condition of
equilibrium is a statement of
translational equilibrium
•  The net external force on the
object must equal zero
•  It states that the translational
acceleration of the object’s
center of mass must be zero
Conditions for Equilibrium
•  The first condition of
equilibrium is a statement of
translational equilibrium
•  The net external force on the
object must equal zero
!
!
!
Fnet = ∑ Fext = ma = 0
•  It states that the translational
acceleration of the object’s
center of mass must be zero
Conditions for Equilibrium
•  The first condition of
equilibrium is a statement of
translational equilibrium
•  The net external force on the
object must equal zero
!
!
!
Fnet = ∑ Fext = ma = 0
•  It states that the translational
acceleration of the object’s
center of mass must be zero
Conditions for Equilibrium
•  The second condition of
equilibrium is a statement of
rotational equilibrium
•  The net external torque on
the object must equal zero
•  It states the angular
acceleration of the object to
be zero
•  This must be true for any axis
of rotation
Conditions for Equilibrium
•  The second condition of
equilibrium is a statement of
rotational equilibrium
•  The net external torque on
the object must equal zero
!
!
!
τ net = ∑τ ext = Iα = 0
•  It states the angular
acceleration of the object to
be zero for any axis of
rotation
Conditions for Equilibrium
•  The second condition of
equilibrium is a statement of
rotational equilibrium
•  The net external torque on
the object must equal zero
!
!
!
τ net = ∑τ ext = Iα = 0
•  It states the angular
acceleration of the object to
be zero for any axis of
rotation
A seesaw consisting of a uniform board of mass mpl and
length L supports at rest a father and daughter with masses
M and m, respectively. The support is under the center of
gravity of the board, the father is a distance d from the
center, and the daughter is a distance 2.00 m from the
center.
A) Find the magnitude of the upward force n exerted by the
support on the board.
B) Find where the father should sit to balance the system at
rest.
A)Findthemagnitudeoftheupward
forcenexertedbythesupportonthe
board.
B)Findwherethefathershouldsitto
balancethesystematrest.
τ net , z = τ d + τ f + τ pl + τ n
= mgd − Mgx + 0 + 0 = 0
mgd = Mgx
2m
⎛m⎞
x = ⎜ ⎟d =
< 2.00 m
M
⎝M ⎠
Fnet , x = ∑ Fext , x = 0
Fnet , y = ∑ Fext , y = 0
τ net , z = ∑ τ ext, z = 0
Axis of Rotation
•  The net torque is about an axis through any
point in the xy plane
•  Does it matter which axis you choose for
calculating torques?
•  NO. The choice of an axis is arbitrary
•  If an object is in translational equilibrium and
the net torque is zero about one axis, then the
net torque must be zero about any other axis
•  We should be smart to choose a rotation axis to
simplify problems
Axis of Rotation
•  The net torque is about an axis through any
point in the xy plane
•  Does it matter which axis you choose for
calculating torques?
•  NO. The choice of an axis is arbitrary
•  If an object is in translational equilibrium and
the net torque is zero about one axis, then the
net torque must be zero about any other axis
•  We should be smart to choose a rotation axis to
simplify problems
Axis of Rotation
•  The net torque is about an axis through any
point in the xy plane
•  Does it matter which axis you choose for
calculating torques?
•  NO. The choice of an axis is arbitrary
•  If an object is in translational equilibrium and
the net torque is zero about one axis, then the
net torque must be zero about any other axis
•  We should be smart to choose a rotation axis to
simplify problems
Axis of Rotation
•  The net torque is about an axis through any
point in the xy plane
•  Does it matter which axis you choose for
calculating torques?
•  NO. The choice of an axis is arbitrary
•  If an object is in translational equilibrium and
the net torque is zero about one axis, then the
net torque must be zero about any other axis
•  We should be smart to choose a rotation axis to
simplify problems
Axis of Rotation
•  The net torque is about an axis through any
point in the xy plane
•  Does it matter which axis you choose for
calculating torques?
•  NO. The choice of an axis is arbitrary
•  If an object is in translational equilibrium and
the net torque is zero about one axis, then the
net torque must be zero about any other axis
•  We should be smart to choose a rotation axis to
simplify problems
B)Findwherethefathershouldsittobalancethesystematrest.
Rota$onaxisO
τ net , z = τ d + τ f + τ pl + τ n
Rota$onaxisP
τ net , z = τ d + τ f + τ pl + τ n
= mgd − Mgx + 0 + 0 = 0
= 0 − Mg (d + x) − m pl gd + nd = 0
mgd = Mgx
− Mgd − Mgx − m pl gd + ( Mg + mg + m pl g )d = 0
2m
⎛m⎞
x = ⎜ ⎟d =
M
⎝M ⎠
mgd = Mgx
P
⎛m
x=⎜
⎝M
O
2m
⎞
d
=
⎟
M
⎠
Sixiden$cal2-metermasslessrodsaresuppor$ngiden$cal
12-Newtonweights.Ineachcase,aver$calforceFisholding
therodsandtheweightsatrest.Therodsaremarkedathalfmeterintervals.
Rankthemagnitudeofthever2calforceFappliedtotherodfromthe
strongesttoweakest.
A.F=E>C>D=A>B
B.F>E>C>D>A>B
C.F>C>E=A>D=B
D. E>F>C>A>D>B
E. Sameforallcases
Sixiden$calmasslessrodsareallsupportedbyafulcrumandare
$ltedatthesameangletothehorizontal.Eachrodismarkedat1meterintervals.
Rankthemagnitudeofthever2calforceFappliedtotheendofthe
rodfromthestrongesttotheweakest.
A.F>E>D>B>C>A
B.D>E>F>A>C>B
C.F>B>E>C>D>A
D. F>E>B>D>C>A
E. Sameforallcases
Horizontal Beam Example
A uniform horizontal beam with a
length of l = 8.00 m and a weight of
Wb = 200 N is attached to a wall by a
pin connection. Its far end is
supported by a cable that makes an
angle of φ = 53° with the beam. A
person of weight Wp = 600 N stands
a distance d = 2.00 m from the wall.
Find the tension in the cable as well
as the magnitude and direction of the
force exerted by the wall on the
beam.
Horizontal Beam Example
•  The beam is uniform
–  So the center of gravity
is at the geometric
center of the beam
•  The person is standing on
the beam
•  What are the tension in
the cable and the force
exerted by the wall on
the beam?
Horizontal Beam Example, 2
•  Analyze
–  Draw a free body
diagram
–  Use the pivot in the
problem (at the wall)
as the pivot
•  This will generally be
easiest
–  Note there are three
unknowns (T, R, θ)
Horizontal Beam Example, 3
•  The forces can be
resolved into
components in the
free body diagram
•  Apply the two
conditions of
equilibrium to obtain
three equations
•  Solve for the
unknowns
Horizontal Beam Example, 3
l
∑ τ z = (T sin φ )(l ) − W p d − Wb ( ) = 0
2
l
W p d + Wb ( )
2 = (600 N )(2m) + (200 N )(4m) = 313N
T=
l sin φ
(8m) sin 53!
Horizontal Beam Example, 3
l
∑ τ z = (T sin φ )(l ) − W p d − Wb ( ) = 0
2
l
W p d + Wb ( )
2 = (600 N )(2m) + (200 N )(4m) = 313N
T=
l sin φ
(8m) sin 53!
∑ Fx = R cosθ − T cos φ = 0
∑ Fy = R sin θ + T sin φ − W p − Wb = 0
Horizontal Beam Example, 3
l
∑ τ z = (T sin φ )(l ) − W p d − Wb ( ) = 0
2
l
W p d + Wb ( )
2 = (600 N )(2m) + (200 N )(4m) = 313N
T=
l sin φ
(8m) sin 53!
∑ Fx = R cosθ − T cos φ = 0
∑ Fy = R sin θ + T sin φ − W p − Wb = 0
W p + Wb − T sin φ
R sin θ
= tan θ =
R cosθ
T sin φ
⎛ W p + Wb − T sin φ ⎞
⎟⎟ = 71.7 !
T sin φ
⎝
⎠
θ = tan −1 ⎜⎜
T cos φ (313N ) cos 53!
R=
=
= 581N
!
cosθ
cos 71.7
Abeamissupportedbyafric$onlesspinsothatitis
horizontal.Amassissuspendedfromtherightendofthe
beam,andtheleSendistetheredbyananchoredcable.Both
casesbelowareiden$cal,exceptfortheloca$onofthe
anchorpointofthecable.
Isthemagnitudeofthetorqueduetothecableaboutthe
pivotpingreaterincaseA,greaterincaseB,orthesamein
bothcases?
A.  GreaterincaseA
B.  GreaterincaseB
C.  Thesameinbothcases
Abeamissupportedbyafric$onlesspinsothatitis
horizontal.Amassissuspendedfromtherightendofthe
beam,andtheleSendistetheredbyananchoredcable.Both
casesbelowareiden$cal,exceptfortheloca$onofthe
anchorpointofthecable.
IsthetensioninthecablegreaterincaseA,greaterincaseB,
orthesameinbothcases?
A.  GreaterincaseA
B.  GreaterincaseB
C.  Thesameinbothcases
Ladder Example
•  A uniform ladder of length
l rests against a smooth,
vertical wall. The mass of
the ladder is m, and the
coefficient of static friction
between the ladder and
the ground is µs = 0.40.
Find the minimum angle θ
at which the ladder does
not slip.
A Problem-Solving Strategy
•  Choose a convenient axis for calculating the net torque
on the object
–  Remember the choice of the axis is arbitrary
•  Choose an origin that simplifies the calculations as
much as possible
–  A force that acts along a line passing through the origin
produces a zero torque
•  Be careful of sign with respect to rotational axis
–  I’ll choose positive if force tends to rotate object in CCW
–  negative if force tends to rotate object in CW
–  zero if force is on the rotational axis
•  Apply the second condition for equilibrium τ net , z = ∑ τ ext , z = 0
Choose an origin O that simplifies the
calculations as much as possible ?
•  A uniform ladder of length l rests against a smooth, vertical
wall. The mass of the ladder is m, and the coefficient of
static friction between the ladder and the ground is µs =
0.40. Find the minimum angle.
B)
A)
O
C)
D)
O
O
mg
O
mg
mg
mg
Choose an origin O that simplifies the
calculations as much as possible ?
•  A uniform ladder of length l rests against a smooth, vertical
wall. The mass of the ladder is m, and the coefficient of
static friction between the ladder and the ground is µs =
0.40. Find the minimum angle.
B)
A)
O
C)
D)
O
O
mg
O
mg
mg
mg
Choose an origin O that simplifies the
calculations as much as possible ?
•  A uniform ladder of length l rests against a smooth, vertical
wall. The mass of the ladder is m, and the coefficient of
static friction between the ladder and the ground is µs =
0.40. Find the minimum angle.
B)
A)
O
C)
D)
O
O
mg
O
mg
mg
mg
LeverArm
Aleverarmistheshortestperpendiculardistance
betweenthelineofac$onofaforceandtheaxisof
rota$on.Itprovidesanalterna$vewayofdeterminethe
torqueandisespeciallyusefulwhentheforcesdonotlie
alongcoordinateaxes.
leverarm=rsinφ
Torque=F*r*sinφ
Torque=F*leverarm
Some5mesitissimpletoreadtheleverarmrightoffofthe
diagram.
Threeforcesofequal
A
magnitudeareappliedto
a3-mby2-mrectangle.
B
Forcesandactat45°
!
anglestothever$calas F1
shown,whileacts
45°
horizontally.
C
IsthenettorqueaboutthepointA
1.clockwise
2.counterclockwise
3.zero
WhataboutpointB?pointC?
!
!
45°
F2
F3
Anotherladderexample
Auniformladder
weighing60Nleans
againstafric$onless
ver$calwall.Whatisthe
minimumcoefficientof
fric$onnecessary
betweentheladderand
theflooriftheladderis
nottoslip?
4m
3m
Anotherladderexample
Auniformladder
weighing60Nleans
againstafric$onless
ver$calwall.Whatisthe
minimumcoefficientof
fric$onnecessary
betweentheladderand
theflooriftheladderis
nottoslip?
Ffloor(fric$on)
Fwall(normal)
Fearth(weight)
4m
3m
Ffloor(normal)
Whatdoweknow?
Auniformladder
weighing60Nleans
againstafric$onless
ver$calwall.Whatisthe
minimumcoefficientof
fric$onnecessary
betweentheladderand
theflooriftheladderis
nottoslip?
Ffloor(fric$on)
Fwall(normal)
Fearth(weight)
4m
3m
Ffloor(normal)
Annotatedsolu$on
Wearetryingtofindthecoefficient Weknowthe
offric5on,µ
forceshaveto
Howisitrelatedtowhatweknow balance…ΣF=0
Fwall(normal)
(theweight)?
Ffloor(fric$on)=Ffloor(normal) µ
SinceΣFver$cal=0,byobserva$onwe
canseethatFfloor(normal) =Fearth(weight)
= 60 Ν
Fearth(weight)
4m
SinceΣFhorizontal=0,byobserva$on
wecanseethatFwall(normal)
=Ffloor(fric$on)= ??
Ffloor(fric$on)
Therearetwounknownsandonly
oneequa5on–weneedanother
3m
equa5on.
Ffloor(normal)
Στz=0…totherescue!
Inthiscase,whatarethebestchoices
fortheaxisofrota$on?
I.  atthetopoftheladder
II.  atthemiddleoftheladder
III. attheboaomoftheladder
A.
B.
C.
D.
E.
I
II
III
IorII
IIorIII
Annotatedsolu$on
Weknowthe
Wearetryingtofindthecoefficientof
torquesaboutany
fric5on,µ
Howisthefric5onaltorquerelatedtothe
axishaveto
weighttorqueaboutthebase?
Fwall(normal)
balance…Στ=0
Ffloor(fric$on)=Ffloor(normal) µ andFfloor(fric$on)=Fwall(normal) SinceΣτbase=0,then Fearth(weight)*(itsleverarm)mustbalance
withFwall(normal) )*(itsleverarm)
Fearth(weight)
4m
Howdoyoufindtheleverarm?
Itistheshortestdistancefromthelineof
ac5onoftheforcetotheaxisofrota5on.
Ffloor(fric$on)
Theleverarmfortheweight=1.5mand
O
theleverarmforthewallnormalforceis4
3m
m.
Ffloor(normal)
Pudngitalltogether
Στbase=0,so
Fearth(weight)*(itsleverarm)mustbalancewith
Fwall(normal) )*(itsleverarm)
Pudngitalltogether
Στbase=0,so
Fearth(weight)*(1.5m)=Fwall(normal)*(4m)
Fwall(normal)=Fearth(weight)*(1.5m)/(4m)
Ffloor(fric$on)=Fwall(normal) = Fearth(weight)*0.375
µ  = Fwall(normal)/Ffloor(normal)
SinceinthiscaseFearth(weight)=Ffloor(normal)
µ  = Fwall(normal)/Fearth(weight)
µ  =[Fearth(weight)*0.375]/Fearth(weight)
µ=0.375
Howdoesthisproblemchangeifthere
isapersonofknownweightstanding
ontheladderataknownposi5on?
Oldexamproblem
A50Nmonkeyclimbsupa
120Nladderandstops1/3of
thewayup.Theupperand
lowerendsoftheladderrest
onfric$onlesssurfaces.The
lowerendisfastenedtothe
wallbyamasslesshorizontal
rope.Findthetensioninthe
ropewhenthemonkeyis1/3
ofthewayuptheladder.
8m
6m
OldExamProblem
AmassofM=225kghangsfromtheendofthe
uniformstrutwhosemassismstrut=45.0kg.Find:
(a)  thetensionTinthecable
(b) horizontalandver$calforcecomponents
exertedonthestrutbythehinge.