12. Static Equilibrium
1. Conditions for Equilibrium
2. Center of Gravity
3. Examples of Static Equilibrium
4. Stability
The Alamillo Bridge in Seville, is the work of architect Santiago Calatrava.
What conditions must be met to ensure the stability of this dramatic?
= =net netF τ 0
12.1. Conditions for Equilibrium
(Mechanical) equilibrium = zero net external force & torque.
Static equilibrium = equilibrium + at rest.
=iF 0
=i τ 0 i i r F
Pivot point = origin of ri .
is the same for all choices of pivot points=iF 0 i τ
Prob 55:
For all pivot points
Example 12.1. Drawbridge
The raised span has a mass of 11,000 kg uniformly distributed over a length of 14 m.
Find the tension in the supporting cable.
=i τ 0
Force Fh at hinge not known.
Choose pivot point at hinge.
g T τ τ 0
1 2sin sin 02
Lm g L T
1 90 30 120
2 180 30 15 165
1
2
sin
2sin
m gT
211,000 9.8 / sin120
2sin165
kg m s
180 kN Another choice of pivot: Ex 15
y
x30
Tension T
Gravity mgHinge force Fh
15
2
1
GOT IT? 12.1.
Which pair, acting as the only forces on the object, results in static equilibrium?
Explain why the others don’t.
(C)
(A): F 0.
(B): 0.
12.2. Center of Gravity
i i τ r F i im r g
Total torque on mass M in uniform gravitational field :
= i im r g
cm M τ r g
Center of gravity = point at which gravity seems to act
cg cmr r for uniform gravitational field
net cg net τ r F
CG does not exist if net is not Fnet .
Conceptual Example 12.1. Finding the Center of Gravity
1st pivot
2nd pivot
Explain how you can find an object’s center of gravity by suspending it from a string.
GOT IT? 12.2.
The dancer in the figure is balanced; that is, she’s in static equilibrium.
Which of the three lettered points could be her center of gravity?
12.3. Examples of Static Equilibrium
All forces co-planar: =iF 0
=i τ 0
2 eqs in x-y plane
1 eq along z-axis
Tips: choose pivot point wisely.
Example 12.2. Ladder Safety
A ladder of mass m & length L leans against a frictionless wall.
The coefficient of static friction between ladder & floor is .
Find the minimum angle at which the ladder can lean without slipping.
Fnet x : 1 2 0n n
Fnet y : 1 0n m g
Choose pivot point at bottom of ladder.
z : 2 sin 180 sin 90 02
LL n m g
2 1n nm g
2 sin cos 02
LL n m g
2
tan2
m g
n
1
2
0 90
y
x
mgn1
fS = n1i
n2
Example 12.3. Arm Holding Pumpkin
Find the magnitudes of the biceps tension & the contact force at the elbow joint.
Fnet x : cos 0c xF T
Fnet y : sin 0c yT F m g M g
Pivot point at elbow.
z : 1 2 3sin 0x T x m g x M g
2 3
1 sin
x m x M gT
x
20.036 2.7 0.32 4.5 9.8 /
0.036 sin80
m kg m kg m s
m
500 N
cosc xF T
sinc yF T m M g
500 cos80N 87 N
2500 sin80 2.7 4.5 9.8 /N kg kg m s 420 N
2 2c c x c yF F F 430 N ~ 10 M g
y
x
Mgmg
T
Fc80
GOT IT? 12.3.
A person is in static equilibrium leaning against a wall.
Which of the following must be true:
(a) There must be a frictional force at the wall but not necessarily at the floor.
(b) There must be a frictional force at the floor but not necessarily at the wall.
(c) There must be frictional forces at both floor and wall.
Need frictional force to balance normal force from wall.
Application: Statue of Liberty
Sculptor Bartholdi : lasting as long as the pyramids.
Deviation from Eiffel’s plan resulted in excessive torque.
Major renovation was required after only 100 yrs.
12.4. Stability
Stable equilibrium: Original configuration regained after small disturbance.
Unstable equilibrium: Original configuration lost after small disturbance.
Stable equilibrium
unstable equilibrium
Stable
Unstable
Neutrally stable
Metastable
Equilibrium: Fnet = 0.
V at global min
V at local max
V = const
V at local min
2
20
d V
d x
2
20
d V
d x
2
20
d V
d x
2
20
d V
d x
0d V
d x
Example 12.4. Semiconductor Engineering
A new semiconductor device has electron in a potential U(x) = a x2 – b x4 ,
where x is in nm, U in aJ (1018 J), a = 8 aJ / nm2, b = 1 aJ / nm4.
Find the equilibrium positions for the electron and describe their stability.
Equilibrium criterion : 0d U
d x
32 4 0a x b x
2 nm
0x
2
ax
bor
2
4
8 /
2 1 /
aJ nm
aJ nm
22
22 12
d Ua b x
d x
2
2
0
2 0x
d Ua
d x
x = 0 is (meta) stable
2
2
/2
4 0x a b
d Ua
d x
x = (a/2b) are unstable
equilibria
Metastable
Saddle Point
, ,0
U x y U x y
x y
Equilibrium condition
2
2
,0
U x y
x
Saddle point
stable
2
2
,0
U x y
y
unstable
stable
unstable