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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.
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
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