Short Version : 12. Static Equilibrium

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

Stable equilibrium : PE at global min

Metastable equilibrium : PE at local min

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