Static Equilibrium & Center of Gravity: Concepts & Examples, Slides of Physics

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Short Version :

12. Static Equilibrium

12.1. Conditions for Equilibrium

(Mechanical) equilibrium = zero net external force & torque.Static equilibrium = equilibrium + at rest.

= i

F 

0^ = i

i^

i

^

r 

F

Pivot point = origin of

r. i^ is the same for all choices of pivot points

= i

F 

^

i

Prob 55:

For all pivot points

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^

mi

^

r 

g

Total torque on mass

M^ in uniform gravitational field :

^



=^

m^ i^ i

^

r^

g

^

 cm^

M

^

τ^

r^

g Center of gravity = point at which gravity seems to act

cg^  cm r^

r^

for uniform gravitational field

net^

cg^

net ^

τ^

r^

F

CG does not exist if

net

is not

^ F

.net ^ = 0 at CG.

12.3. Examples of Static Equilibrium

All forces co-planar:

= i

F 

0^ = i

^ 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

n^

n

^

^

Fnet y

:^

1

n^

m g 

Choose pivot point at bottom of ladder. :^ z^

^

^

^



sin 180 2

sin 90

L 2

L n

m g

 ^

^

 ^

2

1 n^

n 

m g 



sin 2

cos

L 2

L n

m g

^

2

tan^

m g^2 n

^ 

0



^ 

90 

y

x



 m g

n^1

f =S^

 n^1 n^2 i

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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 stableMetastable

Equilibrium:

F net

=^^0 .

V at global min V at local maxV = const V at local min

2 2

d Vd x^

d Vd x^

2 2

d Vd x^

2 2

d Vd x^

d Vd x

Example 12.4. Semiconductor Engineering

A new semiconductor device has electron in a potential

U (

x ) =

a x

2 –^

(^4) b x ,

where

x^ is in nm,

U^

in aJ (

^18 J),

a^ = 8 aJ / nm

2 ,^ b

= 1 aJ / nm

Find the equilibrium positions for the electron and describe their stability.

Equilibrium criterion :

d Ud x



3

2

a x

b x 

2 nm ^ 

x^ 

a 2 x^

b   or

^

aJ^

nm aJ^

nm

2

2

2

d^ U

a^

b x

d x

2 2

(^20)

0 x d^ U

a

d x^

 

^

x = 0 is (meta) stable

2 2

/

4

0

x^

a^ b d^ U

a

d x^

 

 ^

^

x =^ 

(a/2b) are unstable

equilibria Metastable

Saddle Point

^

^

^



,^

,^

U^

x y

U^

x y

x^

y

^

^

Equilibrium condition

^

 2

,^2

U^

x yx ^

Saddle point

stable

^

 2

,^2

U^

x yy ^

unstable

stable

unstable