Lec15_SectionA

8/16/2019 Lec15_SectionA

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Today:

Static equilibrium The tragedy of Romeo and Juliet

Stability of equilibria.

Physics 2A Olga DudkoUCSD Physics

Lecture 15

What keeps the bridge and the house from falling?

1. Support from belowagainst Fgrav => !F = 0.

2. No tendency to tipin any direction => !" = 0.

San Diego-Coronado bridge Fallen Star, UCSD


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Equilibrium. Static equilibrium.

•

A body is in equilibrium w hen the net external forceand torque on the body are both zero:

•

If, in addition, the body isstationary, then it is in staticequilibrium.

!

F! = d !

p

dt = 0

!

! ! = d !

L

dt = 0

COM of rigid bodyhas a=0

no tendency to start rotating

about any point

Fallen Star, UCSD

Center of Gravity (CG)

•

The CG of a rigid bodyis the point at whichthe gravitational forceseems to act.

•

CG = average location of the weight of an object.

•

CG coincides with COM when gravitational field isuniform (i.e., for objects whose size is


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Center of Gravity • A body whose CG is above the area of support willbe stable if a vertical line projected down from CGfalls within the area of support.

• Fnormal (balancing Fg) can only be exerted within areaof contact. If Fg acts beyond that area, a net torquewill topple the object.

area of support area of support area of support

CG CG

CG

Does the Leaning Tower of Pisa

“defy laws of physics”?

•

Currently leans at a4.7° angle to the vertical

•

d = 7 m, height = 55 m

• The center of gravitylies within the base as

long as the angle is < 7.3° • It’s the laws of physicsthat are holding it up!


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How to avoid tipping over:

Have low center of gravity

+ large area of support

N o w e x t

i n c t !

Calculating torques •

When more than one torque is acting on an object,remember to take vector sum of the torques, !".

• The pivot point is the point where rotation isoccurring or where rotation may occur.

•

Sometimes you are free to choosethe pivot point (the ruler) andsometimes you are not (the door).

• When calculating torque values, itis useful to draw “an extendedforce diagram”.

pivot point

biceps

t r i c e p

s


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Extended force diagram

Example: a ladder rests on africtionless vertical wall. The floor isnot frictionless. Draw the extendedforce diagram for the ladder.

Fnormal,floor onladder

Fnormal, wall on ladder

Fgravity, Earth on ladder

Ffriction, floor on ladder

•

Indicate the forces acting on the object AND wherethey act.

CG

40o 40o

50o

50o

• To calculatetorques, we canchoose a pivotpoint at anylocation, sincethe object isnot rotating.

50o

10 m

Solving Equilibrium Problems

1) Choose an appropriate coordinate system. (x,y)

2) Make an extended force diagram.

3) Formulate equilibrium equations to apply.

!Fx = 0 !Fy = 0 !" = 0

4) Choose appropriate pivot point for torquecalculations.

5) Do algebra.


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Static Equilibrium

Example

Romeo is trying to reach Julietby climbing an 8.00m, 200Nuniform ladder which restsagainst a smooth wall andmakes a 50.0o angle with theground. Romeo weighs 800N.The coefficient of staticfriction between the ladderand the ground is 0.600.

Will Romeo reach Juliet beforethe ladder begins to slip?

50°

8 m

moat with hungryalligators

Juliet

The tragedy of Romeo and Juliet(A physicist’s perspective)

Static Equilibrium

Solution •

Define a coordinate system. • Draw extended force diagram:

Fnormal, flooron ladder

Fnormal, wallon ladder

Fgravity,Earth onladder


Fcontact, Romeoon ladder

w h e r e ? ?

?

50°

8 m

+x

+y

•

Wall is smooth => no friction (idealization!).

• If ladder is on the verge of slipping, static frictionforce btw ladder and floor will be at maximum:

F friction = µ s F N, floor

(ladder onthe vergeof slipping)


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Static Equilibrium Solution (cont

d)

•

Static equilibrium:

• #Fy = 0 FN, floor = Fg + FRomeo

FN, floor = 200N + 800N = 1,000N

=> max. static friction (when ladder is about to slip):

= 0 . 600 ( ) 1 , 000 N ( ) = 600 N F friction = µ s F N, floor

!! = 0! F = 0

= FN,floor– Fg – FRomeo

Fnormal,floor onladder

Fnormal, wallon ladder

Fgravity,Earth onladder


Fcontact, Romeoon ladder

• #Fx = 0

FN,wall = Ffriction

= Ffriction – FN,wall

= 600N

Static Equilibrium

Solution (cont

d)

•

Turn to the nettorque.

•

Choose pivotpoint…

FN, floor on ladder

Fgravity,Earth on ladder

Ffriction,floor on ladder

Fcontact,Romeo on ladder

= ?

50°

8 m

P i v o t p o i n t

FN, wall on ladder

at thebase (eliminates! N,floor and ! friction).

! ! = ! wall + ! grav + ! Romeo = 0

50° r wall F N,wall sin + r grav F grav sin 40° - = 0

ccw cw cw

F Romeo sin 40° - r Romeo

= L

= L / 2 = d

d =

L F N,wall sin 50° L / 2 F grav sin 40° -

F Romeo sin 40°

40o

50o

40o

50o


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Romeo’s Fate? Answer

= F f r i

c t

d =8m 600 N 0.766 4 m 200 N -

800 N

0.643

0.643 = 6.15 m

=> Distance that Romeo climbed upthe ladder before it began to slip:

d =

L F N,wall sin 50°

L / 2 F grav sin 40° -

F Romeo sin 40° 50°

8 m

"For never was a story of more woeTan this of Juliet and her Romeo."

- William Shakespeare, Romeo and Juliet, 5.3

=> At 6.15m up the ladder Romeo loses equilibrium !

• What could Romeo have done? Taken PHYS 2A!

Stability of Equilibria

• To examine each, turn to potential energy, U.

•

Force is related to theassociated potential energy as F = -

dU

dx

•

In equilibrium: dU

dx = 0

•

Stable equilibrium:

• Unstable equilibrium:

•

Stability? Look atd

2 U

dx 2

a small deviation fromthe equilibrium creates a restoring force thatdrives the system back to the equilibrium.

a small deviation fromthe equilibrium creates a force that drivesthe system away from the equilibrium.


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•

Equilibrium pointson the U(x) curve:

•

At x = 2: unstableequilibrium.

Stability of Equilibria

F = - dU

dx = 0

• At x = 4: stableequilibrium.

U (

x )

x 0 2 4 6

0

2

4

•

Mathematically: take the second derivative of thepotential energy in order to determine the stability

d 2 U

dx 2 > 0 - stable

equilibrium

d 2 U

dx 2 0

U

x 0

=> x=0 is stable equilibrium(local minimum).

U

x 0

U ( x ) = 1

2 kx

2 -

dU dx = kx ; -

d 2 U

dx 2 = k - x=0 is unstable equilibrium

(local maximum).

Example 2

dU dx

= 0 at x=0 Equilibrium?

Stability?

Equilibrium?

Stability?

dU dx

= 0 at x=0

(k > 0)

(k > 0)


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Three identical uniform rods are each acted on by twoor more forces, all perpendicular to the rods and allequal in magnitude (F). Which of these rods is inequilibrium?

Clicker Question

A)

rod 1.

B) rod 2.

C) rod 3.

D) None of these rods are in equilibrium.

E) All these rods are in equilibrium.

F F

F

F

F

F F

F

rod 1 rod 2 rod 3

For Next Time:

•

Read Chapter 14

• Study AS HARD AS YOU CAN for Quiz 8 (Ch.14)

•

Do homework for Chapter 14

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