PHYS 218sec. 517-520

ReviewChap. 11

Equilibrium and Elasticity

What you have to know

• Conditions for equilibrium for a rigid-body

• How to solve rigid-body equilibrium problems

• Only Sections 11.1 – 11.3 are reviewed. (We skipped the other sections in the class.)

Conditions for equilibrium

For a point-like particle: 0 (Chap. 4 & Chap. 5)F =årr

For an extended body, its CM should have zero acceleration

0 0x y zF F F F= Þ = = =å å å årr

First condition for equilibrium

Second condition for equilibrium

The body must have no tendency to start rotating

0 about any pointt =å

If the body is at rest, it is in static equilibrium.

Center of Gravity

In most equilibrium problems, gravity is one of the forces that make the body rotate.

We can assume that the entire force of gravity is concentrated at the center of gravity (CG) of the body.

If the change in the gravitational acceleration g can be ignored, the center of gravity (CG) of the body is identical to its center of mass (CM).

position vector of the center of mass

1, where CM i i ir m r M m

M= =å å

r r

Center of Gravity

( )Then the total torque isi i i i i

i i i CM CM CM CM

r w r mg mr g

mr g mr g Mr g r Mg r w

t

t t

= ´ = ´ = ´

= = ´ = ´ = ´ = ´ = ´å å å

r r r r r rr

r r r r r r r r r rr r

The total gravitational torque is the same as though the total weight were acting on the CM of the body.

A body supported at several points must have its CG somewhere within the area bounded by the supports.

CG

stable

CG

Not in equilibrium (non-vanishing net torque)

Ex 11.1Wooden plank of length 6.0 m and mass 90 kg.

Two sawhorses are separated by 1.5 m.

If the plabk is to remain at rest, how big can be?

L M

D

m

= =

=

D

mRange of CG for equilibrium

Without , the CG in at O,

With , the CG will move to the right.

If is larger than a certain value,

the CG will be out of the region of supports

and the system is not in equilibrium.

m

m

m

Origin O (geometric center)

( )The CG of the plank is at O 0 and is at / 2,

then the CG of the system is

0 2

2Since should be less than 2,

30 kg2 2

plank mCG

CG

x m x L

M x m x M m L m Lx

M m M m M mx D

m L D Dm M

M m L D

= =

´ + ´ ´ + ´Þ = = =

+ + +

£ Þ £ =+ -

Ex 11.2

2.46 m

The total normal force on the front wheel is 0.53 and that

on the rear wheel is 0.47 . How far in front of the rear axle

is the car's CG?

w

w

w

0.53w0.47w

O

( ) ( ) ( )

( )

Let be the position of the CG.

Then the torque around should vanish, which gives

0.47 0 0.53 0,

0.530.53 2.46 m 1.3 m

cg

cg

cg

x

O

w wx w L

wLx

w

t = ´ - + ´ =

Þ = = ´ =

å

L

Ex 11.3

q

sf

2n

1n

CG of the ladder

Lw

Mw

3L

2

length of the ladder: 5.0 m

weight of the ladder: 180 N

weight of the man: 800 N

=53.1

? ? minimum ?

L

M

s s

L

w

w

n f

q

m

=

=

=

°

= =

( )

1 2

1 2

1

1

First condition for equilibrium

0, 0

and 980 N

Second condition for equilibrium

sin sin sin 02 2 3 2

1cos cos

sin 2 3

x s y L M

s L M

L M

L M

F f n F n w w

f n n w w

L Ln L w w

w wn

p pt p q q q

q qq

= - = = - - =

Þ = = + =

æ ö æ ö÷ ÷ç ç= - - + - + =÷ ÷ç ç÷ ÷ç çè ø è ø

æÞ = +

è

å å

å

1

cos268 N

sin 2 3

268 N

L M

s

w w

f n

qq

ö æ ö÷ ÷ç ç= + =÷ ÷ç ç÷ ÷ç çø è ø

\ = =

2 ,min2

0.27ss s s

ff n

nm m£ Þ = =

2

The contact force on the ladder at the base is

the vector sum of and sf nr r

Ex 11.4

w

B

T

A qxE

yED

L

tension and , ?x yT E E

( )

Second condition for equilibrium (about )

sin 0

sinFirst condition for equilibrium

0

coscos

sin tan

0

y

x x x

x x

y y y

y y

O

Lw DT Lw DT

LwT

D

F T E

Lw Lw LwE T T

D D h

F T E w

L D wLwE w T w

D D

t q

q

qq

q q

= - = - =

Þ =

= - =

Þ = = = = =

= + - =

-Þ = - = - =-

å

å

åO

h

The negative sign means that Ey is downward.

If you don’t know the direction of a force a priori, then choose an arbitrary direction. After calculation, if you get a negative value, then it means that the

direction of the force is opposite to your guess.