StaticConditions of Equilibrium:

0

0

tot

totF

0

0

0

0

0

0

z

y

x

z

y

x

F

F

F

0

0

a

Static Equilibrium:

0

0

v

1m 1m

1kg 2kg Not in equilibrium

2m 1m

1kg 2kg In equilibrium

Example: Equilibrium and torque

221121

21 0

lFlF

221121

21 0

lFlF

When d = 20 cm, the scale

reads 2 lb exactly. If L = 1 m,

How big is your turkey?

Turkey

Light rod Scale

dLf1f2

P

0321

W

01 LfWd

lbm

mlb

d

LfW 10

2.0

121

Example: You bought a turkey and forgot how many pounds it was. All you have is a tiny kitchen scale that can weigh a maximum of 2 lb. But here’s good trick you can use to get an estimation.

How you can do it? Use static equilibrium!

Example: On the figure shown below, one end of a uniform road with a weight of 400 N is attached to a wall. The other end of the rod is

supported by a wire. What is the tension in the wire?

φ =30°

mg

NNmgT

LmgTL

tot

346400

060sin2/30sin

0

273.1

23

φ

φ

θ

θ

θ = φ + φ = 60°

23

21

60sin

30sin

φ =30°

mg

φ

φ

θ

θ

T

TTcos30°

Tsin30°

030cos

030sin

0

mgTF

TF

F

y

x

tot

NmgmgmgTmgF

NTF

y

x

10030cos

1732/

41

23

23

Fx

Fy

Example: A person of mass M climbs a ladder of mass m and length L that leans against a smooth wall. (We can neglect the friction between the wall and the ladder.) The frictional force between the floor and the ladder keeps it from slipping. The angle between the ladder and the wall is φ. Determine how the magnitude of the frictional force depends on φ.

N1

fS

N2

φ

Mg

mg

d

0

0

0

x

y

F

F

S 2

1

2

0

0

sin sin cos 02

f N

N Mg mg

LMgd mg N L

C is a good point to use as axis of rotation because then two of the forces will produce zero torque.

gmMN

fN s

1

2

gmMfNM

m

L

dMg ss

2tan

2

1

21tan

M

m

L

d

M

ms

C

Center of gravity(the point where the gravitational force can be considered to act)

•The torque due to gravity is the same as if all the mass of the object was concentrated at a point called the center of gravity.•It is the same as the center of mass as long as the gravitational force does not vary among different parts of the object.•It can be found experimentally by suspending an object from different points.

torque is not zero

mg

r

torque is zero

mg

r

gMrgrMgrmgmr CMCMiiii

Example: Three boxes are placed on identical inclines. Frictionprevents them from sliding. The boxes are not uniform and theircenter of mass are indicated by the blue dot in each case. In which cases does the box tip over?

A. All B. 2 & 3 C. 3 only

In 3, the CM is not above the base: net ≠ 0

1 2 3

2 m 1 m

40 kg

50 kg

CMplank

CMpig

0.5 m x

all

CM

The center of mass of the plank+pig system must not be beyond

the edge.

Position of the CM relative to the edge:

(40 kg)( 0.5 m) (50 kg) x 0

90 kg

20

x

50 0

2 0.4 m

5

x

x

Example: A 40-kg plank rests on a roof as shown. A 50-kg flying pig is about to land on the unsupported side. How far from the edge of the roof can the pig land without tipping the plank and itself over?

aMh

CM

2d

Example: A truck carries a refrigerator of mass M on a horizontal road. The center of mass of the fridge is at height h from the bed of the truck and the width of the fridge is 2d. What is the maximum acceleration aM that the truck can have without tipping the fridge? (Assume the static friction between the truck and the fridge is large enough for the fridge not to slip).

CM

N

Mg

When the truck has no acceleration (or is at rest), the normal force is right below the CM.

0

0 (trivial)

yF N Mg

When the truck has some acceleration (< aM) to the left, the normal force “moves” to the right to produce enough torque (about the CM) to cancel the torque due to friction (about the CM)!

“Moves” = “The pressure on the ground is redistributed”

When the fridge is just starting to tip over, the normal acts on the bottom right edge of the fridge.

CM

NMga

fS

CM

NMga

fS

S

0

0 0

x

y

Nf

F Ma f Ma

F N Mg

Ndhf s

MgdMah

hgda /

The fridge tips over because the normal force cannot produce a torque larger than Nd.