Static
Conditions of Equilibrium:

Ftot  0

 tot  0
F  0
F  0
F  0
  0
  0
  0

a0

 0
x
y
z
x
y
z
Static Equilibrium:

v 0

 0
Example: Equilibrium and torque
1kg
2kg
1m
1kg
1   2  0
 1   2 F1l1  F2l2
1m
2kg
2m
Not in equilibrium
1m
In equilibrium
1   2  0
1   2
F1l1  F2l2
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!
1   2   3  0
Wd  f1L  0
When d = 20 cm, the scale
reads 2 lb exactly. If L = 1 m,
L
1m
W  f1  2 lb 
 10 lb
d
0.2 m
How big is your turkey?
f1
Scale
L
d
f2
P
Light rod
Turkey
W
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?
 tot  0
φ =30°
φ
θ = φ + φ = 60°
T
θ
θ
sin 30  
φ
sin 60  
mg
T
1
2
3
2
3
2
mg  1.273 400 N  346 N

Ftot  0
φ =30°
φ
Fy
TL sin 30  mg L / 2 sin 60  0
Fx  T sin 30  0
Fx
T
θ
θ
Tcos30°
φ
Tsin30°
mg
Fy  T cos 30  mg  0
Fx  T / 2  173N
Fy  mg  T cos 30  mg 
3
2
mg
3
2
 14 mg  100 N
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.
F  0
F  0
  0
x
y
m
d
Mg  
 L 2M
N2  fs
fS  N2  0
N1  Mg  mg  0
N1  M  m g
N2
L
Mgd sin   mg sin   N2L cos   0
2
φ

 tan   N 2  f s   s M  m g

m  d
m 

tan    s 1   

 M  L 2M 
mg
1
C is a good point to use as axis of rotation because
then two of the forces will produce zero torque.
N1
C
Mg
d
fS
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.
r
r
mg
torque is not zero
mg
torque is zero


 

 

   ri  mi g    mi ri  g  MrCM  g  rCM  Mg 

Example: Three boxes are placed on identical inclines. Friction
prevents them from sliding. The boxes are not uniform and their
center of mass are indicated by the blue dot in each case. In which
cases does the box tip over?
A. All
1
B. 2 & 3
C. 3 only
2
In 3, the CM is not above the base: net ≠ 0
3
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?
The center of mass of the plank+pig system must not be beyond
the edge.
Position of the CMall relative to the edge:
xCM
(40 kg)( 0.5 m)  (50 kg)x

0
90 kg
 20  50x  0
x 
2
 0.4 m
5
CMpig
40 kg
CMplank 0.5 m
2m
x
1m
50 kg
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).
2d
CM
a
h M
When the truck has no acceleration (or is at rest),
the normal force is right below the CM.
F  0
  0
y
N  Mg
(trivial)
CM
Mg
N
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”
CM
a
Mg
N
When the fridge is just
starting to tip over, the
normal acts on the
bottom right edge of
the fridge.
fS
 F  Ma
F  0
  0 
CM
a
Mg
N
fS
fS  Ma
x
Mah  Mgd
N  Mg
y
f
 N  0
f s h  Nd
a  gd / h
The fridge tips over because the normal force cannot produce a torque larger than Nd.