Physics 106 Lesson #2
Static Equilibrium
Dr. Andrew Tomasch
2405 Randall Lab
atomasch@umich.edu
Last Time: Scalars & Vectors
• Scalars: magnitude only (ex: radius R =10 cm)
• Vectors: magnitude and direction
– Arrow length ≡ magnitude of the vector
– Arrow direction ≡ vector direction
Example:
d
Displacement ≡
1m
d
d
= 1 m to the right
magnitude
direction
The Net (Total) Force
• Because the
washer is at rest
the net (total) force
on it is zero
• The net force is the
vector sum of all
the forces acting on
the object
• How do we add
vectors?
Spring Force
Fs= kx
?
Washer
Weight
W = mg
Vector Addition
Caution
Quiz
Ahead
A B ?
• Place vectors “tip to tail” and draw an
arrow from the start (tail) of one vector
to the end (tip) of the other
B
• How about A-B ?
A+B
-B
A
A-B
A
A-B = A+(-B)
B
Caution
Quiz
Ahead
Concept Test #1
Two vectors, one with magnitude 3 m and
the other with magnitude 4m, are added
together. The resultant vector could have a
magnitude as small as: 1 m
4m
A) 1m
B) 3m
C) 5m
D) 7m
3m
The resultant has the
largest magnitude when
the two vectors are
parallel and the smallest
when they are antiparallel.
Concept Test #2
Two vectors have
unequal magnitudes.
Can their sum be zero?
0
• Yes
• No
The smallest possible vector sum
Is when the two vectors are antiparallel.
The only way for this sum to be zero is
If the vectors have equal magnitudes.
Equilibrium Condition for a Body at Rest
• For a body to be in
equilibrium and
remain at rest:
• An example:
F  0
Spring Force
F  0
Fs= kx
Shorthand for
the net force:
“The vector sum
of all forces”
Weight
W = mg
Washer
+ =0
Torque: t
A force which points through the rotation
axis produces no torque because r = 0!
• Torque t (a vector): Magnitude ≡ the
product of a perpendicular force and the
distance r to a rotation axis:
t  t  rF
t  "The magnitude
(length) of vector t "
F is perpendicular to r
r
F
The distance r is also
called the lever arm
Torque can also be defined more generally for
forces that are not perpendicular to r by
 expressing F as the sum of two component
vectors, one parallel to the door and one
perpendicular to it. Only the perpendicular
component vector produces a torque about the

hinge to make the door rotate.
F
F t F r
t
Torque Continued
• Direction (looking down the rotation axis):
– Counterclockwise rotation is “(+)” (positive)
– Clockwise rotation is “(-)” (negative)
– Direction in space by Right Hand Rule (RHR)
• Why a perpendicular force F ?
– Because only a perpendicular force can produce a
rotation
• Units:
N m
t  rF
Caution
Quiz
Ahead
points
t
out of page
F
Arrow
out of
page
Arrow
into
page
Concept Test #3
Both r and F remain
unchanged. F does not
change magnitude or
direction → torque
remains unchanged.
You are trying to open a door that is
stuck by pulling on the knob in a
direction perpendicular to the door. If
you instead tie a rope to the knob and
pull with the same magnitude force in
the same direction, does the torque you
exert increase?
•
•
Yes
No
r
F
Concept Test #4
You are using a wrench to try to tighten a nut.
Which of the arrangements shown is most
effective for tightening the nut? List in order of
decreasing effectiveness. (The “rod” can be
used to extend the wrench handle length). The
force F is the same in each case.
F
F
r
rod
F
F
r
r
rod
r
3
1
2
4
The longest lever arm r
exerts the largest torque
for the same applied force
A. 2 > 1 > 3 > 4
B. 2 > 1 = 4 > 3
C. 4 > 2 > 1 > 3
D. 2 > 3 > 4 > 1
Equilibrium for Rigid Bodies
• Extended objects
which do not change
shape are called rigid
bodies
• For a rigid body to be
in equilibrium:
F

0

t

0

• For a rigid body to
remain stationary in one
place, the net force
acting on it must be zero
• For a rigid body to
remain stationary and
not rotate the net torque
about any axis through
the body must be zero
• The net force/torque is
the vector sum of all
forces/torques acting on
the body (analyze with
Free Body Diagram)
Center of Mass/Gravity
•
•
•
•
•
There is a special geometrical point associated with
any rigid body (extended object) where we can attribute
the body’s mass and weight as acting at that point
This special point is called the center of mass (CM) and
the center of gravity (CG) of the object
The definitions for center of mass and center of gravity
are different, but if the acceleration of gravity g is
constant over the extent of the object, then the center
of mass and the center of gravity are located at the
same geometrical point. For uniform objects this will
be the geometrical center of the object
The motion of any rigid body of mass m can be
described as the trajectory through space of a point
mass m located at its center of mass on which is
superposed rotation of the object about the center of
mass
For the purpose of calculating torques, all the weight of
the extended object can be assumed to act at the center
of gravity
Demo: Dumbbell With Lights
Where Is the Center of
Gravity of This Yummy
Donut?
Center of
Gravity (CG)
The Center of gravity
does not need to be
located inside the
object. Sometimes the
CG can be located in a
surprising place…
It is at the center of
the circular ring,
half way from the
bottom of the
donut - where
there is no dough!
After A Bite…
Center of
Gravity
If it falls within the object, the center of gravity
corresponds to the point where an object balances.
Wine Bottle Balance
• Treat the wine rack and
the bottle as a rigid body
• Two forces:
– Supporting force FN from
table
– Weight at the center of
gravity of the system
• Net torque = 0
→No rotation →
• The center of gravity is
located exactly above the
point where the
supporting force acts
• The supporting force FN
from the table is also
called the “Normal”
(perpendicular) force
CG
W
FN
Axis of rotation
Neither force produces a torque
about the Axis of Rotation since
both forces point through the axis
so that r = 0 for both forces
Wine Bottle Balance
• The wine bottle
and seesaw are
balanced in the
same way!
FN
CG
Axis of rotation
W
CG
W
FN
Axis of rotation