Announcements
• CAPA #11 due this Friday at 10 pm
• Reading: Chapter 9
• Section – this week Lab #4, next week Lab #5 (no prelab)
• Midterm Exam #3 on Tuesday November 8th, 2011
 details given on course web page “exam info”
 practice exam and solutions on CULearn
 formula sheet and info. posted on web page
• Fraction of all clicker questions answered posted
on CULearn. Email me with your clicker ID, name,
student ID if you believe it is incorrect.
Static Equilibrium
A mass m is hanging (statically) from two strings.
The mass m, and the angles α and β are known.
What are the tensions T1 and T2?
Note: No lever arm. Thus no torques.
Two equations with two unknowns,
can solve for T and T after some algebra.
Static Equilibrium Problem: 4 Step Process
y
x
N
CCW= +
Axis of rotation
Mg
mg
#1 – Draw the free body diagram identifying all forces
and exactly where they act.
#2 – Label the coordinate axis and axis of rotation being
considered (and the sign convention)
Static Equilibrium Problem: 4 Step Process
y
x
N
CCW= +
Axis of rotation
Mg
mg
#3 – Write out Fnet = ma and tnet = Ia equations…
Fnet,x = (M+m)ax = 0 (no forces in this direction)
Fnet,y = (M+m)ay = 0 = +N – Mg – mg
tnet = Ia = 0 = +(Mg)D1 – (mg)D2
#4 – Solve…
Clicker Question
Room Frequency BA
We treat the force of gravity (Mg) as
if it acts only on the center-of-mass
position of the object.
Approximately in the middle of the pencil.
Assume pencil of length L and mass M
and with q=60 degrees.
L/2
Mgcos(60)
N
Mgsin(60)
q
Mg
What is the torque around the tip of the pencil at this time?
A) -(N-Mg) sin(60) L
B) -Mg cos(60) L/2 – N cos(60) L
C) -Mg sin(60) L/2
D) -Mg cos(60) L/2
E) -Mg sin(60) L
tnet = r Fperp = -(L/2) (Mgcos(60))
Stability and Balance
If only the gravitational force and the
Normal force are acting (as shown), if
the center-of-mass is not directly over
the point of contact, the system is
unstable (will fall over).
System is in static equilibrium.
Called unstable equilibrium
(slightest change and it falls)
System is in static equilibrium.
Called neutral equilibrium.
Clicker Question
Room Frequency BA
Case 1
A)
B)
C)
D)
E)
Case 2
Which of the following are correct (assume in both cases
the sphere is at rest at this moment)?
Case 1 and Case 2 are in neutral equilibrium
Case 1 is in unstable equilibrium and Case 2 is just unstable
Case 1 is in neutral equilibrium and Case 2 is in neutral eq.
Case 1 is in neutral equilibrium and Case 2 is just unstable
None of the above
Tipping Point
If force of gravity applied at the center-of-gravity (CG)
points outside the point of balance,
the object will fall (unstable).
U.S. Consumer Product Safety Commission finds approximately
2.18 deaths per year from vending machine tipovers.
Clicker Question
Room Frequency BA
The leaning tower of Pisa is
55 m high and leans 4 m off
vertical at the top. Its base is
7 m wide.
4.0 m
55 m
CM
How much further could it
lean horizontally (at the top)
before it would fall over?
A) 1.5 m
B) 2 m
C) 3 m
D) 4 m
7m
E) 6 m
Ignoring any small forces holding it to the ground.
A sign with mass ms is hung from a uniform bar of
mass mb and length L.
The sign is suspended ¾ of the way from the pivot.
The sign is held up with a cable at an angle θ.
How strong a cable is required (i.e. what is tension T)?
Step #1: Force Diagram
Step #2: Coordinate System
y
x
CCW= +
Clicker Question
Room Frequency BA
A sign of mass ms is hung from a uniform horizontal
bar of mass mB as shown.
What is the sign of the x-component of the force
exerted on the bar by the wall?
A) Positive
B) Negative
C) Fwx = 0.
Using Σ Fx = 0, the hinge force must have
a positive x-component, in order to
cancel the negative x-component of the
tension force.
y
x
CCW= +
Step #3: Static Equilibrium condition F=ma=0 and t=Ia=0
Fnet , x  ma x  0  Fwx  T cosq
Fnet, y  may  0  Fwy  T sin q  mB g  ms g
Not enough information to solve for T, Fwx, Fwy
(2 constraint equations and 3 unknowns)
t net  Ia  0  (mB g )( L / 2)  (ms g )(3L / 4)  (T sin q )( L)
t net  Ia  0  (mB g )( L / 2)  (ms g )(3L / 4)  (T sin q )( L)
(2mB  3ms ) g
T
4 sin q
Fnet , x  ma x  0  Fwx  T cosq
Fwx
(2mB  3ms ) g  cosq 
 T cosq 


4
 sin q 
Fnet, y  may  0  Fwy  T sin q  mB g  ms g
Fwy
(2mB  3ms ) g
 (mB  ms ) g  T sin q  (mB  ms ) g 
4