Announcements
• CAPA Set #5 due today (Friday) at 10 pm
• CAPA Set #6 now available, due next Friday
• Next week in Section 
Lab 2: Acceleration due to gravity
Make sure to do the pre-lab and print materials before lab
• Complete reading all of Chapter 4
• Over the weekend, read Chapter 5 Sections 5.1-5.5
Rock Climber hanging by rope
Object of mass m suspended by a cord.
+y
T
m
Fg=mg
Tension in the cord is
equal to the weight
of the object.
Stationary: v = 0 and a = 0.
0 = ma = Fnet
0 = T - mg
T = mg
Clicker Question
+y
T
v
m
mg = Fg
Room Frequency BA
Person of mass m suspended by a rope.
If v=0:  T  mg  Fg
Now, imagine the person is being raised
at a constant velocity v > 0.
What’s the relation between T and mg?
v = constant, a = 0 again, as in the stationary case.
A) T = mg
0  ma  T  Fg  T  mg
B) T > mg
C) T < mg
 T  mg
D) Indeterminate from information given.
Clicker Question
Room Frequency BA
Object of mass m suspended by a cord.
+y
T
a
m
-mg = Fg
Before: v = constant  T  mg  Fg
Now, imagine the object is being raised
with an acceleration a > 0.
What’s the relation between T and mg?
0  ma  T  Fg  T  mg
A) T = mg
B) T > mg
 T  m(a 
C) T < mg
D) Indeterminate from information given.
g)  mg
The Atwood machine was
invented in 1784 by Rev. George
Atwood as a laboratory
experiment to verify the
mechanical laws of motion with
constant acceleration.
http://en.wikipedia.org/wiki/Atwood_machine
Atwood Machine (Pulley)
+x
Two objects with masses
M > m suspended from a
stationary pulley.
M
m
+x
Step #1: Choose a coordinate system.
+x
M
m
Odd coordinate system (curves around pulley).
This choice means that the acceleration a for both
masses will be the same (direction + magnitude).
Step #2: Specify the forces (free-body diagram)
-T
+T
+x
M
m
+x
Fg=+Mg
Ma M  Mg  TM
Fg=-mg
ma m  Tm  mg
Accelerations a must be same for both objects,
otherwise the string would stretch or break.
Room Frequency BA
Clicker Question
Consider the pair of suspended masses.
Assume that the pulley is frictionless and
the cord does not stretch.
M>m
TM
Tm
What is the relationship between the
tension TM in the cord above larger mass
M and the tension Tm above the
smaller mass?
A) TM = Tm
B) TM > Tm
C) TM < Tm
Tension along the
cord is constant.
Step #3: Solve the Equations
FM  MaM  Mg  T
Fm  ma m  T  mg
T  Mg  Ma
a M  am
Two equations.
Two unknowns (a, T)
T  mg  ma
Check if the answer makes sense…..
M m
a
g

M  m
Room Frequency BA
Clicker Question
Frictionless table & pulley.
+x
M1
+T
1)Choose coordinates
2)Identify forces.
3) Write down F = ma
for each object
-T
M2
Fg = M2g
Mass 1: M1a = T
Mass 2: M2a = M2g - T
+x
What is F = ma for mass M2?
(A) M 2 a  T
(B) M 2 a  T  M 2 g
(C) M 2 a  M 2 g  T
(D)M 2 a  M 2 g
Frictionless table & pulley.
+x
M1
+T
1)Choose coordinates
2)Identify forces.
3) Write down F = ma
for each object
-T
M2
Fg = M2g
Mass 1: M1a = T
Mass 2: M2a = M2g - T
+x
M2 a = M2g –M1a
Substitute for
T in the lower
equation
Solve for a
 M2 
a
g

 M1  M 2 
Pulley Advantage
In a system with multiple pulleys
(for which there are various possible arrangements),
we loop a single continuous rope.
There is only one tension in the rope throughout its length.
Clicker Question
Room Frequency BA
4 Ropes
with
Tension T
pulling up
Mg
What is the force (EFFORT) requires to prevent the LOAD from falling?
A) 0 lbs.
B) 25 lbs.
C) 50 lbs.
Hint: Draw a free-body diagram.
D) 100 lbs.
E) 400 lbs.
Demonstration
Friction
Why have you been ignoring me all this time?
Force resisting the movement of two objects in physical
contact past each other.
Force of Friction depends on:
1. Characteristics of materials
2. Vertical force pushing the materials together