Physics 151: Lecture 9
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Announcements
Homework #3 (due this Fri. 9/22/06, 5 PM)
Homework #4 (due Mon. 10/2/06, 5 PM)

Review sessions:
Sept. 25, Mon. 8.10 PM - 9:00 pm in P-36

Today’s Topics:
Review of Newton’s Law 3 - Ch. 5.1-6
Some applications of Newton’s laws - Ch5.7
Friction - Ch5.8
Physics 151: Lecture 9, Pg 1
See text: Ch. 5
Review
Newton’s Laws 1, 2, 3

Isaac Newton (1643 - 1727) proposed three “laws” of motion:
Law 1: An object subject to no external forces is at rest or moves
with a constant velocity if viewed from an inertial reference
frame.
Law 2: For any object, FNET = F = ma
Law 3: Forces occur in pairs: FA ,B = - FB ,A
(For every action there is an equal and opposite reaction.)
Physics 151: Lecture 9, Pg 2
Lecture 9, ACT 1
A book is placed on a chair. Then a videocassette is placed
on the book. The floor exerts a normal force
A.
B.
C.
D.
on all three.
only on the book.
only on the chair.
upwards on the chair and downwards on the
book.
E. only on the objects that you have defined to be
part of the system.
Physics 151: Lecture 9, Pg 3
Lecture 9, ACT 2
Gravity and Normal Forces
A woman in an elevator is accelerating upwards
The normal force exerted by the elevator on the woman is,
A) greater than
B) the same as
C) less than
the force due to gravity acting on the woman
Physics 151: Lecture 9, Pg 4
Lecture 9, ACT 1b
Gravity and Normal Forces
A woman in an elevator is accelerating downwards
The normal force exerted by the elevator on the woman is,
A) greater than
B) the same as
C) less than
the force the woman exerts on the elevator.
Physics 151: Lecture 9, Pg 5
Lecture 9, ACT 1c
Gravity and Normal Forces
A woman in an elevator is downwards at constant speed
The tension on the rope holding the elevator is,
A) greater than
B) the same as
C) less than
the weight of the elevator and the woman ?
Physics 151: Lecture 9, Pg 6
Example-1 with pulley


Two masses M1 and M2 are connected by
a rope over the pulley as shown.
Assume the pulley is massless and
frictionless.
Assume the rope massless.
If M1 > M2 find :
Acceleration of M1 ?
Acceleration of M2 ?
Tension on the rope ?
T1
T2
Video
M2
Animation
M1
a
Free-body diagram for each object
Physics 151: Lecture 9, Pg 7
Example-2 with pulley


A mass M is held in place by a force F.
Find the tension in each segment of the
rope and the magnitude of F.
Assume the pulleys massless and
frictionless.
Assume the rope massless.
T4
T1
We use the 5 step method.
F
Draw a picture: what are we looking for ?
What physics idea are applicable ? Draw
a diagram and list known and unknown
variables.
Newton’s 2nd law : F=ma
T3
T2
<
T5
M
Free-body diagram for each object
Physics 151: Lecture 9, Pg 8
Pulleys: continued

FBD for all objects
T4
T2
T3
T4
T1
T3
T2
5
F
<
T
T5
T3
F=T1
T5
M
T2
M
Mg
Physics 151: Lecture 9, Pg 9
Pulleys: finally

Step 3: Plan the solution (what are the relevant equations)
F=ma , static (no acceleration: mass is held in place)
T5
T5=Mg
M
T2
T1+T2+T3=T4
T4
Mg
F=T1
T3
T2+T3=T5
T3
F=T1
T
5
T2
Physics 151: Lecture 9, Pg 10
Pulleys: really finally!

Step 4: execute the plan (solve in terms of variables)
We have (from FBD):
F=T1
T5=Mg
T2+T3=T5
T1+T2+T3=T4
Pulleys are massless and frictionless
T1=T3
T4
T2=T3
T2+T3=T5 gives T5=2T2=Mg
T1
T2=Mg/2

T2
F
T1=T2=T3=Mg/2 and T4=3Mg/2
T5=Mg
T3
and F=T1=Mg/2
Step 5: evaluate the answer (here,
dimensions are OK and no numerical values)
<
T5
M
Physics 151: Lecture 9, Pg 11
See text: 5.8
New Topic: Friction

What does it do?
It opposes motion!

How do we characterize this in terms we have learned?
Friction results in a force in a direction opposite to the
direction of motion!
j
N
FAPPLIED
ma
fFRICTION
i
mg
Physics 151: Lecture 9, Pg 12
See text: 6-1
Friction...

Friction is caused by the “microscopic” interactions
between the two surfaces:
Physics 151: Lecture 9, Pg 13
See text: 6-1
Friction...

Force of friction acts to oppose motion:
Parallel to surface.
Perpendicular to Normal force.
j
N
F
ma
fF
See figure 6-1
i
mg
Physics 151: Lecture 9, Pg 14
See text: 6-1
Model for Sliding Friction

The direction of the frictional force vector is perpendicular
to the normal force vector N.

The magnitude of the frictional force vector |fK| is
proportional to the magnitude of the normal force |N |.
 |fK| = K | N | ( = K | mg | in the previous example)
The “heavier” something is, the greater the friction will
be...makes sense!

The constant K is called the “coefficient of kinetic friction”.
(This friction is called Kinetic Friction)
Physics 151: Lecture 9, Pg 15
See text: 6-1
Model...

Dynamics:
i:
j:
F  KN = m a
N = mg
F  Kmg = m a
so
j
N
F
ma
K mg
i
mg
Physics 151: Lecture 9, Pg 16
Lecture 9, ACT 4
In a game of shuffleboard (played on a horizontal
surface), a puck is given an initial speed of 6.0
m/s. It slides a distance of 9.0 m before coming to
rest. What is the coefficient of kinetic friction
between the puck and the surface ?
A.
B.
C.
D.
E.
0.20
0.18
0.15
0.13
0.27
Physics 151: Lecture 9, Pg 17
Lecture 9, ACT 4

A box of mass m1 = 1 kg is being pulled by a
horizontal string having tension T = 40 N. It slides
with friction (k= .5) on top of a second box having
mass m2 = 2 kg, which in turn slides on an ice rink
(frictionless).
What is the acceleration of the second box ?
(A) a = 0 m/s2
(B) a = 2.5 m/s2
(C) a = 10 m/s2
T
a=?
m1
m2
slides with friction (k=0.5 )
slides without friction
Physics 151: Lecture 9, Pg 18
Lecture 9, ACT 3
Two-body dynamics

A block of mass m, when placed on a rough inclined plane
( > 0) and given a brief push, keeps moving down the plane
with constant speed.
If a similar block (same ) of mass 2m were placed on the
same incline and given a brief push, it would:
(a) stop
(b) accelerate
m
(c) move with constant speed
Physics 151: Lecture 9, Pg 19
Lecture 9, ACT 3
Solution

Draw FBD and find the total force in the x-direction
FTOT,X = mg sin q Kmg cos q
KN
= ma = 0 (first case)
Doubling the mass will simply
double both terms…net force
will still be zero !
j
N q
Speed will still be constant !
mg
i
q
Animation
Physics 151: Lecture 9, Pg 20
See text: Ch 5.8
Static Friction...

So far we have considered friction acting when something moves.
We also know that it acts in un-moving, “static” systems:

In these cases, the force provided by friction will depend on the forces
applied on the system.
N
F
j
i
fS
mg
Physics 151: Lecture 9, Pg 21
See text: Ch 5.8
Static Friction...

Just like in the sliding case except a = 0.
i:
F fS = 0
j:
N = mg

While the block is static: fS  F ! (unlike kinetic friction)
N
F
j
i
fS
mg
Physics 151: Lecture 9, Pg 22
See text: Ch 5.8
Static Friction...

The maximum possible force that the friction between two
objects can provide is fMAX = SN, where s is the
“coefficient of static friction”.
So fS  S N.
As one increases F, fS gets bigger until fS = SN and
the object “breaks loose” and starts to move.
N
F
j
i
fS
mg
Physics 151: Lecture 9, Pg 23
See text: Ch 5.8
Static Friction...

S is discovered by increasing F until the block starts to
slide:
i:
FMAX SN = 0
j:
N = mg
S  FMAX / mg
N
FMAX
Smg
j
i
mg
Physics 151: Lecture 9, Pg 24
See text: 6-1
Additional comments on Friction:

Since f = N , the force of friction does not depend on the
area of the surfaces in contact.

By definition, it must be true that
system (think about it...).
S > K
for any
Animation
Physics 151: Lecture 9, Pg 25
Example
Problem 5.40 from the book
Three blocks are connected on the table as shown. The table
has a coefficient of kinetic friction of 0.350, the masses are m1
= 4.00 kg, m2 = 1.00kg and m3 = 2.00kg.
m2
m1
a)
b)
T1
m3
What is the magnitude and direction of acceleration on the
three blocks ?
What is the tension on the two cords ?
Physics 151: Lecture 9, Pg 26
T12
T23
m2
T1
T1
2
T23
m1
T12
T12
m3
k m2g
a
m2g
m1g
T23
a
T23
m2
m1
a
m3
m3g
T23 - m3g = m3a
T12 - m1g = - m1a
-T12 + T23 + k m2g = - m2a
SOLUTION: T12 = = 30.0 N ,
T23 = 24.2 N
,a
= 2.31 m/s2 left for m2
Physics 151: Lecture 9, Pg 27
Recap of today’s lecture
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Newton’s Third Law
Friction
Reading for Wed., Ch 6.1, pp. 151-158
Newton’s Laws and Circular Motion
Homework #3 (due Fri. 9/22/06, 5 PM)
Homework #4 (due Mon. 10/2/06, 5 PM)

Review sessions:
Sep. 25, Mon. 8.10 - 9:00 pm in P-38
Physics 151: Lecture 9, Pg 28