Transcript Physics 131: Lecture 9 Notes

Physics 151: Lecture 10

Homework #3 (9/22/06, 5 PM) from Chapter 5
Today’s Topics:
Example with a pulley and kinetic
Static Friction
Circular motion and Newton’s Laws - Ch 6
Physics 151: Lecture 10, Pg 1
Example with pulley and kinetic friction
Problem from the textbook
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 ? a = 2.31 m/s2
What is the tension on the two cords ?
T12 = = 30.0 N , T23 = 24.2 N
Physics 151: Lecture 10, Pg 2
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 10, Pg 3
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 10, Pg 4
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
Physics 151: Lecture 10, Pg 5
Newton’s Laws and Circular Motion
(Chapter 6)
v
aC
R
Centripetal Acceleration
aC = v2/R
What is Centripetal Force ?
FC = maC = mv2/R
Physics 151: Lecture 10, Pg 6
See Example 6.2
Example Problem
I am feeling very energized while I shower. So I
swing a soap on a rope around in a horizontal
circle over my head. Eventually the soap on a
rope breaks, the soap scatters about the shower
and I slip and fall after stepping on the soap. To
decide whether to sue ACME SOAP I think
about how fast I was swinging the soap
(frequency) and if the rope should have
survived. From the manufacturers web site I find
a few details such as the mass of the soap is 0.1
kg (before use), the length of the rope is 0.1 m
and the rope will break with a force of 40 N.
(assume FBS is large versus the weight of the soap)
Physics 151: Lecture 10, Pg 7
See Example 6.2
Example Problem
Step 1: we need to find the frequency of the
soap’s motion that caused the rope to break.
Step 2 : Diagram.
v
T
Step 3 – Solve Symbolically
Step 4 – Numbers
f~ 10 rev./s
Step 5 –It seems that the suit is in trouble. Being able to
twirl your soap safely 10 rev/s is pretty good.
Physics 151: Lecture 10, Pg 8
Lecture 10, ACT 2
Circular Motion Forces

How fast can the race car go ?
(How fast can it round a corner with this radius of curvature ?)
mcar = 1500 kg
S = 0.5 for tire/road
R = 80 m
R
A) 10 m/s
B) 20 m/s
C) 75 m/s
D) 750 m/s
Physics 151: Lecture 10, Pg 9
Lecture 10, ACT 2
Circular Motion Forces

This is just like the soap on a rope problem but friction
replaces the tension in the rope as the centripedal force.
N
F?
mcar = 1500 kg
S = 0.5 for tire/road
R = 80 m
mg
v  (0.5)(10
m
)(80m)  400  20(m /s)
s2
Answer is (B)
A) 10 m/s
B) 20 m/s
C) 75 m/s
D) 750 m/s

Physics 151: Lecture 10, Pg 10
An example before we considered a race car going
around a curve on a flat track.
N
Ff
mg
What’s differs on a banked curve ?
Physics 151: Lecture 10, Pg 11
Banked Corners
Free Body Diagram for a banked curve.
N
ma
Ff
CAR
mg
For small banking angles, you can assume that Ff is
parallel to ma. This is equivalent to the small angle
approximation sinq = tanq. Can you show that ?
Physics 151: Lecture 10, Pg 12
Lecture 10, ACT 3


Because of your physics background, you have been hired as a
member of the team the state highway department has assigned to
review the safety of Connecticut roads. This week you are studying US6, trying to redesign the section from Willimantic to I-384 so fewer folks
die on it. The proposed new road has a curve that is essentially 1/8 of a
circle with a radius of 0.3 miles (1/2 km). The road has been designed
with a banked curve so that the road makes an angle of 4° to the
horizontal throughout the curve. To begin the study, the head of your
department asks that you calculate the maximum speed (in km/hr) for a
standard passenger car (a little more than 2000 lbs or about 1,000 kg )
to complete the turn while maintaining without sliding off the road. She
asks that you first consider the case of a slick, ice and slush covered
road. When you have completed that calculation she wants you to do
the case of a dry, clear road where the coefficient of kinetic friction is
0.50 and the coefficient of static friction is 0.70 between the tires and
the road. This will give her team the two extremes of Connecticut
driving conditions on which to base the analysis.
What is the maximum speed you can drive through the curve for
A) dry road condition, and for B) icy road condition ?
Physics 151: Lecture 10, Pg 13
Lecture 10, ACT 3
Solution
B) For icy road => no friction !

Draw FBD and find the total force in the x-direction
y
x
v = 18.5 m/s ~ 40 mi / hr
r =500m
Q = 4o
k =0.5
s =0.7
M =1,000 kg
mg sin q
q
q
N
mg
Physics 151: Lecture 10, Pg 14
Lecture 10, ACT 3
Remember
Static Friction

While the block is static:
a = 0, fS  F
fS,max  s N
N
Kinetic Friction

The object is moving :
a = 0 , fk < F
fk  k N
N
F
F
fS
mg
fk
mg
dry road => static friction keeps objects from moving
Physics 151: Lecture 10, Pg 15
Lecture 10, ACT 3
Solution
A) dry road => static friction keeps objects from moving
y

Draw FBD and find the total force in the x-direction
v = 61.3 m/s ~ 120 mi / hr !
r =500m
Q = 4o
k =0.5
s =0.7
M =1,000 kg
x
sN
mg sin q
q
N
mg
Physics 151: Lecture 10, Pg 16
Lecture 10, ACT 4

When a pilot executes a loopthe-loop (as in figure on the
right) the aircraft moves in a
vertical circle of radius R=2.70
km at a constant speed of v=225
m/s. Is the force exerted by the
seat on the pilot:
A) Larger
B) Same
C) Smaller
then pilot’s weight (mg) at :
I) the bottom and
II) at the top of the loop.
2.7 km
Physics 151: Lecture 10, Pg 17
Lecture 10, ACT 4
Solution
II)
ANSWER (C)
NII
mg
Fc
A) Larger
B) Same
C) Smaller
Fc
2.7 km
NI
I)
Fc
mg
ANSWER (A)
Physics 151: Lecture 10, Pg 18
Example
Gravity, Normal Forces etc.
Consider a women on a swing:
Animation
1. When is the tension on the rope largest ?
2. Is it :
A) greater than
B) the same as
C) less than
the force due to gravity acting on the woman
(neglect the weight of the swing)
Physics 151: Lecture 10, Pg 19
Recap of today’s lecture

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


Example with pulley
Forces and Circular Motion - text Ch 6.1
Reading for next class: Section 6.2 and 6.3
Homework #3 (due 9/22/06, 5 PM) from Chapter 5
Homework #4 (due 10/02/06, 5 PM) from Chapter 6
Physics 151: Lecture 10, Pg 20