Friction & Inclined Planes

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Transcript Friction & Inclined Planes

Friction & Inclined Planes
AP Physics
2013-2014
TWO types of Friction


Static – Friction that keeps an object at rest
and prevents it from moving
Kinetic – Friction that acts during motion
Force of Friction

The Force of Friction is F f  F N
  constant of proportion ality
directly related to the
Force Normal.
  coefficien t of friction

Mostly due to the fact
that BOTH are surface
forces
F sf   s F N
F kf   k F N
Note: Does friction depend on surface area? Discuss.
The coefficient of
friction is a unitless
constant that is
specific to the
material type and
usually less than
one.
Some Common Coefficients of Friction
The Key to Force Problems
1.
2.
3.
4.
5.
6.
7.
Draw an accurate, labeled FBD
Ask if the system is in equilibrium (∑F = 0 )
or accelerating
Identify the required unknown
Set up your equations, minding signs
Solve for the required unknown
NOW plug in values and evaluate.
Check units and check for reality
Coefficient of Friction Calculation


A 24-kg crate initially at rest on a horizontal floor requires a 75 N
horizontal force to set it in motion. Find the coefficient of static
friction between the crate and the floor.
A: Given
Fs, max = 75 N m = 24 kg
Unknown:
μs = ?
Solve: μs = Fs, max
Fn
μs = Fs, max
mg
= (75N)/(24 kg*9.8 m/s2)
=
0.32 (note: dimensionless)
Friction & Newton’s First Law
If the coefficient of kinetic friction between a 35-kg crate and the floor is
0.30, what horizontal force is required to move the crate to the right at
a constant speed across the floor?
Fa  F f
Fn
Fa
F f   k FN
Fa   k F N
FN  m g
Ff
Fa   k m g
mg
Fa  (0.30)(35)(9.8)
Fa 
102.9 N
Friction & Newton’s Second Law
Suppose the same 35 kg crate was not moving at a constant speed, but
rather accelerating at 0.70 m/s/s. Calculate the applied force. The
coefficient of kinetic friction is still 0.30.
FN ET  m a
Fa  F f  m a
Fn
Fa
Ff
Fa= - Ff
Fa   k F N  m a
Fa   k m g  m a
Fa  m a   k m g
Fa  (35)(0.70)  (0.30)(35)(9.8)
mg
Fa  127.4 N
Warm Up: FBD Practice

You take the elevator to the fourth floor.
What forces are acting on you while in
the elevator?
Warm Up: FBD Practice 2

You push two boxes
down the hall over a
rough floor. What
forces are acting on
box B?
Warm Up: FBD Practice 3

Box B accelerates constantly to the right
while box A sits on top of it. What forces are
acting on box A?
Stranger Than Friction

Students in an introductory physics lab design an experiment to determine
the coefficient of static friction between a 1.8 kg jack-o-lantern and the
tabletop on which it rests. Their setup is that they attach a massless string
to the jack-o-lantern, loop it over a pulley, and then attach the other end to a
pail in which the students will add sand as the experiment progresses.

The pail with sand was observed to weigh 11 N when the jack-o-lantern
began to slide.

A. Draw a free body diagram for this set up.
B. What is known? What are you solving for? What equations can you use?
C. Find the coefficient of static friction.


Stranger Than Friction Continued
Key to solving: The maximum force of static friction is equal
to the force exerted by the pail at the moment the jack-olantern begins to slide.
Fs, max= μsN
μs = Fs, max = W
N mg
After substituting values…..
μs = 0.61
Reality Check: Coefficients of friction are always between 0
and about 1 and are dimensionless
Static and Kinetic Friction Lab

Today: 1. Collect data
2. Analysis questions 1-3

Tomorrow: 1. Graphical analysis
2. Conclusions
Inclines

Ff
FN
mg cos 


mg 
mg sin 


Tips
•Rotate Axis
•Break weight into components
•Write equations of motion or
equilibrium
•Solve
Friction & Inclines
A person pushes a 30-kg shopping cart up a 10 degree incline with a force of 85
N. Calculate the coefficient of friction if the cart is pushed at a constant
speed.
Fa  F f  m g sin 
Fa
Fn
Fa   k F N  m g sin 
F f   k FN
F N  m g cos 
Fa   k m g cos   m g sin 
Fa  m g sin    k m g cos 
m g cos 
k 

Ff
mg
m g sin 

k 
Fa  m g sin 
m g cos 
85  (30)(9.8)(sin 10)
(30)(9.8)(cos 10)
 0.117
Example
A 5-kg block sits on a 30 degree incline. It is attached to string that is thread
over a pulley mounted at the top of the incline. A 7.5-kg block hangs
from the string.
 a) Calculate the tension in the string if the acceleration of the system is 1.2
m/s/s
 b) Calculate the coefficient of kinetic friction.
T
FN
FN E T  m a
m1 g  T  m1a
m2gcos30
30
T
T  ( F f  m 2 g sin  )  m 2 a
m2g
m1
30
m2gsin30
m1g
Ff
F N  m 2 g cos 
Example
FN ET  m a
m1 g  T  m1 a
m1 g  m1 a  T
T  ( F f  m 2 g sin  )  m 2 a
T  F f  m 2 g sin   m 2 a
T   k F N  m 2 g sin   m 2 a
T  m 2 a  m 2 g sin    k F N
(7.5)(9.8)  (7.5)(1.2)  T T  m a  m g sin 
2
2
 k
T  64.5 N
F
F N  m 2 g cos 
N
T  m 2 a  m 2 g sin 
m 2 g cos 
 k
64.5  (5)(1.2)  (5)(9.8)(sin 30)
(5)(9.8)(cos 30)
 k  0.80 N
 k
10/4/13 Exit Ticket
On a half sheet of paper:
Explain why pushing downward on a book as
you push it across a table increases the force
of friction between the table and the book.
(Use vocabulary from this unit to answer this question!)