Download Friction

Friction
Students
•  Recitation homework due Thursday/Friday
•  Exam 1 scores are on D2L
•  There may be a curve for the final score but there are no
individual curves for each exam.
•  Questions on the answers and solutions can be addressed to
professors or TA’s.
•  Questions on
70
Average: 71
your score should
60
be addressed to
Median: 72
Professor Kinney. 50
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Exam scores
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90 1 100
About the exam
•  I believe the exam was a fair representation of what was
covered in lecture, CAPA, and recitation. It was also
similar to the practice exams on D2L.
•  I expect all of you to be able to solve every problem on the
exam.
•  You should go over the solutions to make sure you
understand the problem and where you might have gone
wrong so that you can avoid the same mistakes in the
future.
•  At least two questions from each midterm will appear on
the final
2
Learning assistant info session
•  There will be an info session on the LA program on
Monday, February 23 from 5 to 6:30pm in UMC 235.
•  Refreshments will be served.
•  Great for people interested in teaching and learning
about education. Nearly all participants find it to be
very rewarding. Feel free to chat with your LA’s during
recitation about their experiences.
•  Applications are accepted between 2/23 and 3/9.
•  More info available by contacting Steve Pollock at
[email protected] or by visiting the web
site: laprogram.colorado.edu
3
Friction
Experiments have found two main types of sliding friction
Static friction is the force exerted when the two objects
are not in motion relative to each other (no slipping)
Kinetic friction is the force exerted when the two
objects are in motion relative to each other (slipping)
Force of kinetic friction is fk = µk n.
It depends on µk which depends
on the materials and on the
normal force n which is the force
pushing the materials together.
v
m
µk is the coefficient of kinetic friction
n
fk = µk n
m
Fg = mg
4
Static friction
Formula for static friction is similar: f s ≤ µs n.
The static friction will oppose forces which try to slide one
object across another. However, once the maximum
possible static friction force (µs n) is exceeded the object
will “break loose” and start moving.
If the block is not moving then:
Fext
m
f s = Fext ≤ µs n
n
m
Fext
µs is the coefficient of static friction Fg = mg
5
Static and Kinetic Friction
Start with object at rest feeling a normal force of n
Start applying a force perpendicular to n.
Up to a force of µsn static friction prevents movement
Force of friction
After movement starts, frictional force is reduced to µkn.
fs = µsn
kinetic friction
fk = µk n
Applied force
6
Clicker question 1
Set frequency to BA
A stationary block rests on an inclined plane. The coefficient
of static friction is µs = 0.2. The normal, gravity, and static
friction forces are shown. Which set of relations applies to
this situation?
n y
f
A. n = mg tan θ and f = µsn
x
B. n = mg and f = µs n
mg
θ
C. n = mg cosθ and f = mg sin θ
D. n = mg sin θ and f = mg cosθ
mg cosθ
θ 90 − θ
mg sin θ
mg
θ
Fnet,x = mgsin θ − f = 0 so f = mgsin θ
Fnet,y = n − mg cosθ = 0 so n = mg cosθ
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Propulsion
To walk (or run), we push against the
earth with leg muscles and by the 3rd
law, the earth pushes against us.
Because of this, it is impossible
to walk on a frictionless surface.
A car works the same way with
tires instead of feet.
Is it static friction or kinetic friction that
propels us when walking or driving?
Static. The two surfaces are not sliding.
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Clicker question 2
Set frequency to BA
A thief forgot to bring a safe cracking kit or dolly so is trying to
drag a 200 kg safe up a ramp at an angle of θ=20°. The safe is
currently not moving while he pulls on the rope with a force of 400
N. What is the magnitude and direction of the force of friction on
the safe?
Free clicker question. Full
credit for any answer!
A. Friction force is down the ramp
B. Friction force is up the ramp
C. Friction force is zero
D. Need to know the coefficient of friction
E. Need to know the mass of the thief
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The 200 kg safe is not moving while the thief pulls on the rope
with a force of 400 N. What is the magnitude and direction of the
force of friction on the safe (use g=10 m/s2)?
Always need a free body diagram
T
n
f
Pick a coordinate system
Fg = mg
20°
Decompose forces into x & y components
With our coordinate choice, only Fg,x = mgsin θ
Only now do we
Fg,y = mg cosθ
need to worry about gravity
apply Newton’s
laws and solve. Fnet,x = mgsin θ + f − T and max = 0
mg sin θ + f − T = 0 so that f = T − mg sin θ
Fnet,x = max
Fnet,y = may f = 400 N − 200 kg ⋅10 m/s2 ⋅ sin 20° = −284 N
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It is in the opposite direction!
Clicker question 3
Set frequency to BA
A mass slides down a rough inclined plane with some non-zero
acceleration of a1. The same mass is shoved up the same incline
with a large, brief initial push. As the mass moves up the incline,
its acceleration is a2. How do a1 and a2 compare?
v1
A.  |a2| > |a1|
B.  |a2| < |a1|
C.  |a2| = |a1| in the same direction
D.  |a2| = |a1| but directions are opposite
v2
The force of gravity is the same: mg sin θ down the slope
The force of friction has the same magnitude in both
cases but opposite direction (opposite the velocity)
In 2, both forces are in the same direction so
net force is greater so acceleration is greater.
11
Cornering problem
A car is taking a corner with radius of 100 m at a speed of 56 mph
(25 m/s). On dry pavement the rubber/asphalt coefficient of static
friction is 0.8. Will the car hold the line or start skidding?
Need 3D coordinate
system for these
types of problems.
t
r
May need two free body diagrams
for a single object due to 3D nature.
z
r
There are no forces in the t direction
so only need a rz free body diagram.
n
f
z
mg
r
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Cornering problem
A car is taking a corner with radius of 100 m at a speed of 56 mph
(25 m/s). On dry pavement the rubber/asphalt coefficient of static
friction is 0.8. Will the car hold the line or start skidding?
n
f
mg
z
r
We have the free body diagram so now
we can apply Newton’s 2nd law in r and z.
Sum the forces in z: Fnet,z = n − mg
Acceleration in z is 0 so
Sum the forces in r : Fnet,r = f
so f = marad
n − mg = 0 so n = mg
The acceleration in r is arad
2
v
but remember arad = r
2
mv
so f =
r
13
Cornering problem
A car is taking a corner with radius of 100 m at a speed of 56 mph
(25 m/s). On dry pavement the rubber/asphalt coefficient of static
friction is 0.8. Will the car hold the line or start skidding?
So far we have found
n = mg
2
mv
and f =
r
We also know that for static friction: f s ≤ µs n
The maximum possible friction force is µsn. Is this
sufficient to keep the car going around the curve?
fmax,s = µs n = µs mg = 0.8 ⋅ m ⋅ 9.8 m/s 2 = m ⋅ 7.84 m/s 2 (max friction force)
2
m ⋅ (25 m/s)
mv
f =
=
= m ⋅ 6.25 m/s2
r
100 m
2
(actual friction force needed)
So f ≤ µs n and therefore the car will hold the corner.14