Download Lecture 29: Friction Examples

Introduction to Mechanics
Applying Newton’s Laws
Friction Examples
Lana Sheridan
De Anza College
Mar 1, 2016
Last time
• kinds of forces and problem solving
• normal force as a contact force between objects
• objects accelerated together
• friction
Overview
• friction examples
Friction
There are actually two types of friction:
• kinetic (moving)
• static (stationary)
Kinetic Friction
kinetic friction ∝ normal force
S
n
Motion
S
S
fk
fk = µk n
F
S
mg
µk is the coefficient of kinetic
friction
Some types of forces
Kinetic Friction
The friction force always acts to oppose motion of the surfaces
relative to each other. That means the kinetic friction Fkf
always points opposite to the velocity vector.
Friction
Static Friction
S
max. static friction ∝ normal force
n
S
S
fs
F
fs 6 µ s n
µs is the coefficient of static friction
S
mg
magnitude of the
applied force.
Friction
can breaks free and
accelerates to the right.
S
S
n
n
S
fs
S
fk
F
S
F
S
mg
a
Motion
S
S
mg
b
S
|f |
fs,max
fs
!
F
fk ! mkn
O
c
F
Static region
Kinetic region
Friction Example 1
For waxed wood on wet snow, µs = 0.14 and µk = 0.1. You pull
on a sled of mass 10 kg that is at rest initially. You exert a force of
5 N on the sled. What is the magnitude of the static frictional
force that acts on the sled?
Friction Example 1
For waxed wood on wet snow, µs = 0.14 and µk = 0.1. You pull
on a sled of mass 10 kg that is at rest initially. You exert a force of
5 N on the sled. What is the magnitude of the static frictional
force that acts on the sled?
Fs,max = µs n
= µs mg
= (0.14)(10 kg)g
= 13.7 N
13.7 N > 5 N, so the sled will remain at rest.
If the sled is at rest and remains at rest, it does not accelerate.
Fnet = 0.
Friction Example 1
For waxed wood on wet snow, µs = 0.14 and µk = 0.1. You pull
on a sled of mass 10 kg that is at rest initially. You exert a force of
5 N on the sled. What is the magnitude of the static frictional
force that acts on the sled?
Fs,max = µs n
= µs mg
= (0.14)(10 kg)g
= 13.7 N
13.7 N > 5 N, so the sled will remain at rest.
If the sled is at rest and remains at rest, it does not accelerate.
Fnet = 0.
Fnet,x = 0 = 5 N − Fs
Fs
= 5N
Friction Example 2
For waxed wood on wet snow, µs = 0.14 and µk = 0.1. You pull
on a sled of mass 10 kg that is at rest initially. How much force do
you need to apply to get the sled moving? If you continue to apply
that force, what will the magnitude of sled’s acceleration be once
it is moving?
Sketch.
Hypothesis: 13.7 N, should be the max static friction force we just
worked out; 1 m/s2
Friction Example 2
For waxed wood on wet snow, µs = 0.14 and µk = 0.1. You pull
on a sled of mass 10 kg that is at rest initially. How much force do
you need to apply to get the sled moving? If you continue to apply
that force, what will the magnitude of sled’s acceleration be once
it is moving?
Sketch.
Hypothesis: 13.7 N, should be the max static friction force we just
worked out; 1 m/s2
To get the sled moving Fapp > Fsf
Fsf
= µs n
= (0.14)(10 kg)g
= 13.7 N
Friction Example 2
For waxed wood on wet snow, µs = 0.14 and µk = 0.1. You pull
on a sled of mass 10 kg that is at rest initially. How much force do
you need to apply to get the sled moving? If you continue to apply
that force, what will the magnitude of sled’s acceleration be once
it is moving?
Friction Example 2
For waxed wood on wet snow, µs = 0.14 and µk = 0.1. You pull
on a sled of mass 10 kg that is at rest initially. How much force do
you need to apply to get the sled moving? If you continue to apply
that force, what will the magnitude of sled’s acceleration be once
it is moving?
Fnet,x
Fapp − Fkf
= max
= 13.72 − µk n
= 13.7 − (0.1)(10 kg)g
= 3.92 N
a=
F
3.92 N
=
= 0.39 ms−2
m
10 kg
Friction Example 2
For waxed wood on wet snow, µs = 0.14 and µk = 0.1. You pull
on a sled of mass 10 kg that is at rest initially. How much force do
you need to apply to get the sled moving? If you continue to apply
that force, what will the magnitude of sled’s acceleration be once
it is moving?
Fnet,x
Fapp − Fkf
= max
= 13.72 − µk n
= 13.7 − (0.1)(10 kg)g
= 3.92 N
a=
F
3.92 N
=
= 0.39 ms−2
m
10 kg
Reasonable?: Yes for the force. The acceleration was a bit less
than my guess, but same order of magnitude.
Friction Example 6-2
A trained sea lion slides from rest with constant acceleration down
a 3.0-m-long ramp into a pool of water. If the ramp is inclined at
an angle of 23◦ above the horizontal and the coefficient of kinetic
friction between the sea lion and the ramp is 0.26, how long does
it take for the sea lion to make a splash in the pool?
0
Walker, “Physics”.
force vectors in such a case, we begin by resolving each vector into its x and y
components.
Friction Example 6-2
On an incline, align one axis
the surface, and the other axi
dicular to the surface. That wa
is in the x direction. Since no m
in the y direction, we know
A trained sea lion slides from rest with constant acceleration down
EXAMPLE 6–2 Making a Big Splash
a 3.0-m-long
ramp into a pool of water. If the ramp is inclined at
A trained sea lion slides from rest with constant acceleration down a 3.0-m-long ramp into a pool of water. If the ramp
◦ above
an angleof
of 23°
above
the horizontal
the coefficient and
of kinetic
friction
between the sea
and the ramp is 0.26,
an atdoes
angle
23
the and
horizontal
the
coefficient
oflionkinetic
it take for the sea lion to make a splash in the pool?
friction
between the sea lion and the ramp is 0.26, how long does
Picture the Problem
As is usual
with
we make
choose one
to be parallel
to thepool?
surface and the other to be perpendicular to
it take
for
theinclined
sea surfaces,
lion to
aaxis
splash
in the
sketch, the sea lion accelerates in the positive x direction 1a 7 02, having started from rest, v = 0. We are free to c
0x
x
initial position of the sea lion
to be! x0 = 0. There is no! motion in the y direction, and therefore ay = 0. Finally, we note
!
free-body diagram that N = NyN , fk = - mkNxN , and W = 1mg sin u2xN + 1 - mg cos u2yN .
Sketch:
+y
fk
fk
y
Physical picture
!
N
W
!
+x
mg cos ! !
N
mg sin !
W = mg
Free-b
diagra
x
Strategy
We can use the kinematic equation relating position to time, x = x0 + v0xt + 12axt2, to find the time of the sea lion’s sli
be necessary, however, to first determine the acceleration of the sea lion in the x direction, ax.
0
ToWalker,
find ax we“Physics”.
apply Newton’s second law to the sea lion. First, we can find N by setting ©Fy = may equal to zero (sinc
force vectors in such a case, we begin by resolving each vector into its x and y
components.
On an incline, align one axis
the surface, and the other axi
dicular to the surface. That wa
is in
no m
es from rest with constant acceleration down a 3.0-m-long ramp into a pool of water.
Ifthe
thex direction.
ramp isSince
inclin
in the y direction, we know
aking
a Big SplashExample
Friction
6-2
ve the horizontal and the coefficient of kinetic friction between the sea lion and the ramp is 0.26, how lon
A trained
lion
slides from rest with constant acceleration down
a lion to make
a splashsea
in the
pool?
EXAMPLE 6–2 Making a Big Splash
a 3.0-m-long
ramp into a pool of water. If the ramp is inclined at
A trained sea lion slides from rest with constant acceleration down a 3.0-m-long ramp into a pool of water. If the ramp
◦axis
an angle
of 23°
above
thetohorizontal
and
the
of kinetic
friction
between
theperpendicular
sea
and the ramp
0.26,
ned surfaces,
we
choose
one
be the
parallel
to coefficient
the surface
and
the
other
to be
to it.is In
o
an atdoes
angle
of
23
above
horizontal
and
the
coefficient
oflionkinetic
it take for the sea lion to make a splash in the pool?
celerates in the positive x direction 1ax 7 02, having started from rest, v0x = 0. We are free to choose t
friction the
between the sea lion and the ramp is 0.26, how long does
sea lion
to bePicture
There is no! motion in the y direction, and therefore ay = 0. Finally, we note from t
= 0.Problem
!
! xis0 usual
As
with
inclined
surfaces,to
we make
choose one
to be parallel
to thepool?
surface and the other to be perpendicular to
it
take
for
the
sea
a1axis
splash
in02,N .the
N
Nlion
at N = Ny, fsketch,
Wlion
mksea
Nx
, and
= 1mg
u2xN +
-mg
cos
k = -the
accelerates
in thesin
positive
x direction
having started from rest, v = 0. We are free to c
1a 7u2y
0x
x
initial position of the sea lion
to be! x0 = 0. There is no! motion in the y direction, and therefore ay = 0. Finally, we note
!
free-body diagram that N = NyN , fk = - mkNxN , and W = 1mg sin u2xN + 1 - mg cos u2yN .
Sketch:
+y
+y
fk
fk
y
!
!
N
W
atic
!
+x
y
Physical picture
Physical picture
mg cos ! !
W
!
+x
x
N
mg sin !
fk
mg cos ! !
W
= mg
x
N
mg sin !
Free-bodyFree-b
Wdiagram
= mg
diagra
Strategy
We can use the kinematic equation relating position to time, x = x0 + v0xt + 12axt2, to find the time of the sea lion’s sli
be necessary, however, to first determine the acceleration of the sea lion in the x direction, ax.
0
“Physics”.
equationToWalker,
relating
to time,
to can
find
the
time
of the
w
+ 1First,
axt2, we
x =lawx0to +thevsea
find ax weposition
apply Newton’s
second
find
N by
setting
to zeroIt
(sinc
©Fy sea
= malion’s
0xt lion.
y equalslide.
Friction Example 6-2
Hypothesis: About 5 seconds.
Friction Example 6-2
Hypothesis: About 5 seconds.
Strategy: Use Newton’s 2nd law, find acceleration, use kinematics
equation.
Friction Example 6-2
Hypothesis: About 5 seconds.
Strategy: Use Newton’s 2nd law, find acceleration, use kinematics
equation.
Fnet,y = n − mg cos θ = 0
n = mg cos θ
Fnet,x = mg sin θ − fk
= ma
Friction Example 6-2
Hypothesis: About 5 seconds.
Strategy: Use Newton’s 2nd law, find acceleration, use kinematics
equation.
Fnet,y = n − mg cos θ = 0
n = mg cos θ
Fnet,x = mg sin θ − fk
= ma
mg sin θ − µk n = ma
mg sin θ − µk (mg cos θ) = ma
a = g (sin θ − µk cos θ)
Friction Example 6-2
Hypothesis: About 5 seconds.
Strategy: Use Newton’s 2nd law, find acceleration, use kinematics
equation.
Fnet,y = n − mg cos θ = 0
n = mg cos θ
Fnet,x = mg sin θ − fk
= ma
mg sin θ − µk n = ma
mg sin θ − µk (mg cos θ) = ma
a = g (sin θ − µk cos θ)
a = 1.5 m/s2
Friction Example 6-2
Given: ∆x = 3 m, a = 1.5 m/s2 , vi = 0 m/s.
Asked for: t.
Friction Example 6-2
Given: ∆x = 3 m, a = 1.5 m/s2 , vi = 0 m/s.
Asked for: t.
1
∆x = vi t + at 2
2
t = 2.0 s
Friction Question
Quick Quiz 5.7.1 You are playing with
your daughter in the snow. She sits on
a sled and asks you to slide her across a
flat, horizontal field. You have a choice
of:
(A) pushing her from behind by
applying a force downward on her
shoulders at 30◦ below the horizontal or
(B) attaching a rope to the front of the
sled and pulling with a force at 30◦
above the horizontal.
Which would be easier for you and why?
2
Serway & Jewett, page 132.
30!
S
F
Rubb
Stee
Alum
Glas
Cop
Woo
Wax
Wax
Meta
Teflo
Ice o
Syno
Note:
can e
a
S
F
30!
b
Figure 5.17 (Quick Quiz 5.7)
A father slides his daughter on a
sled either by (a) pushing down
on her shoulders or (b) pulling up
on a rope.
Q u
yo
b
Q u
a
ch
sh
th
(F
Friction Question
Quick Quiz 5.7.1 You are playing with
your daughter in the snow. She sits on
a sled and asks you to slide her across a
flat, horizontal field. You have a choice
of:
(A) pushing her from behind by
applying a force downward on her
shoulders at 30◦ below the horizontal or
(B) attaching a rope to the front of the
sled and pulling with a force at 30◦
above the horizontal.
←
Which would be easier for you and why?
2
Serway & Jewett, page 132.
30!
S
F
Rubb
Stee
Alum
Glas
Cop
Woo
Wax
Wax
Meta
Teflo
Ice o
Syno
Note:
can e
a
S
F
30!
b
Figure 5.17 (Quick Quiz 5.7)
A father slides his daughter on a
sled either by (a) pushing down
on her shoulders or (b) pulling up
on a rope.
Q u
yo
b
Q u
a
ch
sh
th
(F
Friction Question
#5, page
Hopping into your Porsche, you floor it and accelerate at 12 m/s2
without spinning the tires. Determine the minimum coefficient of
static friction between the tires and the road needed to make this
possible.
1
Walker, “Physics”.
Friction Question
#5, page
Hopping into your Porsche, you floor it and accelerate at 12 m/s2
without spinning the tires. Determine the minimum coefficient of
static friction between the tires and the road needed to make this
possible.
Sketch a free body diagram for the car. (What force causes the
car’s forward acceleration?)
1
Walker, “Physics”.
Friction Question
#5, page
Hopping into your Porsche, you floor it and accelerate at 12 m/s2
without spinning the tires. Determine the minimum coefficient of
static friction between the tires and the road needed to make this
possible.
Sketch a free body diagram for the car. (What force causes the
car’s forward acceleration?)
Hypothesis: coefficients of friction are usually between 0 and 1.
Car tires are designed not to slip on asphalt. µs should be high,
but we are looking for the minimum it could be. Guess: 0.5.
1
Walker, “Physics”.
Friction Question
Hopping into your Porsche, you floor it and accelerate at 12 m/s2
without spinning the tires. Determine the minimum coefficient of
static friction between the tires and the road needed to make this
possible.
Strategy: Newton’s 2nd law.
Fnet = ma
Fnet = fs . If we want to find the minimum coefficient of static
friction, assume that we are getting the max possible force from
that coefficient: fs = fs,max = µs n.
µs mg
µs
µs
= ma
a
=
g
=
12 m/s2
9.81 m/s2
= 1.2
Friction Question
Hopping into your Porsche, you floor it and accelerate at 12 m/s2
without spinning the tires. Determine the minimum coefficient of
static friction between the tires and the road needed to make this
possible.
µs
= 1.2
Reasonable?: Woah! This is not only much bigger than my guess,
it is bigger than 1!
This would mean that it requires less force to pick up the entire
Porsche and move it to one side than it does to push it along the
ground starting from rest.
Friction Question
Hopping into your Porsche, you floor it and accelerate at 12 m/s2
without spinning the tires. Determine the minimum coefficient of
static friction between the tires and the road needed to make this
possible.
µs
= 1.2
Reasonable?: Woah! This is not only much bigger than my guess,
it is bigger than 1!
This would mean that it requires less force to pick up the entire
Porsche and move it to one side than it does to push it along the
ground starting from rest.
Research → Apparently, yes! This is the sort of number you can
get for high-performance racing tires. Cool.
Summary
• more friction examples
• pulleys
Homework
Walker Physics:
• Ch 6, onwards from page 177. Questions: 3, 15; Problems: 1,
3, 11, 13, 15, 87