Chapter 6: Friction
• Friction forces are everywhere in our daily life
• Two kinds of friction forces
– Static friction force: fs
– Kinetic friction force: fk
• Direction of frictional force: always opposes the direction
of motion (or intended motion) relative to the surface
fs = µsN
k
(direction OPPOSITE motion)
k
µ : coefficient of friction (either static - or - kinetic)
Sample Problem 5-5
N
T = Ma
M = 3.3 kg, m = 2.1 kg, frictionless surface
H falls as S accelerates to the right
T
(a) What is the acceleration of S?
F = Mg
Acceleration a links the two masses together
Forces on m
mg − T = ma
mg − Ma = ma
m
a=
g
M+m
unbalanced forces on M
T = Ma
Fg = mg
Sample Problem 5-4
N
T = Ma
M = 3.3 kg, m = 2.1 kg, frictional surface
H falls as S accelerates to the right
fs,k
Now let the table top have friction
with µs = 0.2, µk = 0.15.
(b) What is the new acceleration of S?
T
F = Mg
Acceleration a links the two masses together
Forces on m
mg − T = ma
mg − Mµsg − Ma = ma
m − Mµ s
a=
g
M+m
Fg = mg
Forces on M
T − fs = T − µsN = T − µsMg = Ma
==> T = µsMg + Ma
Note: Mµs< m, so
the masses *will* accelerate!
Sample Problem 5-4
M = 3.3 kg, m = 2.1 kg, frictional surface
H falls as S accelerates to the right
fs,k
(b) Now let the table top have friction
with µs = 0.2, µk = 0.15.
(b) What is the new acceleration of S?
Now replace µs with µk to find the
actual acceleration.
N
T = Ma
T
F = Mg
Fg = mg
m − Mµ k
2.1 − (3.3)(0.15)
a=
g=
g = 0.297g
M+m
3.3 + 2.1
Problem 6-20
A loaded penguin sled weighing 80 N rests on a plane inclined at 20o to
the horizontal . Between the sled and the plane, µs = 0.25, µk = 0.15.
(a) What is the minimum magnitude of the force F, parallel to
the plane, that will prevent the sled from slipping down the plane?
Set up free-body force diagram.
F
y N
x
f
mg
θ
Notice that f is pointing with F
because without F the penguin will
slide *down* the plane.
Problem 6-20
A loaded penguin sled weighing 80 N rests on a plane inclined at 20o to the
horizontal . Between the sled and the plane, µs = 0.25, µk = 0.15.
(a) What is the minimum magnitude of the force F, parallel to the plane, that
will prevent the sled from slipping down the plane?
Set up free-body force diagram.
F
y N
x
f
mg
θ
Apply Newton’s Second Law
x: F − mg sinθ + fs
= F − mg sinθ + µsN = 0
y:
N − mg cosθ = 0
Problem 6-20
A loaded penguin sled weighing 80 N rests on a plane inclined at 20o to
the horizontal . Between the sled and the plane, µs = 0.25, µk = 0.15.
(a) What is the minimum magnitude of the force F, parallel to
the plane, that will prevent the sled from slipping down the plane?
Set up free-body force diagram.
F
y N
x
f
mg
θ
Apply Newton’s Second Law
F = mg sinθ − µsmg cosθ
= (80) sin(20) − 0.25(80)cos(20)
= 27.36N − 18.79N = 8.57N
Problem 6-20
A loaded penguin sled weighing 80 N rests on a plane inclined at 20o to
the horizontal . Between the sled and the plane, µs = 0.25, µk = 0.15.
(b) What is the minimum magnitude F that will start the sled
moving up the plane?
Now f points *down* the plane because F
will cause motion *up* the plane
Set up free-body force diagram.
F
y N
x
f
mg
θ
Apply Newton’s Second Law
F = mg sinθ + µsmg cosθ
= (80) sin(20) + 0.25(80)cos(20)
= 27.36N + 18.79N = 46.15N
Problem 6-20, continued
c) What value of F is required to move the sled up the plane
at constant velocity?
Now f points *down* the plane because F
causes motion *up* the plane
Constant velocity => zero acceleration => forces are balanced
F = mg sinθ + µkmg cosθ
= 27.36Ν + 11.28Ν = 38.64Ν will cause the sled to
move up the inclined plane at constant velocity
Aerodynamic Drag
Notice that a dropped object will first accelerate and then keep
descending at a constant velocity.
That means the acceleration decreased to zero, which means there
had to be another force to counteract mg. This force is called “drag”.
|D| = 1/2CρAv2
C = drag coefficient
ρ = air density
A = cross sectional area
v = speed
D − mg = ma = 0
=>
1
2
CρAv 2t − mg = 0
2mg
vt =
CρA
Now let’s have some fun
Real Drag Coefficients
Real baseball trajectories
elevation
sea level: 0 m
Denver: 1,524 m
Mexico City: 2,380 m
Skiing at the Triple Point
Skier Forces
Skiing at the Triple Point
Skiing at the Triple Point
Skiing at the Triple Point
Chapter 06: Circular Motion
Hurricane Frances, September 2004
Chapter 6: Uniform circular motion
•For uniform circular motion, the centripetal
acceleration
a = v2/R
which is caused by a force called centripetal force Fc
Fc = ma = m(v2/R)
direction of Fc: points radially inward
Centripetal force is not a new kind of force
Question: What are the centripetal forces exerted on the
following object which are in uniform circular motion?
A) A ball attached to the end of a string
tension
B) A satellite orbiting the Earth
Gravitational Force
C) A car driving in a circle
friction
D) A car driving in a circle on a tilted track
friction and gravitational force
When you ride in a Ferris wheel at constant speed, what
are the directions of your acceleration a and the normal
force N when you are:
A) at the top of the wheel
N
mg
N + mg = mv2/r =>N = mv2/r − mg
B) at the bottom of the wheel
N − mg = mv2/r =>N = mg + mv2/r
N
mg
C) at which point, is the normal force the largest in
magnitude?
Definitely at the bottom!
tension on the string causes the object to move in a circle
( )
v2
v
− R̂
aC =
R
Example: (Sample problem 6-8) In the
figure, a person is riding the Rotor. Suppose
that the coefficient of static friction µs
between the rider’s clothing and the canvas is
0.4 and the cylinder’s radius R is 2.1m.
What minimum speed v must the cylinder
and rider have if the rider is not to fall
when the floor drops?
v
v2
v2
v
v
aC =
− R̂
N = ma C = m
− R̂
R
R
( )
v2
µS m
= mg
R
Rg
v=
= 7.17 m / s
µS
( )
fS = µSN = mg
fS
N
mg
Aerodynamic cars
Shape of the car causes a downward force (negative lift)
Car rounding a flat curve
Friction is causing the race car to move in a circle.
Car rounding a flat curve
Friction is causing the race car to move in a circle.
N
fs
mg
FL
R
0
Car rounding a flat curve
Friction is causing the race car to move in a circle.
N
fs
mg
FL
R
0
Banked Curves
Can the banked track take the place of friction
for circular motion (i.e., fs = 0)?
Car rounding a banked curve
Friction and mgsinθ (the Normal component)
are causing the car to move in a circle.
for the example in the book
( )
v v
v2
v
f S + FNr = ma C = m
− R̂
R
0
v2
FNr = mg sin θ = m
R
FNy = mg cos θ
FNr v 2
tan θ =
=
FNy gR
Problem 6-53
v = 480 km/h and θ = 40o. What is the radius?
Fl sinθ
Fl cosθ θ
lift
FC = maC = mv2/R
r: Fl sinθ = mv2/R
y: Fl cosθ = mg
mg
Problem 6-53 continued
Problem 6-59
d = L = 1.70m and Tupper = 35N.
Find:
A) Tlower
the ball
B) net force on the ball
D) direction of Fnet,string
C) speed of
Problem 6-59 continued
Free-body force diagram is shown below. Tu is the
tension in the upper cord and Tl is the tension in the
lower cord.
y
Tu
x
θ
θ
Tl
mg
Problem 6-59 continued
Problem 6-59 continued
Problem 6-60
m = 0.040 kg, L = 0.90m, circumference = 0.94 m.
Find:
A) String tension
B) period of motion
Problem 6-60 continued
How does the magnitude of the
tension, T, on the string compare
to the mass’s weight, mg?
1.
2.
3.
4.
T < mg
T > mg
T = mg
none of the above
Problem 6-60 continued
R is the circle radius. Note
R = circumference/2π = 0.94 /2π = 0.15m
R 0.15
sin θ = =
= 0.167
L 0.9
⇒ θ = sin −1 (0.167 ) = 9.59o
Tvertical
mg
= T cos θ = mg ⇒ T =
= 0.40 N
cos θ
θ
R
(> mg )
Problem 6-60 continued
R is the circle radius. Note
R = circumference/2π = 0.94 /2π = 0.15m
R 0.15
sin θ = =
= 0.167
L 0.9
⇒ θ = sin −1 (0.167 ) = 9.59o
θ
mv 2
Thorizontal = T sin θ = FC =
R
mv 2
⇒T=
R sin θ
R
TR sin θ
⇒v=
= 0.49m / s
m
Period = circumference/speed = 0.94 /0.49 = 1.9s