Friction Friction Problem Situations
-
Upload
orlando-summers -
Category
Documents
-
view
140 -
download
17
description
Transcript of Friction Friction Problem Situations
FrictionFriction Problem Situations
PhysicsMontwood High School
R. Casao
Friction
• Friction Ff is a force that resists motion•Friction involves objects in contact
with each other.•Friction must be overcome before
motion occurs.• Friction is caused by the uneven
surfaces of the touching objects. As surfaces are pressed together, they tend to interlock and offer resistance to being moved over each other.
Microscopic Friction
Magnified section of a polished steel surface showing surface bumps about 5 x 10-7 m (500 nm) high, which corresponds to several thousand atomic diameters.
Computer graphic from a simulation showing gold atoms (below) adhering to the point of a sharp nickel probe (above) that has been in contact with the gold surface.
Surface Roughness
Adhesion
Friction
• Frictional forces are always in the direction that is opposite to the direction of motion or to the net force that produces the motion.
• Friction acts parallel to the surfaces in contact.
Types of Friction• Static friction: maximum frictional force
between stationary objects.• Until some maximum value is reached and
motion occurs, the frictional force is whatever force is necessary to prevent motion.
• Static friction will oppose a force until such time as the object “breaks away” from the surface with which it is in contact.
• The force that is opposed is that component of an applied force that is parallel to the surface of contact.
Types of Friction• The magnitude of the static friction force Ffs
has a maximum value which is given by:
• where μs is the coefficient of static friction and FN is the magnitude of the normal force on the body from the surface.
fs s NF F
Types of Friction• Sliding or kinetic friction: frictional force
between objects that are sliding with respect to one another.• Once enough force has been applied to the
object to overcome static friction and get the object to move, the friction changes to sliding (or kinetic) friction.
• Sliding (kinetic) friction is less than static friction.• If the component of the applied force on the
object (parallel to the surface) exceeds Ffs then the magnitude of the opposing force decreases rapidly to a value Fk given by:
where μk is the coefficient of kinetic friction.
k k NF F
The static frictional force keeps an object from starting to move when a force is applied. The static frictional force has a maximum value, but may take on any value from zero to the maximum, depending on
Static FrictionStatic Friction
what is needed to keep the sum of forces zero.
Types of Friction
• From 0 to the maximum value of the static frictional force Fs in the figure, the applied force is resisted by the static frictional force until “breakaway”.
• Then the sliding (kinetic) frictional force Fk is approximately constant.
Types of Friction
• Static and sliding friction are dependent on:• The nature of the surfaces in contact.
Rough surfaces tend to produce more friction.
• The normal force (Fn) pressing the surfaces together; the greater Fn is, the more friction there is.
Friction vs. Area
Question: Why doesn’t friction depend on contact area?
The microscopic area of contact between a box and the floor is only a small fraction of the macroscopic area of the box’s bottom surface.
If the box is turned on its side, the macroscopic area is increased, but the microscopic area of contact remains the same (because the contact is more distributed). Therefore the frictional force f is independent of contact area.
Types of Friction
• Rolling friction: involves one object rolling over a surface or another object.
• Fluid friction: involves the movement of a fluid over an object (air resistance or drag in water) or the addition of a lubricant (oil, grease, etc.) to change sliding or rolling friction to fluid friction.
Coefficient of Friction
• Coefficient of friction (): ratio of the frictional force to the normal force pressing the surfaces together. has no units.
• Static:
• Sliding (kinetic):
n
fss F
Fμ
n
fkk F
Fμ
•The maximum frictional force is 50 N. As the applied force increases from 0 N to 50 N, the frictional force also increases from 0 N to 50 N and will be equal to the applied force as it increases.
•Once the static frictional force of 50 N has been overcome, only a 40 N force is needed to overcome the 40 N kinetic frictional force and produce constant velocity (a = 0 m/s2).
•As the applied force increases beyond 40 N, the kinetic frictional force remains at 40 N and the 100 N block will accelerate.
A Model of Friction
Friction
Static Friction
Kinetic Friction
push k k NF f F
The kinetic frictional force is also independent of the relative speed of the
surfaces, and of their area of contact.
Kinetic Friction and SpeedKinetic Friction and Speed
Rolling Friction
Horizontal Surface – Constant Speed
•Constant speed: a = O m/s2.•The normal force pressing the surfaces together is the weight; Fn = Fw
fx
fx
fx
x
FF
N0FF
amFF
amFΣ
wkfx
wkf
w
f
n
fk
FμFF
FμF
FF
FF
μ
Horizontal Surface: a > O m/s2
wkf
w
f
n
fk
wn
fx
x
fx
FμF
FF
FF
μ
FF
amFF
amFΣ
FF
Horizontal Surface: a > O m/s2
• If solving for:• Fx:
• Ff:
• a:
gmμamF
FμamF
FamF
kx
wkx
fx
amFF xf
mFF
a fx
Horizontal Surface: Skidding to a Stop or Slowing Down (a < O m/s2)
• The frictional force is responsible for the negative acceleration.
• Generally, there is no Fx.
wkf
w
f
n
fk
wn
f
FμF
FF
FF
μ
FF
amF
)ta(vv
)ta5.0()tv(xΔ
)xΔa2(vv
if
2i
2i
2f
Horizontal Surface: Skidding to a Stop or Slowing Down (a < O m/s2)
• Most common use involves finding acceleration with a velocity equation and finding k:
• Acceleration will be negative because the speed is decreasing.
Horizontal Surface: Skidding to a Stop or Slowing Down (a < O m/s2)
• The negative sign for acceleration a is dropped because k is a ratio of forces that does not depend on direction.
• Maximum stopping distance occurs when the tire is rotating. When this happens, a = -s·g.
• Otherwise, use a = -k·g to find the acceleration, then use a velocity equation to find distance, time, or speed.
ga
gmam
FF
FF
μw
f
n
fk
Friction, Cars, & Antilock Brakes The diagram shows forces acting on a car with front-wheel drive. Typically, Fn > Fn’ because the engine is over the front wheels. The largest frictional force fs the tire can exert on the road is µs·Fn. Attempts to make the tire exert a force larger than this causes the tire to “burn rubber” and actually reduces the force, since µk<µs.
Note that while all points on the rolling tire have the same speed v in the reference frame of the car, in the reference frame of the road the bottom of the tire is at rest, while top is moving forward with a speed of 2·v.
Antilock brakes sense the wheel rotation and “ease off” if it close to stopping, maintaining static friction with the road and allowing better control of steering than if the wheels were locked.
Antilock Brakes
Example:The Effect of Antilock Brakes
22 2 0
0 2 and 0, so 2
v
v v a x v xa
A car is traveling at 30 m/s along a horizontal road. The coefficients of friction are ms=0.50 and mk=0.40.(a) What is the braking distance xa with antilock brakes?(b) What is the braking distance xb if the brakes lock?
and a s b ka g a g2 20
2
(30 m/s)91.7 m
2 2 (0.5) (9.81 m/s )
as
vx
g2 20
2
(30 m/s)114.7 m
2 2 (0.4) (9.81 m/s )
bk
vx
g
Example: A Game of Shuffleboard
k k nf F
A cruise-ship passenger uses a shuffleboard cue to push a shuffleboard disk of mass 0.40 kg horizontally along the deck, so that the disk leaves the cue at a speed of 8.5 m/s. The disk then slides a distance of 8.0 m.
What is the coefficient of kinetic friction between the disk and deck?
0 y yF m a
0 n nF m g F m g
x xF m a
k k xf m g m a x ka g
2 2 20 02 0 2 x x x x kv v a x v g x
2 20
2
(8.5 m/s)0.46
2 2 (9.81 m/s ) (8.0 m)
x
k
v
g x
Down an Inclined Plane
Down an Inclined Plane
• Resolve Fw into Fx and Fy.
• The angle of the incline is always equal to the angle between Fw and Fy.
• Fw is always the hypotenuse of the right triangle formed by Fw, Fx, and Fy.
θsinFF
θcosFF
FF
θsinF
Fθcos
wx
wy
w
x
w
y
Down an Inclined Plane
• The force pressing the surfaces together is NOT Fw, but Fy; Fn = Fy.
or
ykf
y
f
n
fk
fx
fx
FμF
FF
FF
μ
mFF
a
amFF
amΣF
sin cos
mass m cancels out
( sin ) ( cos )
( sin )cos
m g m g m a
g g a
g ag
Down an Inclined Plane• If we place an object on an inclined plane and
increase the tilt angle to the point at which the object just begins to slide.
• What is the relation between and the static coefficient of friction µs?
- : cos N wy axis F F
- : sin s s N wx axis f F F
sincos
sintan
cos
s ws
n w
s
f FF F
Down an Inclined Plane• If the object slides down the incline at constant
speed (a = 0 m/s2), the relation between and the kinetic coefficient of friction µk:
θtanμ
θtanθcosθsin
θcosFθsinF
FF
FF
μ
FF
N0FFs
m0mFF
k
w
w
y
x
n
fk
fx
fx
2fx
Down an Inclined Plane
• To determine the angle of the incline:• If moving:
• If at rest:
k1 μtanθ
s1 μtanθ
Example: A Sliding Coin A hardcover book is resting on a tabletop with its front cover facing upward. You place a coin on the cover and very slowly open the book until the coin starts to slide. The angle is the angle of the cover just before the coin begins to slide.
Find the coefficient of static friction µs between the coin and book.
, cos s s N s sf F at so f m g
cos 0 cos y y n NF m a F m g or F m g
sin 0 or sin x x s sF m a m g f f m g
, cos sin tan s sTherefore or
Example:Dumping a file cabinet
A 50.0 kg steel file cabinet is in the back of a dump truck. The truck’s bed, also made of steel, is slowly tilted. What is the size of the static friction force when the truck’s bed is tilted by 20°? At what angle will the file cabinet begin to slide?
Steel on dry steel
Free-body diagram
Example:Dumping a file cabinet
sin sin 0;
cos cos 0;
x s s
y
F w f m g f
F n w n m g
2sin (50.0kg) (9.80m/s ) sin 20 168 N; sf m g
max cos ; s s s sf f n m g
sin sin cos 0; s sm g f m g m g
sintan ; arctan arctan(0.80) 38.7
coss s
File cabinet will begin to slide when:
Non-Parallel Applied Force on Ramp
m·gm·g ·cos
m·g ·sin
fk
N
If an applied force acts on the box at an angle above the horizontal, resolve FA into parallel and perpendicular components using the angle + :
FA ·cos ( + θ) and FA ·sin ( + θ)
FA serves to increase acceleration directly and indirectly: directly by FA ·cos ( + θ) pulling the box down the ramp, and indirectly by FA ·sin ( + θ) lightening the normal support force with the ramp (thereby reducing friction).
FA
FA ·cos( + )
FA ·sin( + )
Non-Parallel Applied Force on Ramp
m·gm·g
·cos
m·g ·sin
fk
N
FA
FA ·cos( + )
FA ·sin ( + )
If FA ·sin( + ) is not big enough
to lift the box off the ramp, there is no acceleration in the perpendicular direction. So, FA ·sin( + ) + FN = m·g·cos.
Remember, FN is what a scale would read if placed under the box, and a scale reads less if a force lifts up on the box. So, FN = m·g ·cos - FA ·sin( + ), which means fk = k ·FN = k ·[m·g ·cos - FA ·sin( + )].
Non-Parallel Applied Force on Ramp
m·g
m·g ·cos
m·g ·sin
fk
N
FA
FA ·cos( + )
FA ·sin( + )
If the combined force of FA ·cos( + ) + m·g ·sin is is enough to move the box: FA ·cos( + ) + m·g·sin
- k ·[m·g·cos - FA ·sin( + )] = m·a
Up an Inclined Plane
Up an Inclined Plane• Resolve Fw into Fx and Fy.
• The angle of the incline is always equal to the angle between Fw and Fy.
• Fw is always the hypotenuse of the right triangle formed by Fw, Fx, and Fy.
θsinFF
θcosFF
FF
θsinF
Fθcos
wx
wy
w
x
w
y
Up an Inclined Plane
• Fa is the force that must be applied in the direction of motion.
• Fa must overcome both friction and the x-component of the weight.
• The force pressing the surfaces together is Fy.
Up an Inclined Plane
ykf
y
f
n
fk
xfa
xfa
x
yn
FμF
FF
FF
μ
mFFF
a
amFFF
amFΣ
FF
•For constant speed, a = 0 m/s2.
Fa = Fx + Ff
•For a > 0 m/s2.Fa = Fx + Ff +
(m·a)
Pulling an Object on a Flat Surface
Pulling an Object on a Flat Surface
•The pulling force F is resolved into Fx and Fy.
θsinFF
θcosFFF
Fθsin
FF
θcos
y
x
y
x
Pulling an Object on a Flat Surface
•Fn is the force that the ground exerts upward on the mass. Fn equals the downward weight Fw minus the upward force Fy from the pulling force.•For constant speed, a = 0 m/s2.
mFF
a
amFF
amFΣ
)FF(μF
FFF
FF
μ
FFF
N0FFF
N0FΣ
fx
fx
x
ywkf
yw
f
n
fk
ywn
wyn
y
Example: Pulling A Sled
2
sin 0 sin
50 9.8 100 sin 40 425.72
y y N N
N
F m a F T m g m or F m g T
mF kg N Ns
Two children sitting on a sled at rest in thesnow ask you to pull them. You pull on the sled’s rope, which makes an angle of 40° withthe horizontal. The children have a combinedmass of 45 kg, and the sled has a mass of 5.0 kg. The coefficients of static and kineticfriction are µs=0.20 and µk=0.15, and the sled is initially at rest.
Find the acceleration of the sled and children if the rope tension is 100 N.
2
cos
0.15 425.72 63.86
100 cos 40 63.86100 cos 40 63.86 50 0.2549
50
x x k x
kk k k N
N
x x
F ma T f m a
ff F N N
F
N N mN N kg a askg
Simultaneous Pulling and Pushing an Object on a Flat Surface
Simultaneous Pulling and Pushing an Object on a Flat Surface
θsinFF
θcosFFF
Fθsin
FF
θcos
y
x
y
x
m
FFFa
amFFF
amFΣ
)FF(μF
FFF
FF
μ
FFF
N0FFF
N0FΣ
fpushx
fpushx
x
ywkf
yw
f
n
fk
ywn
wyn
y
Pushing an Object on a Flat Surface
Pushing an Object on a Flat Surface
•The pushing force F is resolved into Fx and Fy.
sinθFF
cosθFFF
Fsinθ
FF
cosθ
y
x
y
x
Pushing an Object on a Flat Surface
•Fn is the force that the ground exerts upward on the mass. Fn equals the downward weight Fw plus the upward force Fy from the pushing force.•For constant speed, a = 0 m/s2.
mFF
a
amFF
amFΣ
)FF(μF
FFF
FF
μ
FFF
N0FFF
N0FΣ
fx
fx
x
ywkf
yw
f
n
fk
ywn
wyn
y
Pulling and Tension
• The acceleration a of both masses is the same.
Pulling and Tension
• For each mass:
• Isolate each mass and examine the forces acting on that mass.
n2kf2
n1kf1
w2n2
w1n1
FμF
FμF
FF
FF
Pulling and Tension
•m1 = mass
•T1 may not be a tension, but could be an applied force (Fa) that causes motion.
amFTT
amΣF
11f21
1
Pulling and Tension
•m2 = mass
amFT
amΣF
22f2
2
Pulling and Tension
• This problem can often be solved as a system of equations:
• See the Solving Simultaneous Equations notes for instructions on how to solve this problem using a TI or Casio calculator.
amFT
amFTT
22f2
11f21
Revisiting Tension and Friction
Revisiting Tension and Friction
•For the hanging mass, m2:
•The acceleration a of both masses is the same.
•For the mass on the table, m1:
amTgm
amTF
amΣF
gmF
22
22w
2
22w
amT-F
FμF
FF
amΣF
1f
1nf
1w1n
Revisiting Tension and Friction
12
f2
12f2
2f12
mmFgm
a
amamFgm
amFamgm
A block of mass m2 = 5.0 kg has been adjusted so that the block m1 = 7.0 kg is just on the verge of sliding.
(a)What is the coefficient of static friction ms between the table and the block?
Example: A Sliding Block
1 1 10 so y N y NF F m g m a F m g
' 2 2 ' 20 so x xF m g T m a T m g
2 21 2
2 2
5Therefore, ; 0.71
7
s s
m g m kgm g m g
m g m kg
1 10 so x x s N sF T f m a f F m g T
Example: A Sliding Block(b) With a slight push, the blocks move with
acceleration a. Find a if µk = 0.54.
1 1 1
1 1
x x k x
x k
F T f m a so T m g m a
T m a m g
' 2 2 ' 2 2 ' x x xF m g T m a T m g m a
1 1 2 2
1 2 2 1
2 1
1 2
22
T=T, therefore,
( )
( )
9.8 m/s 5.0 kg 0.54 7.0 kg1.0 m/s
7.0 kg 5.0 kg
k
k
k
m a m g m g m a
m a m a m g m g
g m ma
m m
a
' x xa a a
Normal Force Not Associated with Weight.
• A normal force can exist that is totally unrelated to the weight of an object.
applied forcefriction
weightnormal
FN = applied force
Friction is Always Parallel to Surfaces….
•In this case, for the block to remain in position against the wall without moving:
• the upward frictional force Ff has to be equal and opposite to the downward weight Fw.•The rightward applied force F has to be equal ad opposite to the leftward normal force FN.
F
FW
Ff
FN
(0.20)