Physics 121,Lecture 8,Pg 1
Physics 121,Sections 9,10,11,and 12
Lecture 8
Today’s Topics:
? Homework 3,Due Friday Sept,23 @ 6:00PM
? Ch.3,# 64,75,and 81.
? Ch.4,# 4,8,21,25,36,40,and 51.
? Chapter 4,Motion in 2-D
?More examples of FBD’s
? Examples with friction
Physics 121,Lecture 8,Pg 2
An Example
Consider the following two cases
Physics 121,Lecture 8,Pg 3
An Example
The Free Body Diagrams
mg
mg
FB,T= N
Ball Falls For Static Situation
N = mg
Physics 121,Lecture 8,Pg 4
An Example
The action/reaction pair forces
FB,E = -mg F
B,T= N
FE,B = mg
FB,E = -mg
FE,B = mg
FT,B= -N
Physics 121,Lecture 8,Pg 5
Lecture 8,ACT 1
Gravity and Normal Forces
A woman in an elevator is accelerating upwards
The normal force exerted by the elevator on the woman is,
A) greater than
B) the same as
C) less than
the force due to gravity acting on the woman
Physics 121,Lecture 8,Pg 6
Lecture 8,ACT 1b
Gravity and Normal Forces
A woman in an elevator is accelerating upwards
The normal force exerted by the elevator on the woman is,
A) greater than
B) the same as
C) less than
the force the woman exerts on the elevator.
Physics 121,Lecture 8,Pg 7
The Free Body Diagram
? Newton’s 2nd Law says that for an object
F = ma.
? Key phrase here is for an object.
? So before we can apply F = ma to any given
object we isolate the forces acting on this object:
Physics 121,Lecture 8,Pg 8
The Free Body Diagram...
? Consider the following case
?What are the forces acting on the plank?
P = plank
F = floor
W = wall
E = earth
FW,P
FP,W
FP,F FP,E
FF,P
FE,P
Physics 121,Lecture 8,Pg 9
The Free Body Diagram...
? Consider the following case
?What are the forces acting on the plank?
Isolate the plank from
the rest of the world.
FW,P
FP,W
FP,F FP,E
FF,P
FE,P
Physics 121,Lecture 8,Pg 10
The Free Body Diagram...
? The forces acting on the plank should reveal
themselves...
FP,W
FP,F FP,E
Physics 121,Lecture 8,Pg 11
Aside...
? In this example the plank is not moving...
?And it is not accelerating!
?So FNET = ma becomes FNET = 0
?This is the basic idea behind statics.
FP,W + FP,F + FP,E = 0
FP,W
FP,F FP,E
Physics 121,Lecture 8,Pg 12
Example
? Example of a dynamics problem:
A box of mass m = 2kg slides on a horizontal
frictionless floor,A force Fx = 10N pushes on it in
the i direction,What is the acceleration of the
box?
F = Fx i a =?
m
j
i
Physics 121,Lecture 8,Pg 13
Example...
? Draw a picture showing all of the forces
F
FB,F
FF,B FB,E
FE,B
j
i
Physics 121,Lecture 8,Pg 14
Example...
? Draw a picture showing all of the forces.
? Isolate the forces acting on the block.
F
FB,F
FF,B FB,E = mg
FE,B
j
i
Physics 121,Lecture 8,Pg 15
Example...
? Draw a picture showing all of the forces.
? Isolate the forces acting on the block.
? Draw a free body diagram.
F
FB,F
mg
j
i
Physics 121,Lecture 8,Pg 16
Example...
? Draw a picture showing all of the forces.
? Isolate the forces acting on the block.
? Draw a free body diagram.
? Solve Newton’s equations for each component.
? FX = maX
? FB,F - mg = maY
F
FB,F
mg
j
i
Physics 121,Lecture 8,Pg 17
Example...
? FX = maX
? So aX = FX / m = (10 N)/(2 kg) = 5 m/s2.
? FB,F - mg = maY
? But aY = 0
? So FB,F = mg.
? The vertical component of the force
of the floor on the object (FB,F ) is
often called the Normal Force (N).
? Since aY = 0,N = mg in this case.
FX
N
mg
j
i
Physics 121,Lecture 8,Pg 18
Example Recap
FX
N = mg
mg
aX = FX / m
j
i
Physics 121,Lecture 8,Pg 19
Tools,Ropes & Strings
? Can be used to pull from a distance.
? Tension (T) at a certain position in a rope is the
magnitude of the force acting across a cross-
section of the rope at that position.
?The force you would feel if you cut the rope
and grabbed the ends.
?An action-reaction pair.
cut
T
T
T
Physics 121,Lecture 8,Pg 20
Tools,Ropes & Strings...
? The direction of the force provided by a rope is
along the direction of the rope:
mg
T
m
Since ay = 0 (box not moving),
T = mg
Physics 121,Lecture 8,Pg 21
Lecture 8,Act 2,Force and acceleration
? A fish is being yanked upward out of the water
using a fishing line that breaks when the tension
reaches 180 N,
The string snaps when the acceleration of the fish
is observed to be is 12.2 m/s2,What is the mass of
the fish?
m =?a = 12.2 m/s2
snap ! (a) 14.8 kg
(b) 18.4 kg
(c) 8.2 kg
Physics 121,Lecture 8,Pg 22
Tools,Pegs & Pulleys
? Used to change the direction of forces.
?An ideal massless pulley or ideal smooth peg will
change the direction of an applied force without
altering the magnitude:
F1
ideal peg
or pulley
F2
| F1 | = | F2 |
Physics 121,Lecture 8,Pg 23
Tools,Pegs & Pulleys
? Used to change the direction of forces.
?An ideal massless pulley or ideal smooth peg will
change the direction of an applied force without
altering the magnitude:
mg
T
m
T = mg
FW,S = mg
Physics 121,Lecture 8,Pg 24
m m m
(a) 0 lbs,(b) 4 lbs,(c) 8 lbs.
(1) (2)
Lecture 8,Act 3,Force and acceleration
? A block weighing 4 lbs is hung from a rope
attached to a scale,The scale is then attached to
a wall and reads 4 lbs,What will the scale read
when it is instead attached to another block
weighing 4 lbs?
Physics 121,Lecture 8,Pg 25
Example with pulley
? A mass M is held in place by a force F,
Find the tension in each segment of the
rope and the magnitude of F.
?Assume the pulleys massless and
frictionless.
?Assume the rope massless.
<M T5
T4
T3T
2
T1
F? We use the 5 step method.
?Draw a picture,what are we looking for?
?What physics idea are applicable? Draw
a diagram and list known and unknown
variables.
Newton’s 2nd law, F=ma
Free-body diagram for each object
Physics 121,Lecture 8,Pg 26
Pulleys,continued
? FBD for all objects
<M T5
T4
T3T
2
T1
F
T4
F=T1
T2
T3
T2 T3
T
5
M
T5
Mg
Physics 121,Lecture 8,Pg 27
Pulleys,finally
? Step 3,Plan the solution (what are the relevant equations)
?F=ma,static (no acceleration,mass is held in place)
M
T5
Mg
T5=Mg
T2 T3
T
5
T2+T3=T5
T4
F=T1
T2
T3
F=T1
T1+T2+T3=T4
Physics 121,Lecture 8,Pg 28
Pulleys,really finally!
? Step 4,execute the plan (solve in terms of variables)
?We have (from FBD):
T5=MgF=T1 T2+T3=T5 T1+T2+T3=T4
<M T5
T4
T3T
2
T1
F
T2=T3T1=T3
T2=Mg/2
T2+T3=T5 gives T5=2T2=Mg
F=T1=Mg/2
T1=T2=T3=Mg/2 and T4=3Mg/2
T5=Mg and
?Pulleys are massless and frictionless
? Step 5,evaluate the answer (here,
dimensions are OK and no numerical values)
Physics 121,Lecture 8,Pg 29
Exercise,Inclined plane
? A block of mass m slides down a frictionless ramp
that makes angle ? with respect to horizontal,
What is its acceleration a?
?
m a
Physics 121,Lecture 8,Pg 30
Inclined plane...
? Define convenient axes parallel and
perpendicular to plane:
?Acceleration a is in x direction only.
?
m a
i
j
Physics 121,Lecture 8,Pg 31
Inclined plane...
? Consider x and y components separately:
? i,mg sin ? = ma a = g sin ?
? j,N - mg cos ?= 0,N = mg cos ?
mg
N
mg sin ?
mg cos ?
?
ma
i
j
Physics 121,Lecture 8,Pg 32
Angles of an Inclined plane
?
?
f
ma = mg sin ?
??+ f = 90?
mg
N
Physics 121,Lecture 8,Pg 33
New Topic,Friction
? What does it do?
?It opposes motion!
? How do we characterize this in terms we have
learned?
?Friction results in a force in a direction
opposite to the direction of motion!
ma
FAPPLIED
fFRICTION mg
N
i
j
Physics 121,Lecture 8,Pg 34
Friction...
? Force of friction acts to oppose motion:
?Parallel to surface.
?Perpendicular to Normal force.
ma
F
fF mg
N
i
j
Physics 121,Lecture 8,Pg 35
Model for Sliding Friction
? The direction of the frictional force vector is perpendicular to the
normal force vector N.
? The magnitude of the frictional force vector |fF| is proportional to
the magnitude of the normal force |N |.
? |fF| = ?K | N | ( = ?K | mg | in the previous example)
?The,heavier” something is,the greater the friction will
be...makes sense!
? The constant ?K is called the,coefficient of kinetic friction”.
Physics 121,Lecture 8,Pg 36
Model...
? Dynamics:
i, F ??KN = m a
j, N = mg
so F ? ?Kmg = m a
ma
F
mg
N
i
j
?K mg
Physics 121,Lecture 8,Pg 37
Lecture 8,ACT 4
Friction and Motion
? A box of mass m1 = 1 kg is being pulled by a horizontal string
having tension T = 40 N,It slides with friction (?k=,5) on top
of a second box having mass m2 = 2 kg,which in turn slides
on an ice rink (frictionless).
?What is the acceleration of the second box?
(a) a = 0 m/s2 (b) a = 2.5 m/s2 (c) a = 10 m/s2
m2
T m1 slides with friction (?k=0.5 )
slides without frictiona =?
Physics 121,Lecture 8,Pg 38
Inclined Plane with Friction:
? Draw free-body diagram:
?
i
j
mg
N
?KN
ma
?
next
Physics 121,Lecture 8,Pg 39
Inclined plane...
? Consider i and j components of FTOT = ma,
i mg sin ? ? ?KN = ma
j N = mg cos ?
?
i
j
mg
N
?
?KN
ma
mg sin ?
mg cos ?
mg sin ? ? ?Kmg cos ? = ma
a / g = sin ? ? ?Kcos ?
next
Physics 121,Lecture 8,Pg 40
Example
Pulleys and Friction
Three blocks are connected on the table as shown,The table
has a coefficient of kinetic friction of 0.350,the masses are m1
= 4.00 kg,m2 = 1.00kg and m3 = 2.00kg,
a) What is the magnitude and direction of acceleration on the
three blocks?
b) What is the tension on the two cords?
m1
T1m2
m3
Physics 121,Lecture 8,Pg 41
m1
T1m2
m3
m1 m2 m3
N=-m2g
T23T12
m1g
T12
m3g
T23
T1
2
T12 T23
T23
T12 - m1g = - m1a T23 - m3g = m3a
?k m2ga a
a
-T12 + T23 + ?k m2g = - m2a
SOLUTION,T12 = = 30.0 N,T23 = 24.2 N,a = 2.31 m/s2 left for m2
m2g
Physics 121,Lecture 8,Pg 42
Static Friction...
F
mg
N
i
j
fS
? So far we have considered friction acting when
something moves.
?We also know that it acts in un-moving,static”
systems:
? In these cases,the force provided by friction will
depend on the forces applied on the system.
Physics 121,Lecture 8,Pg 43
Static Friction...
? Just like in the sliding case except a = 0.
i, F ? fS = 0
j, N = mg
F
mg
N
i
j
fS
? While the block is static,fS = F (unlike kinetic friction)
Physics 121,Lecture 8,Pg 44
Static Friction...
F
mg
N
i
j
fS
? The maximum possible force that the friction between two
objects can provide is fMAX = ?SN,where ?s is the
“coefficient of static friction”.
?So fS ? ?S N.
?As one increases F,fS gets bigger until fS = ?SN and the
object,breaks loose” and starts to move.
Physics 121,Lecture 8,Pg 45
Static Friction...
? ?S is discovered by increasing F until the block starts to
slide:
i, FMAX ? ?SN = 0
j, N = mg
?S = FMAX / mg
FMAX
mg
N
i
j
?Smg
Physics 121,Lecture 8,Pg 46
? Since f =?N,the force of friction does not depend on the
area of the surfaces in contact.
? By definition,it must be true that ?S > ?K for any
system (think about it...).
? Graph of Frictional force
vs Applied force:
Additional comments on Friction:
fF
FA
fF = FA
fF = ?KN
fF = ?SN
Active Figure
Physics 121,Lecture 8,Pg 47
Lecture 8,ACT 5:
Forces and Motion
? A box of mass m =10.21 kg is at rest on a floor,The static
coefficient of friction between the floor and the box is ?s = 0.4
? A rope is attached to the box and pulled at an angle of ? = 30o
above horizontal with tension T = 40 N.
?Does the box move?
(a) yes (b) no (c) too close to call
T
mstatic friction (?s = 0.4 ) ?
next
Physics 121,Lecture 8,Pg 48
Static Friction:
? We can also consider ?S on an inclined plane.
? In this case,the force provided by friction will depend on the
angle ? of the plane.
?
next
Physics 121,Lecture 8,Pg 49
Static Friction
?
mg
N
ma = 0 (block is not moving)
? The force provided by friction,fF,depends on ?.
?
fF
mg sin ? ? ff = 0
(Newton’s 2nd Law along x-axis)
i
j
next
Physics 121,Lecture 8,Pg 50
Static Friction
? We can find ?s by increasing the ramp angle until the block
slides:
?M mg
N
?SN
In this case:
?
mg sin ?M ? ?Smg cos?M = 0
?S = tan ?M
i
j
mg sin ? ? ff = 0
ff = ?SN = ? ?Smg cos?M
next
Physics 121,Lecture 8,Pg 51
Problem,Box on Truck
? A box with mass m sits in the back of a truck,The static
coefficient of friction between box and truck is ?S
?What is the maximum acceleration a that the truck can
have without the box slipping?
m ?S
a
next
Physics 121,Lecture 8,Pg 52
Problem,Box on Truck
? Draw Free Body Diagram for box:
?Consider case where fF is max...
(i.e,if the acceleration were any
larger,the box would slip).
N
fF = ?SN mg
i
j
next
Physics 121,Lecture 8,Pg 53
Problem,Box on Truck
? Use FTOT = ma for both i and j components
?i ?SN = maMAX
?j N = mg
aMAX = ?S g
N
fF = ?SN mg
aMAX
i
j
next
Physics 121,Lecture 8,Pg 54
Lecture 8,ACT 6
Forces and Motion
? An inclined plane is accelerating with constant acceleration
a,A box resting on the plane is held in place by static
friction,What is the direction of the static frictional force?
?S a
(a)
Ff
Ff
(b) Ff(c)
next
Physics 121,Lecture 8,Pg 55
Problem,Putting on the brakes
? Anti-lock brakes work by making sure the wheels roll
without slipping,This maximizes the frictional force slowing
the car since ?S > ?K,
? The driver of a car moving with speed vo slams on the
brakes,The static coefficient of friction between wheels
and the road is ?S, What is the stopping distance D?
ab
vo
v = 0
D
next
Physics 121,Lecture 8,Pg 56
? Use FTOT = ma for both i and j components
?i ?SN = ma
?j N = mg
a = ?S g
Problem,Putting on the brakes
N
fF = ?SN mg
a
i
j
next
Physics 121,Lecture 8,Pg 57
Problem,Putting on the brakes
? As in the last example,find ab = ?Sg.
? Using the kinematic equation,v2- v02 = 2a( x -x0 )
? In our problem,0 - v02 = ? 2ab( D )
ab
vo
v = 0
D
next
Physics 121,Lecture 8,Pg 58
Problem,Putting on the brakes
? In our problem,0 - v02 = ? 2ab( D )
? Solving for D:
Putting in ab = ?Sg
ab
vo
v = 0
D
b
2
0a2vD =
D v g
s
= 0
2
2 ?
end
Physics 121,Lecture 8,Pg 59
Recap of today’s lecture
? Homework 3,Due Friday Sept,23 @ 6:00PM
?Ch.3,# 64,75,and 81.
?Ch.4,# 4,8,21,25,36,and 40.
? Chapter 4,Motion in 2-D
?More examples of FBD’s
?Examples with friction
Physics 121,Sections 9,10,11,and 12
Lecture 8
Today’s Topics:
? Homework 3,Due Friday Sept,23 @ 6:00PM
? Ch.3,# 64,75,and 81.
? Ch.4,# 4,8,21,25,36,40,and 51.
? Chapter 4,Motion in 2-D
?More examples of FBD’s
? Examples with friction
Physics 121,Lecture 8,Pg 2
An Example
Consider the following two cases
Physics 121,Lecture 8,Pg 3
An Example
The Free Body Diagrams
mg
mg
FB,T= N
Ball Falls For Static Situation
N = mg
Physics 121,Lecture 8,Pg 4
An Example
The action/reaction pair forces
FB,E = -mg F
B,T= N
FE,B = mg
FB,E = -mg
FE,B = mg
FT,B= -N
Physics 121,Lecture 8,Pg 5
Lecture 8,ACT 1
Gravity and Normal Forces
A woman in an elevator is accelerating upwards
The normal force exerted by the elevator on the woman is,
A) greater than
B) the same as
C) less than
the force due to gravity acting on the woman
Physics 121,Lecture 8,Pg 6
Lecture 8,ACT 1b
Gravity and Normal Forces
A woman in an elevator is accelerating upwards
The normal force exerted by the elevator on the woman is,
A) greater than
B) the same as
C) less than
the force the woman exerts on the elevator.
Physics 121,Lecture 8,Pg 7
The Free Body Diagram
? Newton’s 2nd Law says that for an object
F = ma.
? Key phrase here is for an object.
? So before we can apply F = ma to any given
object we isolate the forces acting on this object:
Physics 121,Lecture 8,Pg 8
The Free Body Diagram...
? Consider the following case
?What are the forces acting on the plank?
P = plank
F = floor
W = wall
E = earth
FW,P
FP,W
FP,F FP,E
FF,P
FE,P
Physics 121,Lecture 8,Pg 9
The Free Body Diagram...
? Consider the following case
?What are the forces acting on the plank?
Isolate the plank from
the rest of the world.
FW,P
FP,W
FP,F FP,E
FF,P
FE,P
Physics 121,Lecture 8,Pg 10
The Free Body Diagram...
? The forces acting on the plank should reveal
themselves...
FP,W
FP,F FP,E
Physics 121,Lecture 8,Pg 11
Aside...
? In this example the plank is not moving...
?And it is not accelerating!
?So FNET = ma becomes FNET = 0
?This is the basic idea behind statics.
FP,W + FP,F + FP,E = 0
FP,W
FP,F FP,E
Physics 121,Lecture 8,Pg 12
Example
? Example of a dynamics problem:
A box of mass m = 2kg slides on a horizontal
frictionless floor,A force Fx = 10N pushes on it in
the i direction,What is the acceleration of the
box?
F = Fx i a =?
m
j
i
Physics 121,Lecture 8,Pg 13
Example...
? Draw a picture showing all of the forces
F
FB,F
FF,B FB,E
FE,B
j
i
Physics 121,Lecture 8,Pg 14
Example...
? Draw a picture showing all of the forces.
? Isolate the forces acting on the block.
F
FB,F
FF,B FB,E = mg
FE,B
j
i
Physics 121,Lecture 8,Pg 15
Example...
? Draw a picture showing all of the forces.
? Isolate the forces acting on the block.
? Draw a free body diagram.
F
FB,F
mg
j
i
Physics 121,Lecture 8,Pg 16
Example...
? Draw a picture showing all of the forces.
? Isolate the forces acting on the block.
? Draw a free body diagram.
? Solve Newton’s equations for each component.
? FX = maX
? FB,F - mg = maY
F
FB,F
mg
j
i
Physics 121,Lecture 8,Pg 17
Example...
? FX = maX
? So aX = FX / m = (10 N)/(2 kg) = 5 m/s2.
? FB,F - mg = maY
? But aY = 0
? So FB,F = mg.
? The vertical component of the force
of the floor on the object (FB,F ) is
often called the Normal Force (N).
? Since aY = 0,N = mg in this case.
FX
N
mg
j
i
Physics 121,Lecture 8,Pg 18
Example Recap
FX
N = mg
mg
aX = FX / m
j
i
Physics 121,Lecture 8,Pg 19
Tools,Ropes & Strings
? Can be used to pull from a distance.
? Tension (T) at a certain position in a rope is the
magnitude of the force acting across a cross-
section of the rope at that position.
?The force you would feel if you cut the rope
and grabbed the ends.
?An action-reaction pair.
cut
T
T
T
Physics 121,Lecture 8,Pg 20
Tools,Ropes & Strings...
? The direction of the force provided by a rope is
along the direction of the rope:
mg
T
m
Since ay = 0 (box not moving),
T = mg
Physics 121,Lecture 8,Pg 21
Lecture 8,Act 2,Force and acceleration
? A fish is being yanked upward out of the water
using a fishing line that breaks when the tension
reaches 180 N,
The string snaps when the acceleration of the fish
is observed to be is 12.2 m/s2,What is the mass of
the fish?
m =?a = 12.2 m/s2
snap ! (a) 14.8 kg
(b) 18.4 kg
(c) 8.2 kg
Physics 121,Lecture 8,Pg 22
Tools,Pegs & Pulleys
? Used to change the direction of forces.
?An ideal massless pulley or ideal smooth peg will
change the direction of an applied force without
altering the magnitude:
F1
ideal peg
or pulley
F2
| F1 | = | F2 |
Physics 121,Lecture 8,Pg 23
Tools,Pegs & Pulleys
? Used to change the direction of forces.
?An ideal massless pulley or ideal smooth peg will
change the direction of an applied force without
altering the magnitude:
mg
T
m
T = mg
FW,S = mg
Physics 121,Lecture 8,Pg 24
m m m
(a) 0 lbs,(b) 4 lbs,(c) 8 lbs.
(1) (2)
Lecture 8,Act 3,Force and acceleration
? A block weighing 4 lbs is hung from a rope
attached to a scale,The scale is then attached to
a wall and reads 4 lbs,What will the scale read
when it is instead attached to another block
weighing 4 lbs?
Physics 121,Lecture 8,Pg 25
Example with pulley
? A mass M is held in place by a force F,
Find the tension in each segment of the
rope and the magnitude of F.
?Assume the pulleys massless and
frictionless.
?Assume the rope massless.
<M T5
T4
T3T
2
T1
F? We use the 5 step method.
?Draw a picture,what are we looking for?
?What physics idea are applicable? Draw
a diagram and list known and unknown
variables.
Newton’s 2nd law, F=ma
Free-body diagram for each object
Physics 121,Lecture 8,Pg 26
Pulleys,continued
? FBD for all objects
<M T5
T4
T3T
2
T1
F
T4
F=T1
T2
T3
T2 T3
T
5
M
T5
Mg
Physics 121,Lecture 8,Pg 27
Pulleys,finally
? Step 3,Plan the solution (what are the relevant equations)
?F=ma,static (no acceleration,mass is held in place)
M
T5
Mg
T5=Mg
T2 T3
T
5
T2+T3=T5
T4
F=T1
T2
T3
F=T1
T1+T2+T3=T4
Physics 121,Lecture 8,Pg 28
Pulleys,really finally!
? Step 4,execute the plan (solve in terms of variables)
?We have (from FBD):
T5=MgF=T1 T2+T3=T5 T1+T2+T3=T4
<M T5
T4
T3T
2
T1
F
T2=T3T1=T3
T2=Mg/2
T2+T3=T5 gives T5=2T2=Mg
F=T1=Mg/2
T1=T2=T3=Mg/2 and T4=3Mg/2
T5=Mg and
?Pulleys are massless and frictionless
? Step 5,evaluate the answer (here,
dimensions are OK and no numerical values)
Physics 121,Lecture 8,Pg 29
Exercise,Inclined plane
? A block of mass m slides down a frictionless ramp
that makes angle ? with respect to horizontal,
What is its acceleration a?
?
m a
Physics 121,Lecture 8,Pg 30
Inclined plane...
? Define convenient axes parallel and
perpendicular to plane:
?Acceleration a is in x direction only.
?
m a
i
j
Physics 121,Lecture 8,Pg 31
Inclined plane...
? Consider x and y components separately:
? i,mg sin ? = ma a = g sin ?
? j,N - mg cos ?= 0,N = mg cos ?
mg
N
mg sin ?
mg cos ?
?
ma
i
j
Physics 121,Lecture 8,Pg 32
Angles of an Inclined plane
?
?
f
ma = mg sin ?
??+ f = 90?
mg
N
Physics 121,Lecture 8,Pg 33
New Topic,Friction
? What does it do?
?It opposes motion!
? How do we characterize this in terms we have
learned?
?Friction results in a force in a direction
opposite to the direction of motion!
ma
FAPPLIED
fFRICTION mg
N
i
j
Physics 121,Lecture 8,Pg 34
Friction...
? Force of friction acts to oppose motion:
?Parallel to surface.
?Perpendicular to Normal force.
ma
F
fF mg
N
i
j
Physics 121,Lecture 8,Pg 35
Model for Sliding Friction
? The direction of the frictional force vector is perpendicular to the
normal force vector N.
? The magnitude of the frictional force vector |fF| is proportional to
the magnitude of the normal force |N |.
? |fF| = ?K | N | ( = ?K | mg | in the previous example)
?The,heavier” something is,the greater the friction will
be...makes sense!
? The constant ?K is called the,coefficient of kinetic friction”.
Physics 121,Lecture 8,Pg 36
Model...
? Dynamics:
i, F ??KN = m a
j, N = mg
so F ? ?Kmg = m a
ma
F
mg
N
i
j
?K mg
Physics 121,Lecture 8,Pg 37
Lecture 8,ACT 4
Friction and Motion
? A box of mass m1 = 1 kg is being pulled by a horizontal string
having tension T = 40 N,It slides with friction (?k=,5) on top
of a second box having mass m2 = 2 kg,which in turn slides
on an ice rink (frictionless).
?What is the acceleration of the second box?
(a) a = 0 m/s2 (b) a = 2.5 m/s2 (c) a = 10 m/s2
m2
T m1 slides with friction (?k=0.5 )
slides without frictiona =?
Physics 121,Lecture 8,Pg 38
Inclined Plane with Friction:
? Draw free-body diagram:
?
i
j
mg
N
?KN
ma
?
next
Physics 121,Lecture 8,Pg 39
Inclined plane...
? Consider i and j components of FTOT = ma,
i mg sin ? ? ?KN = ma
j N = mg cos ?
?
i
j
mg
N
?
?KN
ma
mg sin ?
mg cos ?
mg sin ? ? ?Kmg cos ? = ma
a / g = sin ? ? ?Kcos ?
next
Physics 121,Lecture 8,Pg 40
Example
Pulleys and Friction
Three blocks are connected on the table as shown,The table
has a coefficient of kinetic friction of 0.350,the masses are m1
= 4.00 kg,m2 = 1.00kg and m3 = 2.00kg,
a) What is the magnitude and direction of acceleration on the
three blocks?
b) What is the tension on the two cords?
m1
T1m2
m3
Physics 121,Lecture 8,Pg 41
m1
T1m2
m3
m1 m2 m3
N=-m2g
T23T12
m1g
T12
m3g
T23
T1
2
T12 T23
T23
T12 - m1g = - m1a T23 - m3g = m3a
?k m2ga a
a
-T12 + T23 + ?k m2g = - m2a
SOLUTION,T12 = = 30.0 N,T23 = 24.2 N,a = 2.31 m/s2 left for m2
m2g
Physics 121,Lecture 8,Pg 42
Static Friction...
F
mg
N
i
j
fS
? So far we have considered friction acting when
something moves.
?We also know that it acts in un-moving,static”
systems:
? In these cases,the force provided by friction will
depend on the forces applied on the system.
Physics 121,Lecture 8,Pg 43
Static Friction...
? Just like in the sliding case except a = 0.
i, F ? fS = 0
j, N = mg
F
mg
N
i
j
fS
? While the block is static,fS = F (unlike kinetic friction)
Physics 121,Lecture 8,Pg 44
Static Friction...
F
mg
N
i
j
fS
? The maximum possible force that the friction between two
objects can provide is fMAX = ?SN,where ?s is the
“coefficient of static friction”.
?So fS ? ?S N.
?As one increases F,fS gets bigger until fS = ?SN and the
object,breaks loose” and starts to move.
Physics 121,Lecture 8,Pg 45
Static Friction...
? ?S is discovered by increasing F until the block starts to
slide:
i, FMAX ? ?SN = 0
j, N = mg
?S = FMAX / mg
FMAX
mg
N
i
j
?Smg
Physics 121,Lecture 8,Pg 46
? Since f =?N,the force of friction does not depend on the
area of the surfaces in contact.
? By definition,it must be true that ?S > ?K for any
system (think about it...).
? Graph of Frictional force
vs Applied force:
Additional comments on Friction:
fF
FA
fF = FA
fF = ?KN
fF = ?SN
Active Figure
Physics 121,Lecture 8,Pg 47
Lecture 8,ACT 5:
Forces and Motion
? A box of mass m =10.21 kg is at rest on a floor,The static
coefficient of friction between the floor and the box is ?s = 0.4
? A rope is attached to the box and pulled at an angle of ? = 30o
above horizontal with tension T = 40 N.
?Does the box move?
(a) yes (b) no (c) too close to call
T
mstatic friction (?s = 0.4 ) ?
next
Physics 121,Lecture 8,Pg 48
Static Friction:
? We can also consider ?S on an inclined plane.
? In this case,the force provided by friction will depend on the
angle ? of the plane.
?
next
Physics 121,Lecture 8,Pg 49
Static Friction
?
mg
N
ma = 0 (block is not moving)
? The force provided by friction,fF,depends on ?.
?
fF
mg sin ? ? ff = 0
(Newton’s 2nd Law along x-axis)
i
j
next
Physics 121,Lecture 8,Pg 50
Static Friction
? We can find ?s by increasing the ramp angle until the block
slides:
?M mg
N
?SN
In this case:
?
mg sin ?M ? ?Smg cos?M = 0
?S = tan ?M
i
j
mg sin ? ? ff = 0
ff = ?SN = ? ?Smg cos?M
next
Physics 121,Lecture 8,Pg 51
Problem,Box on Truck
? A box with mass m sits in the back of a truck,The static
coefficient of friction between box and truck is ?S
?What is the maximum acceleration a that the truck can
have without the box slipping?
m ?S
a
next
Physics 121,Lecture 8,Pg 52
Problem,Box on Truck
? Draw Free Body Diagram for box:
?Consider case where fF is max...
(i.e,if the acceleration were any
larger,the box would slip).
N
fF = ?SN mg
i
j
next
Physics 121,Lecture 8,Pg 53
Problem,Box on Truck
? Use FTOT = ma for both i and j components
?i ?SN = maMAX
?j N = mg
aMAX = ?S g
N
fF = ?SN mg
aMAX
i
j
next
Physics 121,Lecture 8,Pg 54
Lecture 8,ACT 6
Forces and Motion
? An inclined plane is accelerating with constant acceleration
a,A box resting on the plane is held in place by static
friction,What is the direction of the static frictional force?
?S a
(a)
Ff
Ff
(b) Ff(c)
next
Physics 121,Lecture 8,Pg 55
Problem,Putting on the brakes
? Anti-lock brakes work by making sure the wheels roll
without slipping,This maximizes the frictional force slowing
the car since ?S > ?K,
? The driver of a car moving with speed vo slams on the
brakes,The static coefficient of friction between wheels
and the road is ?S, What is the stopping distance D?
ab
vo
v = 0
D
next
Physics 121,Lecture 8,Pg 56
? Use FTOT = ma for both i and j components
?i ?SN = ma
?j N = mg
a = ?S g
Problem,Putting on the brakes
N
fF = ?SN mg
a
i
j
next
Physics 121,Lecture 8,Pg 57
Problem,Putting on the brakes
? As in the last example,find ab = ?Sg.
? Using the kinematic equation,v2- v02 = 2a( x -x0 )
? In our problem,0 - v02 = ? 2ab( D )
ab
vo
v = 0
D
next
Physics 121,Lecture 8,Pg 58
Problem,Putting on the brakes
? In our problem,0 - v02 = ? 2ab( D )
? Solving for D:
Putting in ab = ?Sg
ab
vo
v = 0
D
b
2
0a2vD =
D v g
s
= 0
2
2 ?
end
Physics 121,Lecture 8,Pg 59
Recap of today’s lecture
? Homework 3,Due Friday Sept,23 @ 6:00PM
?Ch.3,# 64,75,and 81.
?Ch.4,# 4,8,21,25,36,and 40.
? Chapter 4,Motion in 2-D
?More examples of FBD’s
?Examples with friction
