Transcript Lecture07-09

Lecture 7
Applications of
Newton’s Laws
(Chapter 6)
Reading and Review
Going Up I
A block of mass m rests on the floor of
a) N > mg
an elevator that is moving upward at
b) N = mg
constant speed. What is the relationship
between the force due to gravity and the
normal force on the block?
c) N < mg (but not zero)
d) N = 0
e) depends on the size of the
elevator
v
m
Going Up I
A block of mass m rests on the floor of
a) N > mg
an elevator that is moving upward at
b) N = mg
constant speed. What is the relationship
between the force due to gravity and the
normal force on the block?
c) N < mg (but not zero)
d) N = 0
e) depends on the size of the
elevator
The block is moving at constant speed, so it
must have no net force on it. The forces on
v
it are N (up) and mg (down), so N = mg, just
like the block at rest on a table.
m
Going Up II
A block of mass m rests on the
a) N > mg
floor of an elevator that is
b) N = mg
accelerating upward. What is
c) N < mg (but not zero)
the relationship between the
d) N = 0
force due to gravity and the
normal force on the block?
e) depends on the size of the
elevator
m
a
Going Up II
A block of mass m rests on the
a) N > mg

floor of an elevator
that is
b) N = mg
accelerating upward. What is
c) N < mg (but not zero)
the relationship between the
d) N = 0
force due to gravity and the
normal force on the block?
e) depends on the size of the
elevator
The block is accelerating upward, so it
must have a net upward force. The
forces on it are N (up) and mg (down),
so N must be greater than mg in order
to give the net upward force!
Follow-up: What is the normal force if
the elevator is in free fall downward?
N
m
a>0
mg
F = N – mg = ma > 0
→N = mg + ma > mg
Frictional Forces
Friction has its basis in surfaces that are not
completely smooth:
Kinetic friction
Kinetic friction: the friction experienced by surfaces
sliding against one another.
This frictional force is proportional to the contact force
between the two surfaces (normal force):
The constant
is called the coefficient of
kinetic friction.
fk always points in the direction
opposing motion of two surfaces
Frictional Forces
fk
fk
Naturally, for any frictional
force on a body, there is an
opposing reaction force on
the other body
Frictional Forces
fk
fk
when moving, one bumps
“skip” over each other
kinetic friction
fs
fs
when relative motion
stops, surfaces settle into
one another
static friction
Kinetic Friction ≤ Static Friction
Static Friction
The static frictional force tries to keep an object from
starting to move when other forces are applied.
The static frictional force has a maximum value, but
may take on any value from zero to the maximum...
depending on what is needed to keep
the sum of forces to zero.
The maximum static frictional
force is also proportional to
the contact force
Similarities between
→normal forces and
→static friction
• Variable; as strong as necessary to
prevent relative motion either
- perpendicular to surface (normal)
- parallel to surface (friction)
Characteristics of Frictional Forces
• Frictional forces always oppose relative motion
• Static and kinetic frictional forces are independent of
the area of contact between objects
• Kinetic frictional force is also independent of the
relative speed of the surfaces.
• Coefficients of friction are
independent of the mass
of objects, but in (most)
cases forces are not:
(twice the mass
→ twice the weight
→ twice the normal force
→ twice the frictional force)
Coefficients of Friction
Q: what units?
Going Sledding
Your little sister wants
you to give her a ride
on her sled. On level
ground, what is the
easiest way to
accomplish this?
a) pushing her from behind
b) pulling her from the front
c) both are equivalent
d) it is impossible to move the sled
e) tell her to get out and walk
1
2
Going Sledding
Your little sister wants
you to give her a ride
on her sled. On level
ground, what is the
easiest way to
accomplish this?
a) pushing her from behind
b) pulling her from the front
c) both are equivalent
d) it is impossible to move the sled
e) tell her to get out and walk
In case 1, the force F is pushing down
(in addition to mg), so the normal force
1
is larger. In case 2, the force F is
pulling up, against gravity, so the
normal force is lessened. Recall that
the frictional force is proportional to
the normal force.
2
Measuring static coefficient of friction
If the block doesn’t move, a=0.
y
x
N
fs
Wx
W


Wy
at the critical point
Acceleration of a block on an incline
If the object is sliding down -
v
y
x
N
fk
Wx
W


Wy
Acceleration of a block on an incline
If the object is sliding up -
v
y
x
N
Wx
fk
W


Wy
What will happen
when it stops?
A mass m, initially moving with a speed of 5.0 m/s, slides up a 30o ramp.
If the coefficient of kinetic friction is 0.4, how far up the ramp will the
mass slide?
If the coefficient of static friction is 0.6, will the mass eventually slide
down the ramp?
A mass m, initially moving with a speed of 5.0 m/s, slides up a 30o ramp.
If the coefficient of kinetic friction is 0.4, how far up the ramp will the
mass slide?
If the coefficient of static friction is 0.6, will the mass eventually slide
down the ramp?
yˆ : 0  N  mg cos  N  mg cos 
xˆ : Fx  mg sin   k N   mg sin   k mg cos 
 ax 
Fx
  g sin   k g cos   g  sin    k cos 
m
 9.81 sin 30o  0.4cos30o   8.30 m s 2
v 2f  0  v02  2ax  x f  x0 
 x f  x0  x  v 2ax  5.0 2  8.30   1.51m
2
0
2
When mass stops:
?
Fx  mg sin    s N  mg sin    s mg cos
  mg   sin    s cos   mg   sin 30o   s cos30o 
 mg  0.02   0
 mass will remain at rest
Sliding Down II
A mass m is placed on an
inclined plane ( > 0) and
slides down the plane with
constant speed. If a similar
block (same ) of mass 2m
were placed on the same
incline, it would:
m
a) not move at all
b) slide a bit, slow down, then stop
c) accelerate down the incline
d) slide down at constant speed
e) slide up at constant speed
Sliding Down II
A mass m is placed on an
inclined plane ( > 0) and
slides down the plane with
constant speed. If a similar
block (same ) of mass 2m
were placed on the same
incline, it would:
a) not move at all
b) slide a bit, slow down, then stop
c) accelerate down the incline
d) slide down at constant speed
e) slide up at constant speed
The component of gravity acting down
N
f
the plane is double for 2m. However, the
normal force (and hence the friction
force) is also double (the same factor!).
This means the two forces still cancel to
give a net force of zero.
Wy

Wx
W

Translational Equilibrium
“translational equilibrium” = fancy term for not accelerating
= the net force on an object is zero
example: book on a table
example: book on a table in an elevator
at constant velocity
Tension
When you pull on a string or rope, it becomes
taut. We say that there is tension in the string.
Note: strings are “floppy”, so force from a string
is along the string!
Tension in a chain
Tup = Tdown when W = 0
Tup
W
Tdown
Massless rope
The tension in a real rope will vary along its
length, due to the weight of the rope.
T3 = mg + Wr
In this class: we will assume
that all ropes, strings, wires,
etc. are massless unless
otherwise stated.
T2 = mg + Wr/2
T1 = mg
Tension is the same
everywhere in a
massless rope!
m
Idealization: The Pulley
An ideal pulley is one that only changes the
direction of the tension
along with a rope: useful class of
problems of combined motion
distance box moves =
distance hands move
speed of box =
speed of hands
acceleration of box =
acceleration of hands
Tension in the rope?
Translational equilibrium?
Translational equilibrium?
T
T
2.00 kg
W
W
Tension in the rope?
m1 : x :
y:
m2 : y :
Three Blocks
Three blocks of mass 3m, 2m, and
a) T1 > T2 > T3
m are connected by strings and
b) T1 < T2 < T3
pulled with constant acceleration a.
c) T1 = T2 = T3
What is the relationship between
d) all tensions are zero
the tension in each of the strings?
e) tensions are random
a
3m
T3
2m
T2
m
T1
Three Blocks
Three blocks of mass 3m, 2m, and
a) T1 > T2 > T3
m are connected by strings and
b) T1 < T2 < T3
pulled with constant acceleration a.
c) T1 = T2 = T3
What is the relationship between
d) all tensions are zero
the tension in each of the strings?
e) tensions are random
T1 pulls the whole set
of blocks along, so it
a
must be the largest. T2
pulls the last two
masses, but T3 only
pulls the last mass.
3m
T3
2m
T2
m
T1
Follow-up: What is T1 in terms of m and a?
Over the Edge
In which case does block m
a) case (1)
experience a larger acceleration?
b) acceleration is zero
In case (1) there is a 10 kg mass
c) both cases are the same
hanging from a rope and falling.
In case (2) a hand is providing a
d) depends on value of m
constant downward force of 98 N.
e) case (2)
Assume massless ropes.
m
m
10 kg
a
a
F = 98 N
Case (1)
Case (2)
Over the Edge
In which case does block m
a) case (1)
experience a larger acceleration?
b) acceleration is zero
In case (1) there is a 10 kg mass
c) both cases are the same
hanging from a rope and falling.
In case (2) a hand is providing a
d) depends on value of m
constant downward force of 98 N.
e) case (2)
Assume massless ropes.
98 N due to the hand. In
case (1) the tension is
m
m
In case (2) the tension is
10 kg
a
less than 98 N because
a
F = 98 N
the block is accelerating
down. Only if the block
were at rest would the
tension be equal to 98 N.
Case (1)
Case (2)
Elevate Me
You are holding your 2.0 kg
a) in freefall
physics text book while
b) moving upwards with a constant
velocity of 4.9 m/s
standing on an elevator.
Strangely, the book feels as if
it weighs exactly 2.5 kg. From
this, you conclude that the
elevator is:
c) moving down with a constant velocity
of 4.9 m/s
d) experiencing a constant acceleration
of about 2.5 m/s2 upward
e) experiencing a constant acceleration
of about 2.5 m/s2 downward
Use Newton’s 2nd law! the apparent weight:
and the sum of forces:
give a positive acceleration ay