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
an elevator that is moving upward at
constant speed. What is the
relationship between the force due to
gravity and the normal force on the
block?
a) N > mg
b) N = mg
c) N < mg (but not zero)
d) N = 0
e) depends on the size of the elevator
m
v
The block is moving at constant speed, so
it must have no net force on it. The
forces on it are N (up) and mg (down), so
N = mg, just like the block at rest on a
table.
Going Up I
A block of mass m rests on the floor of
an elevator that is moving upward at
constant speed. What is the
relationship between the force due to
gravity and the normal force on the
block?
a) N > mg
b) N = mg
c) N < mg (but not zero)
d) N = 0
e) depends on the size of the elevator
m
v
a) N > mg
b) N = mg
c) N < mg (but not zero)
d) N = 0
e) depends on the size of the elevator
m a
A block of mass m rests on the
floor of an elevator that is
accelerating upward. What is
the relationship between the
force due to gravity and the
normal force on the block?
Going Up II
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!
a) N > mg
b) N = mg
c) N < mg (but not zero)
d) N = 0
e) depends on the size of the elevator
F = N – mg = ma > 0→N = mg + ma > mg
m a > 0
mg
N
A block of mass m rests on the
floor of an elevator that is
accelerating upward. What is
the relationship between the
force due to gravity and the
normal force on the block?
Going Up II
Follow-up: What is the normal force if the elevator is in free fall downward?
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
fs
fs
when relative motion stops, surfaces settle
into one another
static frictionkinetic friction
Kinetic Friction ≤ Static Friction
The static frictional force tries to keep an object from starting to move when other forces are
applied.
Static Friction
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.
(twice the mass → twice the weight → twice the normal force → twice the frictional force)
• Coefficients of friction are independent of the mass of objects, but in (most) cases forces are not:
Coefficients of Friction
Q: what units?
Going Sledding
1
2
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
Your little sister wants
you to give her a ride
on her sled. On level
ground, what is the
easiest way to
accomplish this?
Going Sledding
1
2
In case 1, the force F is pushing down
(in addition to mg), so the normal
force 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.
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
Your little sister wants
you to give her a ride
on her sled. On level
ground, what is the
easiest way to
accomplish this?
Measuring static coefficient of friction
N
W
fs
x
y
Wx
Wy
If the block doesn’t move, a=0.
at the critical point
Acceleration of a block on an incline
N
W
fk
x
y
Wx
Wy
If the object is sliding down -
v
Acceleration of a block on an incline
N
W
fk
x
y
Wx
Wy
If the object is sliding up -
v
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?
2
2 20 0
2 20 0
?
ˆ : 0 cos cos
ˆ : sin sin cos
sin cos sin cos
9.81 sin30 0.4cos30 8.30
0 2
2 5.0 2 8.30 1.51
When mass stops:
s
x k k
xx k k
o o
f x f
f x
x
y N mg N mg
x F mg N mg mg
Fa g g gm
m s
v v a x x
x x x v a m
F mg
in sin cos
sin cos sin30 cos30
0.02 0
mass will remain at rest
s s
o os s
N mg mg
mg mg
mg
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
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:
Sliding Down II
The component of gravity acting down
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.
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:
W
Nf
Wx
Wy
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
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
W
Tup
Tdown
Tup = Tdown when W = 0
Massless rope
The tension in a real rope will vary along its length, due to the weight of the rope.
In this class: we will assume that all ropes, strings, wires, etc. are massless unless otherwise stated.
T1 = mg
m
T3 = mg + Wr
T2 = mg + Wr/2
Tension is the same everywhere in a massless rope!
Idealization: The PulleyAn ideal pulley is one that only changes the
direction of the tension
along with a rope: useful class of problems of combined
motiondistance box moves = distance hands move
speed of box = speed of hands
acceleration of box = acceleration of hands
Tension in the rope?
Translational equilibrium?
2.00 kg
Tension in the rope?
Translational equilibrium?
W
W
TT
y :m1 : x :
m2 : y :
Three Blocks
T3 T2 T13m 2m m
a
a) T1 > T2 > T3
b) T1 < T2 < T3
c) T1 = T2 = T3
d) all tensions are zero
e) tensions are random
Three blocks of mass 3m, 2m, and
m are connected by strings and
pulled with constant acceleration a.
What is the relationship between
the tension in each of the strings?
T1 pulls the whole set
of blocks along, so it
must be the largest.
T2 pulls the last two
masses, but T3 only
pulls the last mass.
Three Blocks
T3 T2 T13m 2m m
a
a) T1 > T2 > T3
b) T1 < T2 < T3
c) T1 = T2 = T3
d) all tensions are zero
e) tensions are random
Three blocks of mass 3m, 2m, and
m are connected by strings and
pulled with constant acceleration a.
What is the relationship between
the tension in each of the strings?
Follow-up: What is T1 in terms of m and a?
Over the Edge
m
10 kg a
m
a
F = 98 N
Case (1) Case (2)
a) case (1)
b) acceleration is zero
c) both cases are the same
d) depends on value of m
e) case (2)
In which case does block m
experience a larger acceleration?
In case (1) there is a 10 kg mass
hanging from a rope and falling.
In case (2) a hand is providing a
constant downward force of 98
N. Assume massless ropes.
In case (2) the tension is
98 N due to the hand.
In case (1) the tension is
less than 98 N because
the block is accelerating
down. Only if the block
were at rest would the
tension be equal to 98
N.
Over the Edge
m
10 kg a
m
a
F = 98 N
Case (1) Case (2)
a) case (1)
b) acceleration is zero
c) both cases are the same
d) depends on value of m
e) case (2)
In which case does block m
experience a larger acceleration?
In case (1) there is a 10 kg mass
hanging from a rope and falling.
In case (2) a hand is providing a
constant downward force of 98
N. Assume massless ropes.
Use Newton’s 2nd law! the apparent weight:
You are holding your 2.0 kg
physics text book while
standing on an elevator.
Strangely, the book feels as if
it weighs exactly 2.5 kg. From
this, you conclude that the
elevator is:
Elevate Mea) in freefall
b) moving upwards with a constant velocity of 4.9 m/s
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
and the sum of forces:
give a positive acceleration ay