Transcript Lecture-07-09

Lecture 7
Newton’s Laws of Motion
Midterm Test #1 - Thursday!
-
21 multiple-choice problems
- A calculator will be needed.
- CHECK YOUR BATTERIES!
- NO equations or information may be stored in
your calculator. This is part of your pledge on the
exam.
- Scratch paper will be provided, to be turned in at
the end of the exam.
- A sign-in sheets will be used, photographs of the
class will be taken, all test papers will be collected
A weight on a string...
if I pull the bottom string
down, which string will
break first?
Ftop
a) top string
b) bottom string
Fbot
W
c) there is not enough information to answer
this question
How quickly is the string pulled? A sudden, strong
tug is resisted by the inertia of the mass, protecting
the top string. A gradual pull forces the top string
to keep the system in equilibrium.
Translation 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
Before practicing his routine on the rings, a 67-kg gymnast stands
motionless, with one hand grasping each ring and his feet touching the
ground. Both arms slope upward at an angle of 24° above the
horizontal.
(a) If the force exerted by the rings on each arm has a magnitude of 290
N, and is directed along the length of the arm, what is the magnitude
of the force exerted by the floor on his feet?
(b) If the angle his arms make with the horizontal is greater that 24°,
and everything else remains the same, is the force exerted by the
floor on his feet greater than, less than, or the same as the value
found in part (a)? Explain.
Before practicing his routine on the rings, a 67-kg gymnast stands
motionless, with one hand grasping each ring and his feet touching the
ground. Both arms slope upward at an angle of 24° above the
horizontal.
(a) If the force exerted by the rings on each arm has a magnitude of 290
N, and is directed along the length of the arm, what is the magnitude
of the force exerted by the floor on his feet?
(b) If the angle his arms make with the horizontal is greater that 24°,
and everything else remains the same, is the force exerted by the
floor on his feet greater than, less than, or the same as the value
found in part (a)? Explain.
N
a)
b) if the angle is larger and everything else remains the same,
the applied forces are more vertical. With more upward
force from the arms, LESS normal force is required for zero
acceleration
Lecture 6
Applications of
Newton’s Laws
(Chapter 6)
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
8
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),
N
m
a>0
mg
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?
F = N – mg = ma > 0
Thus, N =mg+ma > mg
9
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
1
10
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
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
The 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 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
Static Friction
A block sits on a flat table. What is
the force of static friction?
a) zero
b) infinite
c) you need to tell me stuff, like the
mass of the block, μs, and what
planet this is happening on
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?
Measuring static coefficient of friction
If the block doesn’t move, a=0.
Given the “critical angle” at
which the block starts to slip,
what is μs?
y
x
N
fs
Wx
at the critical point

W

Wy
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?
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
m
2
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
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.
N
f
Wy
Wx

W

2
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
In this class: we will assume
that all ropes, strings, wires,
etc. are massless unless
otherwise stated.
Tdown
Tension is the same
everywhere in a
massless rope!
Massive vs. 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
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
2
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?
2
Tension
Force is always along a rope
Ty
T
T
T
T
W
Idealization: The Pulley
An ideal pulley is one that simply changes the
direction of the tension
distance box moves =
distance hands move
speed of box =
speed of hands
acceleration of box =
acceleration of hands
Tension in the rope?
Tension in the rope?
T
T
2.00 kg
W
W
m1 : x :
y:
μk
m2 : y :
μk
fk
μk
Over the Edge
In which case does block m experience a
a) case (1)
larger acceleration? In case (1) there is a
10 kg mass hanging from a rope and
b) acceleration is zero
falling. In case (2) a hand is providing a
c) both cases are the same
constant downward force of 98 N.
Assume massless rope and frictionless
d) depends on value of m
table.
e) case (2)
m
m
10 kg
a
a
F = 98 N
Case (1)
Case (2)
3
35
Over the Edge
In which case does block m experience a
a) case (1)
larger acceleration? In case (1) there is a
b) acceleration is zero
10 kg mass hanging from a rope and
falling. In case (2) a hand is providing a
c) both cases are the same
constant downward force of 98 N. Assume
d) depends on value of m
massless rope and frictionless table.
e) case (2)
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
Case (1)
Case (2)
were at rest would the
tension be equal to 98 N.
3
36
Springs
Hooke’s law for springs states that the
force increases with the amount the spring
is stretched or compressed:
The constant k is called the spring constant.
Springs
Note: we are discussing the force of the
spring on the mass. The force of the spring
on the wall are equal, and opposite.
Springs and Tension
S1
S2
A mass M hangs on spring 1, stretching it length L1
Mass M hangs on spring 2, stretching it length L2
Now spring1 and spring2 are connected end-to-end,
and M1 is hung below. How far does the combined
spring stretch?
a) (L1 + L2) / 2
b) L1 or L2, whichever is smaller
c) L1 or L2, whichever is bigger
d) depends on which order the
springs are attached
e) L1 + L2
Springs and Tension
A mass M hangs on spring 1, stretching it length L1
Mass M hangs on spring 2, stretching it length L2
Now spring1 and spring2 are connected end-to-end,
and M1 is hung below. How far does the combined
spring stretch?
S1
S2
Fs=T
W
a) (L1 + L2) / 2
b) L1 or L2, whichever is smaller
c) L1 or L2, whichever is bigger
d) depends on which order the
springs are attached
e) L1 + L2
Springs and Tension
A mass M hangs on spring 1, stretching it length L1
Mass M hangs on spring 2, stretching it length L2
Now spring1 and spring2 are connected end-to-end,
and M1 is hung below. How far does the combined
spring stretch?
S1
S2
Fs=T
W
a) (L1 + L2) / 2
b) L1 or L2, whichever is smaller
c) L1 or L2, whichever is bigger
d) depends on which order the
springs are attached
e) L1 + L2
Spring 1 supports the weight.
Spring 2 supports the weight.
Both feel the same force, and
stretch the same distance as before.