Lecture 7 Newton’s Laws of Motion. Midterm Test #1 - Thursday! 21 multiple-choice problems - A...
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Transcript of Lecture 7 Newton’s Laws of Motion. Midterm Test #1 - Thursday! 21 multiple-choice problems - A...
![Page 1: Lecture 7 Newton’s Laws of Motion. Midterm Test #1 - Thursday! 21 multiple-choice problems - A calculator will be needed. - CHECK YOUR BATTERIES! -](https://reader035.fdocuments.us/reader035/viewer/2022062515/56649f4f5503460f94c7161d/html5/thumbnails/1.jpg)
Lecture 7
Newton’s Laws of Motion
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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
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Ftop
FbotW
A weight on a string...if I pull the bottom string down, which string will break first?
a) top string
b) bottom string
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.
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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
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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.
![Page 6: Lecture 7 Newton’s Laws of Motion. Midterm Test #1 - Thursday! 21 multiple-choice problems - A calculator will be needed. - CHECK YOUR BATTERIES! -](https://reader035.fdocuments.us/reader035/viewer/2022062515/56649f4f5503460f94c7161d/html5/thumbnails/6.jpg)
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
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Lecture 6
Applications of Newton’s Laws
(Chapter 6)
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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
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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 > 0Thus, 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?
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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
10
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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
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Frictional Forces
Friction has its basis in surfaces that are not completely smooth:
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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
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Frictional ForcesFrictional Forces
fk
fk
Naturally, for any frictional force on a body, there is an opposing reaction force on the other body
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Frictional ForcesFrictional Forces
fk
fk
when moving, one bumps “skip” over each other
fs
fs
when relative motion stops, surfaces settle
into one another
static friction
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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
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A block sits on a flat table. What is the force of static friction?
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
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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:
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Coefficients of Friction
Q: what units?
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Measuring static coefficient of friction
N
W
fs
x
y
Wx
Wy
If the block doesn’t move, a=0.
at the critical point
Given the “critical angle” at which the block starts to slip, what is μs?
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Acceleration of a block on an incline
N
W
fk
x
y
Wx
Wy
If the object is sliding down -
v
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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?
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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
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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
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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!
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Tension in a chainTension in a chain
W
Tup
Tdown
Tup = Tdown when W = 0
In this class: we will assume that all ropes, strings, wires, etc. are massless unless otherwise stated.
Tension is the same everywhere in a massless rope!
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Massive vs. 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!
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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?
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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?
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TensionTension
Force is always along a rope
W
TTTy
TT
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Idealization: The PulleyAn 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
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Tension in the rope?
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2.00 kg
Tension in the rope?
W
W
TT
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fk
y :m1 : x :
m2 : y :
μk
μkμk
μk
μk
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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 rope and
frictionless table.
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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 rope and
frictionless table.
36
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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.
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SpringsNote: we are discussing the force of the spring on the mass. The force of the spring on the wall are equal, and opposite.
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Springs and TensionA 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
a) (L1 + L2) / 2b) L1 or L2, whichever is smallerc) L1 or L2, whichever is biggerd) depends on which order the springs are attachede) L1 + L2
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Springs and TensionA 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
W
Fs=T
a) (L1 + L2) / 2b) L1 or L2, whichever is smallerc) L1 or L2, whichever is biggerd) depends on which order the springs are attachede) L1 + L2
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Springs and TensionA 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?
Spring 1 supports the weight.Spring 2 supports the weight.Both feel the same force, and stretch the same distance as before.
S1
S2
W
Fs=T
a) (L1 + L2) / 2b) L1 or L2, whichever is smallerc) L1 or L2, whichever is biggerd) depends on which order the springs are attachede) L1 + L2