Transcript Slide 1

DYNAMICS
1. Newton’s Three Laws
Newton’s First Law
Existence of inertial systems of reference
In inertial system of reference, any object acted by no net force remains
at rest or continues its motion along straight line with constant velocity
Newton’s Second Law
Newton’s Third Law
Units:


F  ma


FAB  FBA
F   1N  1kg  m / s 2
Note: these two forces
act on different objects.
Never add these forces!
(1 lb = 4.448 N)
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1.1 Newton’s First Law (examples)
In inertial system of reference, any object acted by no net force remains
at rest or continues its motion along straight line with constant velocity
Example 1 (Snapped string):
A small ball attached to the end of
a string moves in circles as shown
below. If the string snaps, what will
be the trajectory of the ball?
A
B
C
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Example 2: A book is lying at rest on a table. The book will remain there at
rest because:
A)
B)
C)
D)
E)
there is a net force but the book has too much inertia
there are no forces acting on it at all
it does move, but too slowly to be seen
there is no net force on the book
there is a net force, but the book is too heavy to move
There are forces acting on the book, but the only forces acting are in the
y-direction. Gravity acts downward, but the table exerts an upward force
that is equally strong, so the two forces cancel, leaving no net force.
Example 3: A hockey puck slides on ice at constant
velocity. What is the net force acting on the puck?
A)
B)
C)
D)
E)
more than its weight
equal to its weight
less than its weight but more than zero
depends on the speed of the puck
zero
The puck is moving at a constant velocity, and therefore it is not
accelerating. Thus, there must be no net force acting on the puck.
Follow-up: Are there any forces acting on the puck? What are they?
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1.2 Newton’s Second Law (examples)


F  ma
Example 1: Car brakes provide a force F for 5 s. During this time, the car
moves 25 m, but does not stop. If the same force would be applied for 10 s,
how far would the car have traveled during this time?
1)
2)
3)
4)
5)
100 m
50 m < x < 100 m
50 m
25 m < x < 50 m
25 m
Acceleration:
aF m
In the first 5 s, the car has still moved 25 m.
However, since the car is slowing down,
in the next 5 s, it must cover less distance.
Therefore, the total distance must be
more than 25 m but less than 50 m.
d1  25m
t 2  2t1
t1  5s
at12
d1  v0 t1 
2
2
at22
a2t1 
d 2  v0 t 2 
 v0 2t1 
 2d1  at12  2d1
2
2
t 2  10s
d2  ?
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Example 2: A force F acting on a mass m1 results in an acceleration a1.
The same force acting on a mass m2 results in an acceleration a2 = 2a1.
a1
F
a2 = 2a1
m1
F
m2
If both masses are put together and the same force is applied to
the combination, what is the resulting acceleration?
a?
F
m1
A.
2/3 a1
B.
3/2 a1
C.
3/4 a1
m2
F
m1 =
a1
m1
F
F
m2 =
=
=
a2 2a1
2
F
F
2 F
2
a=
=
=
 a1
1
m1 + m2 m1 + 2 m1 3 m1 3
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1.3 Newton’s Third Law (examples)
For every force, or action force, there is an equal but opposite force,
or reaction force.

FBA


FAB  FBA
A
B

FAB
If object A exerts a force on object B (an “action”), then
object B exerts a force on body A (a “reaction”).
These two forces have the same magnitude but opposite direction.
Note: these two forces act on different objects.
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Example: Two carts are put back-to-back on a track. Cart A has a springloaded piston; cart B, which has twice the mass of cart A, is entirely passive.
When the piston is released, it pushes against cart B, and the carts move
apart. Which of the two forces exerted by the two carts on each other has a
larger magnitude?
It’s a third
1. The force exerted by A.
law pair!!
2. The two forces have equal magnitude.
3. The force exerted by B.
B
A
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Application of Newton’s Laws (Ropes and tension)
Example 1: You tie a rope to a tree and you pull on the rope with a force of 100 N.
What is the tension in the rope?
The tension in the rope is the force that the rope “feels” across any section of it
(or that you would feel if you replaced a piece of the rope). Since you are pulling
with a force of 100 N, that is the tension in the rope.
Example 2: Two tug-of-war opponents each pull with a force of 100 N on opposite
ends of a rope. What is the tension in the rope?
This is literally the identical situation to the previous question. The tension is not
200 N !! Whether the other end of the rope is pulled by a person, or pulled by a
tree, the tension in the rope is still 100 N !!
Example 3: You and a friend can each pull with a force of 200 N.
If you want to rip a rope in half, what is the best way?
1) You and your friend each pull on opposite ends of the rope
2) Tie the rope to a tree, and you both pull from the same end
3) It doesn’t matter -- both of the above are equivalent
Take advantage of the fact that the tree can pull with almost any force
(until it falls down, that is!). You and your friend should team up on one
end, and let the tree make the effort on the other end.
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2. Free Body Diagram
Example: Book on Table – The full story
NBT
WBE
NTE
WTE
WET
Action-Reaction Pairs:
Normal force between book and table
NBT = –NTB
Gravitational force between book and earth
WBE = –WEB
Normal force between table and earth
NTE = –NET
NTB
Gravitational force between table and earth
WEB
The book does not accelerate WBE+NBT=0
NET
WTE = –WET
The table does not accelerate WTE+NTB+NTE=0
Does the earth accelerate?
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3. Weight (Force due to gravity)
Fg  mg
3a) Apparent weight
(The force the body exerts on a support)
Example: You see two cases: a student pulling or pushing a sled with a force
F which is applied at an angle . In which case is the normal force greater?
1)
2)
3)
4)
5)
case 1
case 2
it’s the same for both
depends on the magnitude of the force F
depends on the ice surface
In Case 1, the force F is pushing down
(in addition to mg), so the normal force
needs to be larger.
In Case 2, the force F is pulling up, against
gravity, so the normal force is lessened.
Case 1
Case 2
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Question 1: What can you say about the force
of gravity Fg acting on a stone and a feather?
1)
2)
3)
4)
5)
Fg is greater on the feather
Fg is greater on the stone
Fg is zero on both due to vacuum
Fg is equal on both always
Fg is zero on both always
The force of gravity (weight) depends on the mass of the
object!! The stone has more mass, therefore more weight.
Question 2: The force of gravity is acting on the stone and the feather
falling in vacuum. Which one has greater acceleration?
1)
2)
3)
4)
5)
The feather
The stone
Both acceleration are zero due to vacuum
Both acceleration are zero always
They have the same acceleration
The acceleration is given by F/m The force of gravity (weight) is F=mg,
then we end up with acceleration g for both objects.
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Example 1: A box with a mass of 100 kg is given an upward acceleration
of 2.2 m/s² by a cable. What is the tension in the cable?
m=100kg
a=2.2m/s2
T=?
T
Newton’s equation:
T-mg=ma
T=m(g+a)
m
mg
T=(100kg)(9.8 m/s2 +2.2 m/s2 )= 1200 N
Note: this tension is bigger than the box’s weight: w=mg=(100 kg)( 9.8 m/s2)= 980 N
Example 2: The same box as in example 1 is given an downward
acceleration of 2.2 m/s² by a cable. What is the tension in the cable?
m = 100 kg
a = -2.2 m/s2
T=?
T=m(g+a)
T=(100 kg)(9.8 m/s2 -2.2 m/s2 )= 760 N
Note: this tension is smaller than the box’s weight.
Compare examples 1 & 2: the tension depends on acceleration & is independent from
velocity. The tension is equal to the weight if there is no acceleration (a=0).
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