Download Laws of Motion

Laws of Motion
Causes and Rules for Motion
Dynamics
 Causes of, or changes in motion
 Eq of motion only describe results
Dynamics (Laws of Motion)
deals with
A. Results of motion
B. Causes of motion
& changes in
motion
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Laws of Motion
 Aristotle
 Galileo
 Newton
Aristotle
 Celestial & terrestrial motion different
 Falling & horizontal motion different
Aristotle
 Horizontal motion
 Movement requires a Mover in constant
contact
 Falling Motion
 Heavier objects fall faster
Aristotle
 Horizontal motion
 Without contact w/mover motion stops
 Objects fall straight down
 Thrown objects?
Aristotle
 Falling motion
 Heavier objects fall faster
 The feather and the rock
Aristotle’s physics
A. All objects fall at
same rate
B. Heavier objects
fall faster
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Aristotle’s physics:
A. Objects require a
Mover to keep
moving
B. Objects will keep
moving on their
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Galileo
 Starts ball rolling (literally)
to modern view of motion
 Was a premier
experimentalist
 1st experimental physicist
 Used models that
simplified problem
Galileo
 Horizontal motion
 Inclined plane to study motion
 Smaller angle of plane --> further ball rolls
 Inertia
 Moving object keeps moving unless disturbed
 “natural state” is no change in motion
Galileo
 Horizontal motion
 No Mover
 Inertia
 Dropped object keeps horizontal motion
Galileo
 Horizontal motion
 Inertia
 Dropped object keeps horizontal motion
 Adds falling motion: hits at base of mast
Galileo
 Falling motion
 Used inclined plane to slow down falling
 During equal time periods falling body
increases speed by equal amounts m/s
s
 Acceleration
 Falling bodies accelerate
Galileo
 Falling motion
 Couldn’t create vacuum
 Increased density of medium
 Compared falling in water with air
 Speculated on falling in air vs. vacuum
 Medium w/o resistance (vacuum)
 All bodies fall with same acceleration
Galileo
 Projectile motion
 Combines horizontal & falling motion
 Result is parabola
 Superposition principle
 Motion in one direction (x) doesn’t affect
motion in other direction (y)
 Can solve motion eqs separately
Galileo
 Same laws falling & horizontal motion
 Unification
 Acceleration key to falling motion
 Same rate for all bodies w/o resistance
 Inertia
 Motion doesn’t change w/o influence
 No natural motion
 Superposition
Newton’s Laws
 Newton (1642 – 1727)
Newton’s Laws of Motion
Newton’s Laws of Motion
I.
An object won’t change its motion unless
a force acts on it: Inertia
Newton’s Laws of Motion
II.
Force is needed to change the state of
motion: F = ma
Newton’s Laws of Motion
III.
When you push on something, it pushes
back on you. Action & Reaction are
forces
Newton’s Laws of Motion
I. An object won’t change its motion
unless a force acts on it: Inertia
II. Force is needed to change the state of
motion:
F = ma
III. When you push on something, it
pushes back on you. Action & Reaction
are forces
Inertia was introduced to
motion by
A. Aristotle
B. Galileo
C. Newton
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Who determined that falling
objects accelerate?
A. Newton
B. Galileo
C. Aristotle
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There are four forces
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Stronger
Nuclear strong
Electromagnetism
Nuclear weak
Gravity
Weaker
Weight and Mass
 Mass
 Amount of matter in object
 Measure of inertia
 scalar
 Weight
 Force mass of object times acceleration of gravity
 Varies with gravity
 Vector
Second Law Problems


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F = ma applies to entire system
m refers to mass (kg) being accelerated
F is net force doing the accelerating
a  F a  1/m
Box on frictionless table
 m1 = 4 kg
 m2 = 7 kg
Box on frictionless table
 Accelerating force is m2 x g
 Accelerated mass is m1 + m2
Box on frictionless table
 Accelerating force is m2 x g
 Accelerated mass is m1 + m2
 Because accelerated mass is greater
than m2 the acceleration is less than
9.8 m/s2
Example
 Crate has mass of 3 kg. What is weight?
Example
 Crate has mass of 3 kg. What is weight?
 F = ma --> W = mg
Weight of the crate?
A. 3 kg
B. 9.8 m/s2
C. 29.4 N
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Example
 Crate has mass of 3 kg. What is weight?
 F = ma --> W = mg
 W = (3 kg)(9.8 m/s2)
= 29.4 N
 Direction of weight?
Example
 Crate has mass of 3 kg. What is weight?
 F = ma --> W = mg
 W = (3 kg)(9.8 m/s2)
= 29.4 N
 Direction of weight?
Example
 280 N crate on 20 N dolley. System pushed with
30 N force Find acceleration
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F = ma  a = F/m
m = W/g
m = (300 N)/(9.8 m/s2)
a = (20 N)/(30.6 kg)
a = 0.654 m/s2
Box on frictionless table
 m1 = 4 kg
 m2 = 7 kg
Box on frictionless table
What will accelerate?
A. Just the 4 kg
B. Just the 7 kg
C. Both (11 kg)
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Box on frictionless table
What produces the acceleration?
A. 39.2 N
B. 68.6 N
C. 107.8 N
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Box on frictionless table
What is the acceleration?
A. 0.16 m/s2
B. 6.24 m/s2
C. 9.8 m/s2
D. 17.15 m/s2
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Friction
 Force: Obeys Newton’s Laws
 Always opposes motion or net force
 Contact force
 Surface roughness
 Interatomic forces (bonding)
Friction
 Force: Obeys Newton’s Laws
 Always opposes motion or net force
 Contact force
 Surface roughness
 Interatomic forces (bonding)
 Two kinds
 Kinetic: sliding taking place
 Static: no sliding  rolling
Friction
 Static friction > kinetic friction
 Bonding takes place
 Anti-lock brakes
 Amount depends on
 Surfaces  coefficient of friction 
 Force between surfaces  normal force
 often n = mg
Friction
 Static friction
 f ≤ sn
 Magnitude can vary up to limit (sn)
 Kinetic friction
 f = kn
Friction
 In general use
 f = n
 f = mg
Example
 2 kg box slides across a table.
Coefficient of friction is 0.32
 What is frictional force?
Example
 2 kg box slides across a table.
Coefficient of friction is 0.32
 What is frictional force?
 f = kn = k(mg)
Example
 2 kg box slides across a table.
Coefficient of friction is 0.32
 What is frictional force?
 f = kn
 f = (0.32)(2 kg x 9.8 m/s2) = 6.27 N
Example
 2 kg box slides across different table.
Frictional force is 4 N
 Find coefficient of friction
Example
 2 kg box slides across different table.
Frictional force is 4 N
 Find coefficient of friction
 f = kn  k = f/n = f/(mg)
Example
 2 kg box slides across different table.
Frictional force is 4 N
 Find coefficient of friction
 f = kn  k = f/n = f/(mg)
 k = (4N)/(2 kg x 9.8 m/s2) = 0.204
Example
 Worker loading crates finds that a 20 kg
crate needs 75 N force to start it moving
 What is coefficient of friction?
 Which coefficient?
 In this case f ≤ sn becomes f = sn
 s = f/n
 s = (75 N)/(20 kg x 9.8 m/s2) = 0.383
Example
 Worker loading crates finds that a 20 kg
crate needs 75 N force to start it moving
 Calculate the coefficient of friction
Which type of friction must be
overcome to start motion?
A. Kinetic friction
B. Static friction
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What force is needed to keep
the crate moving?
A. Less force
B. Same force
C. More force
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Box on table with friction
 m1 = 4 kg
 m2 = 7 kg
 𝜇 = 0.3
Box on table with friction
What produces the acceleration?
A. The net force
B. The maximum
force
C. The frictional force
D. The normal force
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Box on table with friction
What is the net force on the box?
A. 68.6 N
B. 107.8 N
C. 56.8 N
D. 11.8 N
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Box on table with friction
What is the acceleration?
A. 5.16 m/s2
B. 6.24 m/s2
C. 7.3 m/s2
D. 9.8 m/s2
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Train Example
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A locomotive pulls 10 freight cars
Each car is 30,000 kg
The locomotive is 50,000 kg
The acceleration of the train is 0.05 m/s2
Train Example
 What is the force applied by the
locomotive?
Train Example
 What is the force applied by the
locomotive?
 What is the force on the first coupler?
Train Example
 What is the force applied by the
locomotive?
 What is the force on the first coupler?
 What is the force on the last coupler?
Train Example
 FL = 1.75 x 104 N
 F1 = 1.5 x 104 N
 F10 = 1.5 x 103 N