1. Newton’s Laws of Motion (continued)
Chapters 4 & 5
Newton’s Laws of Motion
(continued)
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2. Newton’s Second Law of Motion
Chapter 4
Newton’s Second Law of Motion
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3. Newton’s Second Law The Law of Acceleration
Forces cause acceleration
Net force must be greater than zero
Masses resist acceleration due to inertia
This is, in part, why it is harder to start something moving than to keep it moving
Hence we say that acceleration is directly proportional to net force and inversely proportional to mass
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4. Newton’s Second Law The Law of Acceleration
The acceleration of an object is in the direction of the force applied.
Acceleration is directly proportional to the force applied.
The harder you push an object the faster it goes
Acceleration is inversely proportional to the mass of the object.
The heavier the object, the less affect a push has.
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5. Sample Problem 1
What force is required to accelerate a 10 kg object horizontally at 6 m/s2?
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6. Solution
F = ma = (10 kg)(6 m/s2) = 60 Newtons
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7. Friction
Forces always come in pairs, hence the Normal force, which is perpendicular to the contact surface, has a companion force that is parallel to the contact surface, this force is friction
Friction always opposes motion
Friction depends upon two things:
The nature of the contact between two objects
How strong the force of contact is (The Normal Force)
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8. Friction
Friction also occurs in gases and liquids both of which are referred to in physics as fluids.
In fluids we call friction drag and in air we refer to it specifically in air as air resistance.
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9. Sample Problem 2
If a 200 Newton force is applied to a box that undergoes a 100 Newton resistive force (friction). What is the net force on the box? If it is a 30 kg box, what is its acceleration?
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10. Solution Fnet = 200 N – 100 N = 100 N Fnet = ma so a = Fnet/m
= 100 N/30 kg
= 3.3 m/s2
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11. Mass: A Measure of Inertia
Mass is measured in kilograms
Mass is not Weight
Mass is a “built in” property of matter
Just because you leave earth, you don’t change your mass, but you do change your weight
Weight is an force caused by the acceleration due to gravity on the mass
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12. Weight Mathematically Weight = mass x gravity Or Fw = mg
So the force of weight on one kilogram of mass on planet earth is given by:
Fw = (1 kg)(9.8 m/s2) = 9.8 Newtons
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13. Sample Problem 3 What is the weight of a 10 kg object a) on earth
b) on the moon (g = 1/6 that of earth)
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14. Terminal Velocity
When an object is in free fall where there is no atmosphere, there is no friction, hence no opposing force and it will continue to accelerate until it reaches the ground.
However, when there is atmosphere, there is friction – called drag which opposes the motion.
When the force of the drag = force of gravitation (weight) then Fnet = 0, and the acceleration of the object becomes zero. This is called terminal velocity.
Terminal velocity - is then the highest speed reached by a falling object in the presence of air resistance. It is different for every object based on its mass and shape.
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15. Review Questions
Pg 61 # 1, 5, 9, 11, 15, 19,
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16. Chapter 5
Vector Addition
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17. Vectors and Scalars Vector vs. Scalars
Vectors have both magnitude and direction
Displacement, Velocity, Acceleration, Force, and Momentum
Scalars have only magnitude
Mass, Time, and Temperature
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18. Mathematical Addition
Mathematical Addition of Vectors Requires a basic knowledge of Geometry – You must know:
Sin θ = Opposite/Hypotenuse
Cos θ = Adjacent/Hypotenuse
Tan θ = Sin θ/ Cos θ =Opposite/Adjacent
ArcSin, ArcCos, ArcTan
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19. Mathematical Addition
Mathematical addition requires that you be able to
Draw a rough sketch of the original vectors
Draw a parallelogram
Draw the Resultant
Use Triangle geometry to find the magnitude and direction of the resultant
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20. Example
What is the result of adding two vectors if the first vector is 100 km to the west and the second vector is 50 km to the north?
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21. Mathematical Addition
Step One: Sketch the vectors Step Two: Put vectors on the same axis and draw the parallelogram +
100 Km West
50 Km North
Step One
Step Two
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22. Mathematical Addition
Step Three: Draw the Resultant
Step Four:
Since this is a right triangle you can find the resultant using Pythagorean theorem
Pythagorean Theorem: C2 = A2 + B2
50 Km
100 Km
111.8 Km
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23. Components of Vectors
Often we need to change a single vector into two vectors in order to make addition easier. These two vectors will be a right angles to each other and are called component vectors or just components for short.
The breaking down of a vector this way is called resolution.
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24. Vector Resolution Step One: Draw Original Vector 45º
100 Km 45° North of West
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25. Step Two: Draw parallelogram
Vector Resolution
Step Two: Draw parallelogram
100 Km 45° North of West
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26. Vector Resolution Step Three: Label and measure components
Sin 45° = N/100
100 Sin 45 ° = N
North component = 70.7 km
45°
Cos 45° = W/100
100 Cos 45° = West
West component = 70.7 km
100 Km 45° North of West
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27. Chapter 10
Part I: Projectiles
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28. Projectile Motion A projectile is any object thrown or launched
Uses the same kinematic equations you have already learned
Requires that you use two sets of equations; one in the horizontal and one in the vertical
These two sets are related only by time
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29. Projectile Motion
A projectile which is dropped or thrown horizontally has an initial velocity in the Y direction (V0y) = 0
Acceleration in the horizontal direction (ax) = 0; hence Velocity in the horizontal direction (Vx) is constant
This means that only one kinematic equation applies in the horizontal direction:
Xx = Vxt
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30. Projectile Motion
While it is moving forward, it is also falling just as if you had dropped it.
So in the Y direction:
x = ½ at2
Where a is the acceleration due to gravity.
The one thing that both motions have in common is time. When the projectile hits the ground it stops moving in both directions!
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31. Projectile Motion
A ball is thrown horizontally off a building which is 200 m high with a velocity of 10 m/s
How long does it take to reach the ground?
How far from the building will it land?
What is its velocity just before it hits the ground (remember magnitude and direction)
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32. Chapter 5
Newton’s Third Law
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33. Newton’s Third Law Action - Reaction
For every action there is an equal and opposite Reaction.
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34. Herriman High Physics

35. Sample Problem
If a 0.4 kg shotgun shell undergoes a 100 Newton force when it is fired, what is its acceleration?
If it was fired from a 2 kg shotgun what is the recoil acceleration of the shotgun?
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36. Solution Since F = ma; a = F/m So: a = 100 N/0.4 kg = 250 m/s2
Since the recoil force is equivalent to the firing force according to Newton’s Third Law the same equation applied however now you use the mass of the shotgun:
a = 100 N/2 kg = 50 m/s2
Demonstrating that it is the mass of the shotgun that keeps it from doing the same damage as the bullet
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