1. without being touched- That’s Newton’s third law!
Chapter 4:Dynamics: Newton’s Law of Motion No homework for next week since you have the first exam on Monday, Sept. 25
You cannot touch
without being touched-
That’s Newton’s third law!

2. Problem 51
51. (II) A ball is thrown horizontally from the top of a cliff with initial speed v0 (at t=0). At any moment, its direction of motion makes an angle θ to the horizontal (Fig. 3–47). Derive a formula for θ as a function of time, t , as the ball follows a projectile’s path.

3. Solving Problems Involving Projectile Motion
Read the problem carefully, and choose the object(s) you are going to analyze.
Draw a diagram.
Choose an origin and a coordinate system.
Decide on the time interval; this is the same in both directions, and includes only the time the object is moving with constant acceleration g.
Examine the x and y motions separately.
6. List known and unknown quantities. Remember that vx never changes, and that vy=0 at the highest point.
7. Plan how you will proceed. Use the appropriate equations; you may have to combine some of them.

4. Relative velocity Earth=A=stationary reference frame
Reference Frames y
bug
Lemon car
vBA
vCA x
Earth=A=stationary reference frame
vBA=velocity of the bug, B, relative to the earth, A
vCA= velocity of the lemon car, C, relative to the earth, A

5. Relative velocity vBA=velocity of the bug B, relative to the earth , A
car y A B C
vCA
vBA x
Earth=A=stationary reference frame
We can add reference frames to the bug, B and to the Lemon car, C
vBC=vBA-vCA
Or vBC=vBA+vAC
vBA=velocity of the bug B, relative to the earth , A
VCA=-VAC
Notice how the outer subscripts on the right side of the equation correspond with those on the left, and how the inner subscripts are the same but do not exist on the left. Think of them as canceling.

6. Relative Velocity
Here, vWS is the velocity of the water in the shore frame, vBS is the velocity of the boat in the shore frame, and vBW is the velocity of the boat in the water frame.
The relationship between the three velocities is:

7. Chapter 4: Forces: Defining Force
Units:
Two kinds of forces:
A push or a pull
Newtons = kg.m/s2
Long range:
Contact:
Gravitational force
Electromagnetic force
Everything else
Whenever two objects touch

8. Chapter 4: Measuring Force
We can use a calibrated spring scale to measure force
Force is a vector!
Net Force = the vector sum of all forces acting on an object

9. Ah, but force is a vector.
Images:

10. Kinds of forces and direction
Gravitational Force: (W) or Fg attraction between earth and an object.
Forces on the
Most Common Contact Forces: N
Friction: (f)
parallel to contact surface T f
Perpendicular to contact surface
Normal: (N) W
Tension: (T)
along rope or cord or…
Gravitational force (Fg) or Weight (W): force due to the earth= Fg= mg

11. Mass
Mass is the measure of inertia of an object; mass is a measure of an object’s resistance to change its velocity. In the SI system, mass is measured in kilograms.
Mass is not weight.
Mass is a property of an object. Weight is the force exerted on that object by gravity.
If you go to the Moon, whose gravitational acceleration is about 1/6 g, you will weigh much less. Your mass, however, will be the same.

12. Problem 3
3. (I) What is the weight of a 68kg astronaut (a) on Earth, (b) on the Moon g=1.7m/s2 , (c) on Mars (g=3.7m/s2) , (d) in outer space traveling with constant velocity?

13. Question
A UFO is hovering, stationary, 2000m above the earth. The net force on the UFO is
1) zero
2) due east
3) upwards
4) downwards

14. Question
An airplane is flying due East at a constant velocity of 590 mph. The net force on the airplane is 1) zero 2) due east 3) upwards 4) downwards

15. Question
The Earth travels around the Sun with a constant speed. The net force on the Earth is
zero
nonzero

16. Question
A skydiver is falling toward the Earth at terminal velocity, that is, at constant speed. The net force on the skydiver is
zero
nonzero

17. Newton’s First Law of Motion
This is Newton’s first law, which is often called the law of inertia:
Every object continues in its state of rest, or of uniform velocity in a straight line, as long as no net force acts on it.
Figure 4-3. Caption: F represents the force applied by the person and Ffr represents the force of friction.

18. Newton’s First Law of Motion
Inertial reference frames:
Newton’s first law does not hold in every reference frame, such as a reference frame that is accelerating or rotating.
An inertial reference frame is one in which Newton’s first law is valid. This excludes rotating and accelerating frames.
How can we tell if we are in an inertial reference frame? By checking to see if Newton’s first law holds!

19. Equilibrium
When the net force acting on an object is zero, then it is in equilibrium.
and
and
An object in motion remains in motion
And An object at rest remains at rest
if the net external force acting on the object is zero.
a=0, v=constant

20. Newton’s Second Law of Motion
Newton’s second law is the relation between acceleration and force.
Acceleration is proportional to force and inversely proportional to mass.
It takes a force to change either the direction or the speed of an object. More force means more acceleration; the same force exerted on a more massive object will yield less acceleration.
Figure 4-5. Caption: The bobsled accelerates because the team exerts a force.

21. Newton’s Second Law of Motion
Force is a vector, so is true along each coordinate axis.
The unit of force in the SI system is the newton (N).
Note that the pound is a unit of force, not of mass, and can therefore be equated to Newtons but not to kilograms.

22. Question Which of the following always stays the same: A) Your mass
B) Your weight
C) Your apparent weight
D) A and B
E) B and C

23. Problem 7
7. (II) Estimate the average force exerted by a shot-putter on a 7.0-kg shot if the shot is moved through a distance of 2.8 m and is released with a speed of 13 m/s.

24. Newton’s Second Law of Motion
Example 4-2: Force to accelerate a fast car.
Estimate the net force needed to accelerate (a) a 1000-kg car at ½ g; (b) a 200-g apple at the same rate.
Example 4-3: Force to stop a car.
What average net force is required to bring a 1500-kg car to rest from a speed of 100 km/h within a distance of 55 m?
Figure 4-6.
4-2. Use Newton’s second law: acceleration is about 5 m/s2, so F is about 5000 N for the car and 1 N for the apple.
4-3. First, find the acceleration (assumed constant) from the initial and final speeds and the stopping distance; a = -7.1 m/s2. Then use Newton’s second law: F = -1.1 x 104 N.

25. Free Body Diagram Practice
Picture
Lawnmower
Rock
Climber
Auto

26. Problem 12
(II) How much tension must a cable withstand if it is used to accelerate a 1200kg car vertically upward at 0.70m/s2

27. Problem 28
28. (I) Draw the free-body diagram for a basketball player (a) just before leaving the ground on a jump, and (b) while in the air. See Fig. 4–34.

28. Problem 37

29. Problem 40

30. The inclined plane problem
Problem: A box of oranges with a weight of 100 N slides down an incline of 25° above the horizontal with a constant velocity. What is the normal force? The friction force?
25°

31. Solving Problems with Newton’s Laws: Free-Body Diagrams
Example 4-16: Box slides down an incline.
A box of mass m is placed on a smooth incline that makes an angle θ with the horizontal. (a) Determine the normal force on the box. (b) Determine the box’s acceleration. (c) Evaluate for a mass m = 10 kg and an incline of θ = 30°.
Figure Caption: Example 4–16.
(a) Box sliding on inclined plane.
(b) Free-body diagram of box.
Answer: On an incline (or any surface), the normal force is perpendicular to the surface and any frictional forces are parallel to the surface.
(a) The normal force is equal to the component of the weight perpendicular to the incline, or mg cos θ.
(b) The force causing the acceleration is the component of the weight parallel to the incline; therefore the acceleration is g sin θ.
(c) The normal force is 85 N and the acceleration is 4.9 m/s2.

32. Newton’s Third Law of Motion
You cannot touch
without being touched-
That’s Newton’s third law!
From “Conceptual Physics for Everyone”, Paul G. Hewitt, Addison Wesley, 2002.

33. Newton’s Third Law of Motion
Any time a force is exerted on an object, that force is caused by another object.
Newton’s third law:
Whenever one object exerts a force on a second object, the second exerts an equal force in the opposite direction on the first.
Figure 4-8. Caption: If your hand pushes against the edge of a desk (the force vector is shown in red), the desk pushes back against your hand (this force vector is shown in a different color, violet, to remind us that this force acts on a different object).

34. Newton’s Third Law: interaction pairs
Newton’s third Law: for every action, there is an equal and opposite reaction. We call these two equal and opposite forces interaction partners. Note: they operate on different objects Note: they involve the same interaction :eg. Gravity, contact…

35. Newton’s Third Law of Motion
A key to the correct application of the third law is that the forces are exerted on different objects. Make sure you don’t use them as if they were acting on the same object.
Figure 4-9. Caption: An example of Newton’s third law: when an ice skater pushes against the wall, the wall pushes back and this force causes her to accelerate away.

36. Newton’s Third Law of Motion
Rocket propulsion can also be explained using Newton’s third law: hot gases from combustion spew out of the tail of the rocket at high speeds. The reaction force is what propels the rocket.
Note that the rocket does not need anything to “push” against.
Figure Caption: Another example of Newton’s third law: the launch of a rocket. The rocket engine pushes the gases downward, and the gases exert an equal and opposite force upward on the rocket, accelerating it upward. (A rocket does not accelerate as a result of its propelling gases pushing against the ground.)

37. Newton’s Third Law of Motion
Conceptual Example 4-5: Third law clarification.
Michelangelo’s assistant has been assigned the task of moving a block of marble using a sled. He says to his boss, “When I exert a forward force on the sled, the sled exerts an equal and opposite force backward. So how can I ever start it moving? No matter how hard I pull, the backward reaction force always equals my forward force, so the net force must be zero. I’ll never be able to move this load.” Is he correct?
Figure Caption: Example 4–5, showing only horizontal forces. Michelangelo has selected a fine block of marble for his next sculpture. Shown here is his assistant pulling it on a sled away from the quarry. Forces on the assistant are shown as red (magenta) arrows. Forces on the sled are purple arrows. Forces acting on the ground are orange arrows. Action–reaction forces that are equal and opposite are labeled by the same subscripts but reversed (such as FGA and FAG) and are of different colors because they act on different objects.
Answer: No – in order to see whether the sled will accelerate, we need to consider only the forces on the sled. The force that the sled exerts on the assistant is irrelevant to the sled’s acceleration.

38. Problem Solving Strategy
1) Draw a FBD for the object (or objects). Labeling all forces with simple vector symbols. 2) Pick a coordinate system 3) Add up all of the forces in the x-direction. 4) Decide whether or not this sum is equal to zero or ma 5) Add up all of the forces in the y-direction 6) Decide whether or not this sum is equal to zero or ma 7) Solve these two simultaneous equations for your unknown(s).