1. Warm-Up: Solve for F1 and F2, then calculate the Fnet and any resulting acceleration of this crate.

2. Types of Forces
A force is a push or pull acting upon an object as a result of its interaction with another object.
Contact Forces Non-Contact Forces
Frictional Force Gravitational Force Tension Force Electrical Force Normal Force Magnetic Force Air Resistance Force Applied Force Spring Force

3. Newton’s First Law

4. Applications of Newton’s First Law
Blood rushes from your head to your feet while quickly stopping when riding on a descending elevator.
To dislodge ketchup from the bottom of a ketchup bottle, it is often turned upside down and thrusted downward at high speeds and then abruptly halted.
Headrests are placed in cars to prevent whiplash injuries during rear-end collisions.
While riding a skateboard (or wagon or bicycle), you fly forward off the board when hitting a curb or rock or other object which abruptly halts the motion of the skateboard.

5. Newton’s Second Law

6. This verbal statement can be expressed in equation form as follows: a = Fnet / m The above equation is often rearranged to a more familiar form as shown below. The net force is equated to the product of the mass times the acceleration Fnet = m * a

7. The acceleration is directly proportional to the net force; Do not use the value of merely "any 'ole force" in the above equation. It is the net force which is related to acceleration.
The net force is the vector sum of all the forces. If all the individual forces acting upon an object are known, then the net force can be determined.

8. The Fnet = m • a equation is often used in algebraic problem-solving
The Fnet = m • a equation is often used in algebraic problem-solving. The table below can be filled by substituting into the equation and solving for the unknown quantity. Fill in the following table.
Net Force
(N)
Mass
(kg)
Acceleration
(m/s/s) 1. 10 2 2. 20 3. 4 4. 5 5.

9. Compare rows 1 and 2
A doubling of the net force results in a doubling of the acceleration (if mass is held constant).
Compare rows 2 and 4
A halving of the net force results in a halving of the acceleration (if mass is held constant).
Acceleration is directly proportional to net force.

10. Compare rows 2 and 3
Doubling the mass results in a halving of the acceleration (if force is held constant).
Compare rows 4 and 5
Halving the mass results in a doubling of the acceleration (if force is held constant).
Acceleration is inversely proportional to mass.

11. Applications of Newton’s Second Law
The direction of the net force is in the same direction as the acceleration. If the direction of the acceleration is known, then the direction of the net force is also known.
Consider the two diagrams below for an acceleration of a car. From the diagram, determine the direction of the net force which is acting upon the car
Figure 1
Figure 2

12. Instructions for Round Table
Pass ONE paper and ONE pencil/pen around the table clockwise.
When the paper comes to you, solve the sample problem while explaining your reasoning OUT LOUD to your team
The person to your right will be your coach
The person to your left will be your accuracy checker
The person at the diagonal is your encourager

13. Warm-Up 10/27/09
Remember, next week is the first ever College In Colorado College Application Week. It will be held from November 1 – November 8, 2009, and is a week during which Colorado high school seniors can apply to college for free! Log on to to apply.

14. Sample Problems – Round Table
Determine the accelerations which result when a 12-N net force is applied to a 3-kg object and then to a 6-kg object.
2. A net force of 15 N is exerted on an encyclopedia to cause it to accelerate at a rate of 5 m/s2. Determine the mass of the encyclopedia.
3. Suppose that a sled is accelerating at a rate of 2 m/s2. If the net force is tripled and the mass is doubled, then what is the new acceleration of the sled?
4. Suppose that a sled is accelerating at a rate of 2 m/s2. If the net force is tripled and the mass is halved, then what is the new acceleration of the sled?

15. Solutions
A 3-kg object experiences an acceleration of 4 m/s/s. A 6-kg object experiences an acceleration of 2 m/s/s.
Use Fnet= m * a with Fnet = 15 N and a = 5 m/s/s. So (15 N) = (m)*(5 m/s/s) And m = 3.0 kg
Answer: 3 m/s/s -- The original value of 2 m/s/s must be multiplied by 3 (since a and F are directly proportional) and divided by 2 (since a and m are inversely proportional)
Answer: 12 m/s/s -- The original value of 2 m/s/s must be multiplied by 3 (since a and F are directly proportional) and divided by 1/2 (since a and m are inversely proportional) 

16. Finding Acceleration a = Fnet / m
Three major equations
net force: Fnet = m*a
gravitational force: Fgrav = m*g
frictional force: Ffrict = μ*Fnorm
To solve for acceleration, the mass and the net force must be known

17. Sample Problem 5: Individual Forces
A rightward force is applied to a 6-kg object to move it across a rough surface at constant velocity. The object encounters 15 N of frictional force. Use the diagram to determine the gravitational force, normal force, net force, and applied force. (Neglect air resistance.)

18. Solution to Sample 5 Fnet = 0 N; Fgrav = 58.8 N;
Fnorm = 58.8 N; Fapp = 15 N
When the velocity is constant, a = 0 m/s/s and Fnet = 0 N
Since the mass is known, Fgrav can be found: Fgrav = m • g = 6 kg • 9.8 m/s/s = 58.8 N
Since there is no vertical acceleration, the normal force equals the gravity force.
Since there is no horizontal acceleration, Ffrict = Fapp = 15 N

19. Sample Problem 6
A 5-kg object is sliding to the right and encountering a friction force which slows it down. The coefficient of friction ("mu") between the object and the surface is 0.1. Determine the force of gravity, the normal force, the force of friction, the net force, and the acceleration. (Neglect air resistance.)
START BY SKETCHING A FREE BODY DIAGRAM TO REPRESENT THE PROBLEM.

20. Solution to Sample 6 Fgrav = 49 N Fnorm = 49 N Ffrict = 4.9 N
Fnet = 5 N, left
a = 0.98 m/s/s, left
Fgrav = m • g = (5 kg) • (9.8 m/s/s) = 49 N. Since there is no vertical acceleration, the normal force equals the gravity force. Ffrict can be found using the equation Ffrict ="mu"• Fnorm. The Fnet is the vector sum of all the forces: 49 N, up plus 49 N, down equals 0 N. And 4.9 N, left remains unbalanced; it is the net force. Finally, a = Fnet / m = (4.9 N) / (5 kg) = 0.98 m/s/s.

21. Finding Individual Forces
Fnet = m * a
If the numerical value for the net force and the direction of the net force is known, then the value of all individual forces can be determined.
Remember: The net force is the sum of all the forces acting on an object

22. Sample Problem 7 Free-body diagrams for four situations are shown below. The net force is known for each situation. However, the magnitudes of a few of the individual forces are not known. Analyze each situation individually and determine the magnitude of the unknown forces.

23. Solution to Sample Problem 7
A = 50 N (the horizontal forces must be balanced) B = 200 N (the vertical forces must be balanced) C = 1100 N (in order to have a net force of 200 N, up) D = 20 N (in order to have a net force of 60 N, left) E = 300 N (the vertical forces must be balanced) F = H = any number you wish (as long as F equals H) G = 50 N (in order to have a net force of 30 N, right)

24. Inclined Planes: a tilted surface
An object placed on a tilted surface will often slide down the surface.
The rate at which the object slides down the surface is dependent upon how tilted the surface is; the greater the tilt of the surface, the faster the rate at which the object will slide down it.
Objects are known to accelerate down inclined planes because of an unbalanced force

25. Diagram of the Inclined Plane
at least two forces acting upon any
object on an inclined plane
force of gravity acts downward
normal force acts in a direction perpendicular to the surface
Usually, any force directed at an angle to the horizontal is resolved into horizontal and vertical components BUT NOT IN THIS CASE!

26. Resolve the weight vector (Fgrav) into two perpendicular components, one directed parallel to the inclined surface and the other directed perpendicular to the inclined surface.
The equations for the parallel and perpendicular components are:

27. The perpendicular component of the force of gravity balances the normal force
The parallel component of the force of gravity is not balanced by any other force.

28. The parallel component of the force of gravity is the net force which causes this acceleration.
In the absence of friction and other forces (tension, applied, etc.)…
acceleration of an object on an incline is the value of the parallel component
a= gsin
(in the absence of friction and other forces)

29. In the presence of friction or other forces (applied force, tensional forces, etc…
As in all net force problems, the net force is the vector sum of all the forces

30. TNS – Tilted Neck Syndrome
All inclined plane problems can be simplified through a useful trick known as "tilting the head."
tilt your head in the same direction that the incline was tilted. Or better yet, merely tilt the page of paper

31. Example 1: The free-body diagram shows the forces acting upon a 100-kg crate which is sliding down an inclined plane. The plane is inclined at an angle of 30 degrees. The coefficient of friction between the crate and the incline is Determine the net force and acceleration of the crate.
Find the force of gravity acting upon the crate and the components of this force parallel and perpendicular to the incline.
Now the normal force can be determined.
The force of friction can be determined from the value of the normal force and the coefficient of friction
The net force is the vector sum of all the forces.
The acceleration is 2.35 m/s/s (Fnet/m = 235 N/100 kg).

32. Solution to Example 1
The force of gravity is 980 N and the components of this force are Fparallel = 490 N (980 N • sin 30 degrees) and Fperpendicular = 849 N (980 N • cos30 degrees).
Now the normal force can be determined to be 849 N (it must balance the perpendicular component of the weight vector).
; Ffrict is 255 N (Ffrict = "mu"*Fnorm= 0.3 • 849 N).
Ffrict is 255 N (Ffrict = "mu"*Fnorm= 0.3 • 849 N).
The net force is 235 N (490 N N).
The acceleration is 2.35 m/s/s (Fnet/m = 235 N/100 kg).

33.  Example 2:
The two diagrams below depict the free-body diagram for a 1000-kg roller coaster on the first drop of two different roller coaster rides. Use the above principles of vector resolution to determine the net force and acceleration of the roller coaster cars. Assume a negligible affect of friction and air resistance. When done, click the button to view the answers.

34. Solution to Example 2
Fgrav = m • g = (1000 kg) • (9.8 m/s/s) = 9800 N The parallel and perpendicular components of the gravity force can be determined from their respective equations: Fparallel = m • g • sin (45 degrees) = 6930 N Fperpendicular = m • g • cos (45 degrees) = 6930 N Fnorm = Fperpendicular Fnorm = 6930 N Fnet = 6930 N, down the incline a = Fnet / m = (6930 N) / (1000 kg) a = 6.93 m/s/s, down

35. Ticket out the Door
Edwardo applies a 4.25-N rightward force to a kg book to accelerate it across a table top. The coefficient of friction between the book and the tabletop is Determine the acceleration of the book.
In a physics lab, Kate and Rob use a hanging mass and pulley system to exert a 2.45 N rightward force on a kg cart to accelerate it across a low-friction track. If the total resistance force to the motion of the cart is 0.72 N, then what is the cart's acceleration?
A rightward force is applied to a 6-kg object to move it across a rough surface at constant velocity. The object encounters 15 N of frictional force. Use a free body diagram to determine the gravitational force, normal force, net force, and applied force. (Neglect air resistance.)

36. Ticket out the Door, cont.
A rightward force is applied to a 10-kg object to move it across a rough surface at constant velocity. The coefficient of friction between the object and the surface is 0.2. Use the diagram to determine the gravitational force, normal force, applied force, frictional force, and net force. (Neglect air resistance.)

37. Warm-Up 11/02/09
What is the maximum weight of a truck that can park on a hillside whose slope is 18.5o without beginning to slide. (use the coefficient of static friction for rubber on dry concrete – p. 124) 
Fgy
Fgx
Fg (weight)
X axis
Y axis
Fgy =Fg cos
Fgx =Fg sin FN

38. Ticket out the Door, cont.
Lee Mealone is sledding with his friends when he becomes disgruntled by one of his friends comments. He exerts a rightward force of 9.13 N on his 4.68-kg sled to accelerate it across the snow. If the acceleration of the sled is m/s/s, then what is the coefficient of friction between the sled and the snow
In a Physics lab, Ernesto and Amanda apply a 34.5 N rightward force to a 4.52-kg cart to accelerate it across a horizontal surface at a rate of 1.28 m/s/s. Determine the friction force acting upon the cart.

39. Free Fall and Air Resistance
Why do objects which encounter air resistance ultimately reach a terminal velocity?
In situations in which there is air resistance, why do more massive objects fall faster than less massive objects?

40. Falling Without Air Resistance
Objects which are said to be undergoing free fall, are not encountering a significant force of air resistance; they are falling under the sole influence of gravity. Under such conditions, all objects will fall with the same rate of acceleration, regardless of their mass.
Consider the free-falling motion of a
1000-kg baby elephant and a
1-kg overgrown mouse.

41. The gravitational field strength is a property of the Earth's gravitational field and not a property of the baby elephant nor the mouse. All objects placed within Earth's gravitational field will experience this amount of force (9.8 N) upon every 1 kilogram of mass within the object. All objects free fall at the same rate regardless of their mass.

42. Falling With Air Resistance
Air resistance is the result of collisions of the object's leading surface with air molecules. The actual amount of air resistance encountered by the object is dependent upon a variety of factors.
speed of the object : Increased speeds result in an increased amount of air resistance.
cross-sectional area of the object: Increased cross-sectional areas result in an increased amount of air resistance.

43. Suppose that an elephant and a feather are dropped off a very tall building from the same height at the same time. We will assume the realistic situation that both feather and elephant encounter air resistance. Which object - the elephant or the feather - will hit the ground first?

45. Sample Problems – Round Table In the diagrams below, free-body diagrams showing the forces acting upon an 85-kg skydiver (equipment included) are shown. For each case, use the diagrams to determine the net force and acceleration of the skydiver at each instant in time.

46. Solutions
The Fnet = 833 N, down
a = (Fnet / m) = (833 N) / (85 kg)
= 9.8 m/s/s down
2. The Fnet = 483 N, down
a = (Fnet / m) = (483 N) / (85 kg)
= 5.68 m/s/s down
3. The Fnet = 133 N, down
a = (Fnet / m) = (133 N) / (85 kg)
= 1.56 m/s/s down
4. The Fnet = 0 N
a = (Fnet / m) = (0 N) / (85 kg) = 0 m/s/s

47. Terminal Velocity As an object falls, it picks up speed.
The increase in speed leads to an increase in the amount of air resistance.
Eventually, the force of air resistance becomes large enough to balances the force of gravity and the object will stop accelerating and fall with constant velocity.
The object is said to have reached a terminal velocity.

48. Individual Practice: Bronco Billy and Skydiving
Study the diagrams on p. 17 in your packet
Solve for acceleration at each stage of the skydive
Answer the questions 1-8

49. Newton’s Third Law
If object 1 exerts force on object 2, then object 2 exerts an equal and opposite force on object 1.
The two forces are acting on two different objects, therefore even though the two forces are equal and opposite, they do not necessarily cancel each other.
Often called the law of action and reaction

50. 4.4 Newton’s Third Law of Motion
For every force (action), there is an equal and opposite force (reaction).
Note that the action and reaction forces act on different objects.
This image shows how a block exerts a downward force on a table; the table exerts an equal and opposite force on the block, called the normal force N.

51. Example 4.4
A large truck collides head-on with a small car and causes a lot of damage to the small car. Explain why there is more damage to the small car than to the large truck.

52. Identify at least six pairs of action-reaction force pairs in the following diagram.
The elephant's feet push backward on the ground; the ground pushes forward on its feet. The right end of the right rope pulls leftward on the elephant's body; its body pulls rightward on the right end of the right rope. The left end of the right rope pulls rightward on the man; the man pulls leftward on the left end of the right rope. The right end of the left rope pulls leftward on the man; the man pulls rightward on the right end of the left rope. The tractor pulls leftward on the right end of the left rope; the left end of the left rope pulls rightward on the tractor.

53. Homework:

54. P : 44, 48, 52, 53, 55, 60, 61, 62, 63, 65, 71, 74, 78

55. Homework ConcepTest Ch. 4 Newton’s Laws Practice Problems
Applications of Newton’s Laws Quiz Tuesday, 11/11/08