1. 4-4 everyday forces and application of Newton’s Laws
Objectives
Explain the difference between mass and weight.
Find the direction and magnitude of the normal force.
Describe air resistance as a form of friction.
Use coefficients of friction to calculate frictional force.

2. Type of forces We are concerned with four types of forces:
Gravity (weight)
Normal force
Friction force
Tension force
Applied force

3. Gravity Force (Weight) Fgrav
The force of gravity is the force with which the earth, moon, or other massively large object attracts another object towards itself. By definition, this is the weight of the object. All objects upon earth experience a force of gravity that is directed "downward" towards the center of the earth. The force of gravity on earth is always equal to the weight of the object as found by the equation:
Fgrav = m • g
where g = 9.81 N/kg (on Earth) and m = mass (in kg)
Note: g is different at different locations

4. Comparing Mass and Weight
The force of gravity.
Vector, its direction is downward.
W = mg
The weight of an object (measured in Newton) will vary according to where in the universe the object is.
Mass
The mass of an object refers to the amount of matter that is contained by the object;
Scalar, has no direction
The mass of an object (measured in kg) will be the same no matter where in the universe that object is located.

5. The direction of gravity force is always downward

6. The slope of weight vs. mass
Weight = mg
Slope is Gravitational acceleration g
mass

7. practice
Complete the following table showing the relationship between mass and weight. 1.
Different masses are hung on a spring scale calibrated in Newtons.
The force exerted by gravity on 5 kg = ______ N.
The force exerted by gravity on _______ kg = 98 N.
The force exerted by gravity on 70 kg = ________ N.
When a person diets, is their goal to lose mass or to lose weight? Explain.
Object
Mass (kg)
Weight (N)
melon 1
apple
0.98
Joe Samall 25
Fred
980

8. Normal Force (FN )
The normal force is the support force exerted upon an object that is in contact with another stable object (usually a surface).
The direction of the normal force is perpendicular to the surface, from the surface toward the object and on the object.

9. Practice- indicate FN on each box with an arrowFg Fg Fg FN FN FN Fg Fg Fg

10. What is the direction of normal force?

11. The magnitude of normal force
The magnitude of normal force can be determined using Newton’s Laws.
Example: a car is moving along a horizontal road. FN
Because the car is not moving up or down, mg

12. Alert!
Normal Force
Net Force

13. Friction Force (Ff)
The friction force is the force exerted by a surface as an object moves across it or makes an effort to move across it.
Friction results from the two surfaces being pressed together closely, causing intermolecular attractive forces between molecules of different surfaces.
Friction force can be reduced by lubricant such as oil.

14. The direction and magnitude of friction force:
The friction force often opposes the motion of an object.
Friction depends upon the nature of the two surfaces (μ, dimensionless) and upon the degree to which they are pressed together(FN) .
Ff = μFN

15. Kinetic versus Static Friction
kinetic friction results when an object moves across a surface.
Ffrict = μk • Fnorm
The symbol μ represents the coefficient of kinetic friction between the two surfaces. The coefficient value is dependent primarily upon the nature of the surfaces that are in contact with each other. It does not depends on area of contact, the angle of the area, or the temperature, etc.
Static friction results when the surfaces of two objects are at rest relative to one another and a force exists on one of the objects to set it into motion relative to the other object.
The static friction force balances the force that you exert on the box such that the stationary box remains at rest.
Static friction can change.
Ffrict-static ≤ μs• Fnorm

16. Alert: which box has more friction?
A and B have the friction. A B

17. Practice- indicate Ff on each box with an arrowv FN FN FN Ff Ff v Ff v Fg Fg Fg Ff FN Ff Ff Ff FN FN v v v Fg Fg Fg

19. Example
A 24 kg crate initially at rest on a horizontal floor requires a 75 N horizontal force to set it in motion. Find the coefficient of static friction between the crate and the floor.

20. Class work Text book page 145 Practice 4C #1-3 Answers 0.23
a. 1.5; b. 1.3
a. 870 N, 670 N; b. 110 N, 84 N; c N, 500 N; d. 5N, 2 N

21. Air Resistance Force (Fair )
The air resistance is a special type of frictional force that acts upon objects as they travel through the air. The force of air resistance is often observed to oppose the motion of an object. This force will frequently be neglected due to its negligible magnitude.

22. Tension Force (FT )
The tension force is the force that is transmitted through a string, rope, cable or wire when it is pulled tight by forces acting from opposite ends. The tension force is directed along the length of the wire and pulls equally on the objects on the opposite ends of the wire.

23. Spring Force (Fspring )
The spring force is the force exerted by a compressed or stretched spring upon any object that is attached to it. An object that compresses or stretches a spring is always acted upon by a force that restores the object to its rest or equilibrium position – directed toward equilibrium position.
Fs is the force exerted by the spring on the object, k is spring constant and x is displacement from equilibrium position

24. Newton’s Laws Application
Drawing Free-Body Diagrams
Determining the Net Force
Apply Newton’s Laws

25. Drawing Free-Body Diagrams
Free-body diagrams are used to show the relative magnitude and direction of all forces acting upon an object in a given situation.
The size of the arrow in a free-body diagram reflects the magnitude of the force. The arrow shows the direction that the force is acting.
Each force arrow in the diagram is labeled to indicate the exact type of force.
It is generally customary to draw the force arrow from the center of the box outward in the direction that the force is acting.

26. FN Fg A block of wood is sitting motionless on a table.
What forces are acting on it?
Normal Force is a
REACTION
force that any
object exerts
when pushed on
Normal FN
Weight is the force of gravity
pulling an object toward the
CENTER OF THE
EARTH Fg
Weight

27. Class work
Practice on free body diagrams

28. Finding the External Net Force and acceleration
Fnet is the vector sum of all the individual forces. The three major equations that will be useful are
Fnet = m•a,
Fg = m•g,
Ff = μ•FN

29. Example #1
A man pushes a 50 kilogram crate across a frictionless surface with a constant force of 100 Newtons.
What is the normal force that pushes on the crate?
Draw a free-body diagram of the crate.
What is the net force on the crate?
What is the crate’s acceleration?
What is the weight of the crate?
Fnet will only be
the 100N horizontal
force FA FN Fg
Fg = mg
Fg = (50 kg)(9.81 m/s2)
Fg = N
FN = Fg
FN = N
a = Fnet / m
a = (100 N) / (50 kg)
a = 2 m/s2

30. Example #2
A horse pulls a 500 kilogram sled with a constant force of 3,000 Newtons. The force of friction between the sled and the ground is 500 Newtons.
What is the normal force that pushes on the sled?
Draw a free-body diagram of the sled.
What is the net force on the sled?
What is the sled’s acceleration?
What is the weight of the sled?
Fnet = ΣFx
Fnet = 3000 N – 500 N
Fnet = 2500 N
Fg = mg
Fg = (500 kg)(9.81 m/s2)
Fg = 4905 N FA FN Fg Ff
FN = Fg
FN = 4905 N
a = Fnet / m
a = (2500 N) / (500 kg)
a = 5 m/s2

31. Example #3
the object is moving horizontally. Use the diagram to determine the normal force, the net force, the mass, and the acceleration of the object. (g =10 m/s2)
80 N
8 kg
5 m/s2 right
40 N right

32. Example #4
Edwardo applies a 4.25-N rightward force to a kg book to accelerate it across a tabletop. The coefficient of friction between the book and the tabletop is Determine the acceleration of the book.

33. Example #5
Lee Mealone is sledding with his friends when he becomes disgruntled by one of his friend's 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?

34. Class work
Application of Newton’s Laws – pages 1&2

35. Free Fall and Air Resistance
Objects that are said to be undergoing free fall, are
not encountering air resistance;
falling under the sole influence of gravity. All objects will fall with the same rate of acceleration, regardless of their mass. This is due to that the acceleration is The ratio of force to mass (Fnet/m)
Falling with air resistance
As an object falls through air, it usually encounters some degree of air resistance - the result of collisions of the object's leading surface with air molecules.
The two most common factors that have a direct affect upon the amount of air resistance are
the speed of the object: Increased speeds result in an increased amount of air resistance.
the cross-sectional area of the object: Increased cross-sectional areas result in an increased amount of air resistance.

36. Falling with air resistance – 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. At this instant in time, the net force is 0 Newton; the object will stop accelerating. The object is said to have reached a terminal velocity.

37. Net Force Problems in 2D
When forces acting at angles to the horizontal, Newton’s 2nd law still applies:
∑F = ma
Force is a vector quantity. Adding forces in 2 dimensions follows the rules for adding vectors.

38. Determine the Fnet mathematically
Resolve the vectors at an angle into x and y components.
Add all the x components together
Add all the y components together – usually the
Use Pythagorean Theorem to find the resultant (hypotenuse) Resultant2 = x2 + y2
Use trigonometric function to determine the direction: tanθ = opp / adj

39. Example 1 - Pulling on an Angle
A block of 10 kg mass is pushed along a frictionless, horizontal surface with a force of 100 N at an angle of 30° above horizontal.
This applied force (FA)
can be broken into
COMPONENTS
FAx = 100cos(30o) = 87 N
FAy = 100sin(30o) = 50 N
X Y
FAX FAY Fg FN
Total = FAX Total = 0 FN
The total vertical force must
be 0, so
Ry = FN + FAY –Fg = 0
FN = Fg – FAY FA
FAY
30˚
FAX
R = Rx = Fax
Acceleration depends only on
FAX Fg

40. Example 2
A man pulls a 40 kilogram crate across a smooth, frictionless floor with a force of 20 N that is 45˚ above horizontal.
What is the net force on the sled?
How could the acceleration be increased?
Fnet = FA cos θ
Fnet = (20 N)(cos 45°)
Fnet = N
Pushing at a smaller angle will make Fnet greater and
therefore increase acceleration.
What is the crate’s acceleration?
a = Fnet / m
a = (14.14 N) / (40 kg)
a = 0.35 m/s2

41. This applied force (FA) The total vertical force must
Pushing on an Angle
A block is pushed along a frictionless, horizontal surface with a force of 100 newtons at an angle of 30° below horizontal.
This applied force (FA)
can be broken into
COMPONENTS
X Y
FAX FAY Fg FN
Total = FAX Total = 0 FN
The total vertical force must
be 0, so
FN = Fg + FAY
FAX
FAY
-30˚ FA
Acceleration depends only on
FAX Fg

42. Example 3
A girl pushes a 30 kilogram lawnmower with a force of 15 Newtons at an angle of 60˚ below horizontal.
Assuming there is no friction, what is the
acceleration of the lawnmower?
Fnet = FA cos θ
Fnet = (15 N)(cos 60°)
Fnet = 7.5 N
a = Fnet / m
a = (7.5 N) / (30 kg)
a = 0.25 m/s2
What could she do to reduce her acceleration?
Push at an greater angle

43. Example 4 – determine Fnet and a
The vertical forces are balanced (Fgrav, Fy, and Fnorm add up to 0 N),
The horizontal forces add up to 29.3 N, right
The net force is 29.3 N, right
a = Fnet / m = 29.3 N / 10 kg = 2.93 m/s2, right

44. Sample problem 4D
A student moves a box of books by attaching a rope to the box and pulling with a force of 90.0 N at an angle of 30.0 degrees. The box of books has a mass of 20.0 kg, and the coefficient of kinetic friction between the bottom of the box and the sidewalk is find the acceleration of the box.

45. Class work
Newton’s 2nd law practice – page 3 & 4

46. Equilibrium and Static
When all the forces that act upon an object are balanced, then the object is said to be in a state of equilibrium.
An object at equilibrium is either ...
at rest and staying at rest, or
in motion and continuing in motion with the same speed and direction.
"static equilibrium." refers to an object at rest

47. Example
A frame is shown with the given tension. Determine the weight of the frame.
Ax = 50cos(150o) = -43 N
Ay = 50sin(150o) = 25 N
Bx = 50cos(30o) = 43 N
By = 50sin(30o) = 25 N A B
30o
Rx = Ax + Bx + Cx = 0
-43 N + 43 N – Cx =0
Cx = 0
C = ?
Ry = Ay + By + Cy = 0
25 N + 25 N + Cy = 0
Cy = -50 N
C2 = Cx2+ Cy2
R = 50. N

48. example
A sign is shown with the given mass of 5 kg. Determine the tension of each cable. (g = 10 m/s2)
Tsin140o
Tsin40o
A = T
B = T
40o
40o
Tcos140o
Tcos40o
Ry = Ay + By + Cy = 0
Tsin40o + Tsin140o – Fg = 0
T = 38 N
C = Fg
= 50 N

49. An important principle
As the angle with the horizontal increases, the amount of tensional force required to hold the sign at equilibrium decreases.
Fg = 10 N

50. Down the Slope Inclined Planes
A tool used to move objects from one height to another.
Allows for the movement of an object without lifting it directly against gravity.
Down the Slope
Inclined Planes
The object accelerate downward due to the component gravity that is parallel to the plane.

51. Fg on Inclined Plane

52. Forces on an Incline Calculations
Consider forces:
Perpendicular
F┴ = Fg cos θ
Cancel out Normal (FN )
Parallel
F// = Fg sin θ
Could be in the same or opposite of Friction (Ff ) θ
Tilt you head method θ

53. Essential Knowledge
What happens to the component of weight that is perpendicular to the plane as the angle is increased?
Decreases – Fg perpendicular
What happens to the component of weight that points ALONG the plane as the angle is increased?
Increases – Fg parallel
What happens to the normal force as the angle is increased?
Decreases – depends on Fg perpendicular
What happens to the friction force as the angle is increased?
Decreases – depends on normal force

54. The net force is the vector sum of all the forces.
All the perpendicular components (including the normal force) add to 0 N.
All the parallel components (including the friction force) add together to yield the net force. Which should directed along the incline.

55. In the absence of friction
Fnet = F//
mgsinθ = ma
a = gsinθ

56. With friction - Object is at equilibrium – at rest or moving with constant velocityFf
Fnet = 0
Horizontal:
F// = Ff
mgsinθ = μFN
mgsinθ = μ∙mgcosθ
tanθ = μ
Vertical:
F┴ = FN
mgcosθ = FN

57. Example What is the magnitude of the normal force?
Fg = 50N
30°
What is the magnitude of the normal force?
FN = Fg perpendicular = Fg cos θ = 43.3 N
If the box is sliding with a constant velocity, what is the magnitude of the friction force?
Ff = Fg parallel = Fg sin θ = 25 N

58. Example 1
The free-body diagram shows the forces acting upon a 100-kg crate that 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 0.3. Determine the net force and acceleration of the crate.
F┴ = Fgrav∙cos30o = 850 N
F// = Fgrav∙sin30o = 500 N
In perpendicular direction:
Fnorm = F┴ = 850 N
In parallel direction:
Fnet = F// - Ff
Fnet = 500 N - µFnorm
Fnet = 235 N
a = Fnet / m = 2.35 m/s2

59. practice

60. Class work

61. Lab 10 – weight vs. mass
Purpose: determine the relationship between mass and weight
Material: spring scale, known masses
Data table: include heading in data table
Data analysis: graph weight vs. mass
Conclusion:
What is the relationship between weight and mass?
Determine the slope and indicate the meaning of the slope in the graph.

62. Double Trouble (a.k.a., Two Body Problems)
Two body-problems can typically be approached using one of two basic approaches.
One approach is the system analysis, the two objects are considered to be a single object moving (or accelerating) together as a whole.
Another approach is the individual object analysis, either one of the two objects is isolated and considered as a separate, independent object.

63. Example - system analysis
A 5.0-kg and a 10.0-kg box are touching each other. A 45.0-N horizontal force is applied to the 5.0-kg box in order to accelerate both boxes across the floor. Ignore friction forces and determine the acceleration of the boxes and the force acting between the boxes.
m = 15 kg
Fnet = 45 N
a = Fnet / m = 3 m/s2

64. Example - individual analysis
In vertical direction:
FN = Fg = (5 kg) (9.81 m/s2) = 49 N
In vertical direction:
FN = Fg = (10 kg) (9.81 m/s2) = 98 N
In horizontal direction:
Fnet = Fapp - Fcontact
(5 kg)a = 45 N - Fcontact
In horizontal direction:
Fnet = Fcontact
(10 kg)∙a = Fcontact
5a = 45 – 10a
a = 3 m/s2

65. Example: system analysis
A 5.0-kg and a 10.0-kg box are touching each other. A 45.0-N horizontal force is applied to the 5.0-kg box in order to accelerate both boxes across the floor. The coefficient of kinetic friction is Determine the acceleration and the contact force.
In vertical direction:
FN = Fg = (15 kg) (9.81 m/s2) = 147 N
In horizontal direction:
Fnet = Fapp - Ffrict = 45 N - μ•Fnorm
Fnet = 15.6 N
a = Fnet / m = (15.6 N/15.0 kg) = 1.04 m/s2
However, in order to find the contact force between the objects, we must make individual analysis.

66. Example: individual analysis
In vertical direction:
FN = Fg = (10 kg) (9.81 m/s2) = 98 N
In horizontal direction:
Fnet = Fcontact - Ff
(10 kg)∙(1.04 m/s2) = Fcontact - μ•Fnorm
10.4 = Fcontact – (0.2)(9.8)
Fcontact = 8.44 N

67. Lab 10 – weight vs. mass
Purpose: determine the relationship between mass and weight
Material: spring scale, known masses
Data table: include heading in data table
Data analysis: graph weight vs. mass
Conclusion:
What is the relationship between weight and mass?
What does the slope in the graph mean?