1. Forces &
Free-body Diagrams
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2. A force is any influence that can change the velocity of a body.
What is a force?
A force is any influence that can change the velocity of a body.
Forces can act either through the physical contact of two objects (contact forces: push or pull) or at a distance (field forces).
Contact Forces
Action-at-a-Distance Forces
Frictional Force
Gravitational Force
Tensional Force
Electrical Force
Normal Force
Magnetic Force
Air Resistance Force
Applied Force
Spring Force

3. Spring Scales are used to measure forces.
How do we measure forces?
Spring Scales are used to measure forces.
The spring inside the scale stretches an amount proportional to the force applied and can be calibrated (scaled)

4. Weight = mass x gravity Newton = kg x m/s2 1 N = 1 kgm/s2 W = m x g
The weight of a body is the gravitational force with which the Earth attracts the body.
Weight (a vector quantity) is different from mass (a scalar quantity). The weight of a body varies with its location near the Earth (or other astronomical body), whereas its mass is the same everywhere in the universe.
The weight of a body is the force that causes it to be accelerated downward with the acceleration of gravity g.
Weight = mass x gravity
W = m x g
Newton = kg x m/s2
1 N = 1 kgm/s2

5. Newton (N) Pound (lb) What units are used for FORCES? In Metric…
is the amount of force required to give a 1-kg mass an acceleration of 1 m/s2.
In English…
is the amount of force required to give a 1-slug mass an acceleration of 1 ft/s2.
Pound (lb)
Length
Time
Mass
Force
English
Foot
(ft)
Second
(s)
Slug
(sl)
Pound
(lb)
Metric
Meter
(m)
Kilogram (kg)
Newton
(N)

6. 1 pound = 4.45 Newtons 1 kilogram weighs 9.80 N = 2.2 lbs
Conversions or Equivalents
1 pound = 4.45 Newtons
1 kilogram weighs 9.80 N = 2.2 lbs

7. a magnitude & a direction
How do we describe forces?
Forces have
a magnitude & a direction

8. Free-body Diagrams
Free-body diagrams are diagrams used to show the relative magnitude and direction of all forces acting upon an object in a given situation.
The size of the arrow reflects the magnitude of the force.
The direction of 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.
In a free-body diagram the object is represented by a box or dot

9. FT (T) force due to tension: a rope or cord
Complete the free body diagram showing all of the forces acting on the mass M. Be sure to show the direction of each force as an arrow and label each force clearly!
FT (T) force due to tension: a rope or cord
FN (N) force Normal: acting perpendicular to a surface
Ff (f) force of friction: opposes motion
Fg (W) gravitational force or weight, always downward
Fa applied force: push or pull
Fs force exerted by a spring

10. Example: (with friction)FN Fa Ff Fg

11. Fs Fg

13. FN1
FN2
(without friction) Fa FT FT
Fg1
Fg2

14. Example: (with friction)FN Fa Ff Fg

15. Universal Gravitation
Weight vs Mass and
Newton’s Law of
Universal Gravitation

16. Universal Gravitation
Newton’s Law of
Universal Gravitation
Any two objects in the Universe exert an attractive force on each other -called the gravitational force- whose strength is proportional to the product of the objects’ masses and inversely proportional to the square of the distance between them. If we let G be the universal gravitational constant, then the strength of the gravitational force is given by the equation:
Units: Newtons (N)

17. Henry Cavendish determined the first reasonably accurate numerical value for G more than one hundred years after Newton’s Law was published.
To three decimal places, the currently accepted value is:
G = 6.67 x10-11 N.m2/kg2

18. Example 1: What is the force of gravity acting on a 2000 kg spacecraft when it orbits x 10 7 m from the Earth’s center? (this is about 2 Earth Radii)
m1 = 2000 kg
ME = 5.98x1024 kg
rE = 1.276x107 m
= 4899 N

19. The equation for Universal Gravitation is an inverse square law
If the force of attraction is 320 N
What happens if the distance is quadrupled?
20 N
What happens if the distance is halved?
1280 N

20. Example 2 a. Derive the expression for g from the Law of Universal Gravitation.
Fg = FUG

21. b. Estimate the value of g on top of the Everest (8848 m) above the Earth’s surface.
mE = 5.98x1024 kg
RE = 6.38x106 m
RT = x106 = 6.388x106 m
= 9.77 m/s2

23. Newton’s Laws Problems
No friction!
Dog Inertia

24. Newton's First Law of Motion:
" If no net force acts on it, a body at rest remains at rest and a body in motion remains in motion at constant speed in a straight line."
Isaac Newton
( )

25. MASS and INERTIA
The property a body has of resisting any change in its state of motion is called inertia. The inertia of a body is related to the amount of matter it contains.
A quantitative measure of inertia is mass.
The SI unit of mass is the kilogram (kg).

26. Newton’s
2nd Law

27. Fnet = ma SECOND LAW OF MOTION
According to Newton's Second Law of Motion, the net force acting on a body equals the product of the mass and the acceleration of the body. The direction of the force is the same as that of the acceleration.
In equation form:
Fnet = ma

28. Example 3 A 60.0 g tennis ball approaches a racket at 15 m/s, is in contact with the racket for s, and then rebounds at 20 m/s. Find the average force exerted by the racket.
(-) 420 N, away from racket.

29. Example 4 The brakes of a 1000 kg car exert 3000 N.
a. How long will it take the car to come to a stop from a velocity of 30 m/s?
t = 10 sec
b. How far will the car travel during this time?
d = 150 m

30. Example 5 A 5. 0-kg object is to be given an upward acceleration of 0
Example 5 A 5.0-kg object is to be given an upward acceleration of 0.3 m/s2 by a rope pulling straight upward on it. What must be the tension in the rope?
Ft = 50.5 N

31. Example 6 A 200-N wagon is to be pulled up a 30 incline at constant speed. How large a force parallel to the incline is needed if friction is negligible?
Fa = 100 N

32. Newton’s 3rd Law

33. THIRD LAW OF MOTION Hammer hits post Post hits hammer
 According to Newton's third law of motion, when one body exerts a force on another body, the second body exerts on the first an equal force in opposite direction.
Hammer hits post
The Third Law of Motion applies to two different forces on two different objects: "The action force one object exerts on the other, and the equal but opposite reaction force the second object exerts on the first."
Post hits hammer

34. Forces occur in pairs…

35. When we make decisions about the motion of an object we look at the forces on an object in isolation
Tension pair
Friction pair
Who exerts the greater force?
Who will experience the greater affect from the applied force?
Free body of forces
on girl
10 N
200 N

36. The old bug on the windshield problem…
Fcar = Fbug
a a
m = m

37. Newton’s Laws . . . With Friction!

38. THE NORMAL FORCE
A normal force is a force exerted by one surface on another in a direction perpendicular to the surface of contact.
Note: The gravitational force and the normal force are not an action-reaction pair.

39. Friction results from the mutual contact of irregularities in the surfaces. (The atoms cling together at many points of contact. As sliding occurs, the atoms snap apart or are torn from one surface to the other.)

40. TYPES OF FRICTION
Static Friction the resistance force that must overcome to start an object in motion.
Kinetic or Sliding Friction the resistance force between two surfaces already in motion
THINGS TO REMEMBER ABOUT FRICTION
The direction of friction is ALWAYS in a direction opposing motion.
Sliding friction is always less than starting friction.

41. μ = Ff FN COEFFICIENT OF FRICTION The frictional force between two
surfaces depends on the normal
force pressing them together and on the natures of the surfaces. The latter factor is expressed quantitatively in the coefficient of friction,  (mu) whose value depends on the materials in contact.   μ
μ = Ff FN
Tables of coefficients of friction are printed in references and textbooks.

42. COEFFICIENT OF FRICTION
   
Static Friction force = μs Normal force
Sliding (kinetic) Friction force = μk Normal force
Tables of coefficients of friction are printed in references and textbooks.

43. μ = Ff FN = Ff mg = 50N (10kg)(9.8m/s2) μ = 0.51
Example 7: If a force of 50 N is applied to a block of mass 10 kg and it does not move, what is the coefficient of friction for the block and table. Fg FN FA fs
Givens:
Fa = Ff = 50 N
FN = Fg = mg
μ = Ff FN
= Ff mg
= N
(10kg)(9.8m/s2)
μ = 0.51

44. ΣFy = 0 FN – Fg = 0 FN = Fg FN = mg ΣFx = 0 FN = 60kg(9.8m/s2)
Example 8: A horizontal force of 140 N is needed to pull a 60.0 kg box across the horizontal floor at constant speed. What is the coefficient of friction between floor and box? Fa Ff
Givens:
Fa = 140 N
m = 60 kg Fg
ΣFy = ma
ΣFy = 0
FN – Fg = 0
FN = Fg
FN = mg
FN = 60kg(9.8m/s2)
FN = 588 N
ΣFx = 0
Fa – Ff = 0
Fa = Ff = 140 N N
= 0.24 N

45. What happens when we put everything on an incline?

46. Gravity is ALWAYS directed DOWNWARD So our weight now has
component forces; Fgx and Fgy.
Fgy is the part of the weight that acts on the incline.
Fgx is the part of the weight that acts down the incline.

47. Example 9: A 500 N trunk sits on an inclined plane that forms a 16 with the horizontal.
What is the coefficient of friction μ ?

48. Friction isn’t always bad
Traction
Brakes

49. Problem Solving Strategy
Draw a sketch and label all information.
Draw a free-body diagram.
Find components of all forces (+ and -).
Apply First Condition for Equilibrium:
SFx= 0 ; SFy= 0
5. Solve for unknown forces or angles.

50. Example 10 A 50 N sign hangs by a single wire as shown
Example 10 A 50 N sign hangs by a single wire as shown. What is the tension in the wire? T2 T1
T2Y
T1Y
T1X
T2X
BEARS
Step 1. Sketch
Step 4 Find the sum of x and y components
Step 3 Find components
Step 2. FBD
Step 5 Solve for T FW
ΣFy = 0
T1Y + T2Y – Fw = 0
T1sinΘ +T2sinΘ – Fw = 0
T1sin40 + T2sin40 – 50N = 0
T1sin40 + T1sin40 – 50N = 0
2T1sin40 – 50N = 0
T1 = 50N
2sin40
T1 =T2 = 38.9 N
ΣFx = 0
T1X - T2X = 0
T1cosΘ - T2cosΘ = 0
T1cos40 - T2cos40 = 0
T1cos40 = T2cos40
T1 = T2

51. Example 11 A 50 kg crate is hanging on a 4 m long rope which sags 5
Example 11 A 50 kg crate is hanging on a 4 m long rope which sags 5.0 cm. What is the tension in the rope?
ΣFx = 0
T1X - T2X = 0
T1 = T2 T1 T2
T1Y
0.05m
T2Y Θ
T2X
T1X 2m
ΣFy = 0
T1Y + T2Y – Fw = 0
T1sinΘ +T2sinΘ – Fw = 0
2T1sinΘ – 490N = 0
2T1sin1.43 – 490N = 0
T1 = 490N
2sin1.43
T1 =T2 = N Fw
How do I find the angle?
tan Θ = opp/adj
tan Θ = .05m/2m
Θ = 1.43
Fw = 50kg(9.8m/s2) = 490N

52. Apparent Weight
Apparent weight is a force that acts in opposition to gravitational force in order to prevent a body from going into freefall.
When you stand on the floor, the floor pushes up on your feet with a force equal to your apparent weight

53. When you are in an elevator, your actual weight (mg) never changes.
You feel lighter or heavier during the ride because your apparent weight increases when you are accelerating up, decreases when you are accelerating down, and is equal to your weight when you are not accelerating at all.

54. So you feel heavier FN + a - Fg
Example 12: An 85-kg person is standing on a bathroom scale in an elevator. What is the person’s apparent weight
a) When the elevator accelerates upward at 2.0 m/s2?
So you feel heavier FN Fg a + -

55. b) when the elevator is moving at constant velocity between floors?
c) when the elevator begins to slow at the top floor at 2.0 m/s2?
So you feel lighter FN Fg - a +

56. - a + Fg So you feel weightless
OH NO… the cable breaks…
The acceleration is downward so the net force is downward - a + Fg
So you feel weightless

57. Example 14 A cord passing over a frictionless pulley has a 7
Example 14 A cord passing over a frictionless pulley has a 7.0 kg mass hanging from one end and a 9.0-kg mass hanging from the other. (This arrangement is called Atwood's machine).
Find the acceleration of the masses.
B. Find the tension in the cord
m1 = 7 kg
m2 = 9 kg

58. Magic Pulleys on a flat table
Magic pulleys bend the line of action of the force without affecting tension. The pulley is “magic” – no mass, no friction, and no effect on the tension. T
m2g N
m1g -x x m1
Frictionless table m2

59. Example 13 An Atwood off the table and friction An object m1 = 25 kg rests on a tabletop. A rope attached to it passes over a light frictionless pulley and is attached to a mass m2 = 15 kg. If the coefficient of friction is 0.20 between the table and block A, how far will block B drop in the first 3.0 s after the system is released?
m1 = 25 kg
m2 = 15 kg
μ = 0.2
t = 3 s

60. WOW is this going to be fun!