1. Newton’s Laws

2. Newton's Laws & Classical Mechanics
Classical mechanics describes the motion of object and the forces acting on them.
Classical mechanics is very accurate as long as we do not try to study something as small as an atom, or something moving close to the speed of light.
Classical mechanics describes the motion of object and the forces acting on them.

3. EQ: What is a force?
A force is an action exerted on an object which may change the object’s state of rest or motion.
What happened to the door?
Forces can cause a massive object to accelerate.
Forces can act through contact or at a distance.
What happened to the door?
The SI unit of force is the newton, N.

4. Newton’s Contributions
Calculus
Light is composed of rainbow colors
Reflecting Telescope
Your objectives are to learn.
Three Laws of Motion
Theory of Gravitation

5. What is a force? A push A pull A stretch A squeeze A catch A twist
Any change in motion
Here are some ways of describing forces:
A push
A pull
A stretch
A squeeze
A catch
A twist
We can’t see forces but we can see the effects of a force.
Acceleration
Tension
Compression-Deformation
De-acceleration
Torque
Acceleration
Tension
Compression
Deformation
De-acceleration
Torque

6. Forces can make things:
Speed up
Slow down
Change direction
Change shape

7. Forces can act through contact or at a distance.
Forces can be labeled into two categories: field force and contact force.

8. Force = Mass*Acceleration: UnitsX =
Mass vs. weight:
Do you have the same amount of mass on the Moon as you do the Earth?
Derived units:
Mass vs. weight
195lb
N= kg * m s2
lb = slug * ft s2

9. Do you have the same amount of mass on the Moon as you do the Earth?
Mass vs. weight:
Do you have the same amount of mass on the Moon as you do the Earth?
Mass- a measure in the amount of matter in an object.
Weight: a measure of the amount of gravitational force acting on the mass of an object.
EX:
Wf= the weight of an object due to a gravitational field
m = mass of the object
g = acceleration due to gravity
Wf = mg  m = Wf g
Wf = mg  m = Wf g
Wf = 195lb
gE = 32.2ft/s2
m = ?
Wf = mg  m = 195lb * s2
32.2ft
= but what are the units?

10. Weight W = mg Units weight is a force
weight = mass x acceleration due to gravity
Units
N = kg x m/s2
weight in Newtons = mass in kg x m/s2
a 1 kg mass weighs 9.81 N

11. Derived Units : Pause for a Cause Your Turn: Calculate your mass
Wf = mg  m = 195lb * s2
32.2ft
NOTE: MASS and WEIGHT are NOT the same thing. MASS never changes
When an object moves to a different planet.
Pause for cause:
m = 195slug * ft * s2
s ft
Your Turn: Calculate your mass

12. Pause for a Cause
What is the weight of an 85.3-kg person on earth? On Mars=3.2 m/s2?

13. Force Diagrams
The effect of a force depends on both magnitude and direction. Thus, force is a vector quantity.
Diagrams that show force vectors as arrows are called force diagrams, or Free Body Diagrams.
Force diagrams that show only the forces acting on a single object are called free-body diagrams.

14. Force Diagrams Force Diagram Free-Body Diagram
A free-body diagram shows only the forces acting on the object of interest—in this case, the car.
In a force diagram, vector arrows represent all the forces acting in a situation.
The sum of all the forces is called the Net Force.
The sum of all the forces is called the Net Force.

15. Free Body Diagrams The key to analyzing problems:
pictorial representation of forces complete with labels. FN
Weight(mg) – Always drawn from the center, straight down
Force Normal(FN) – A surface force always drawn perpendicular to a surface.
Tension(T or FT) – force in ropes and always drawn AWAY from object.
Friction(Ff)- Always drawn opposing the motion. T Ff T
W1,Fg1 or m1g
m2g

16. Free Body Diagrams Fn Ff mg θ
mg sinθ
Practice A
mg cosθ
Practice A

17. Pause for a Cause soh cah toa
A women is pulling on her suitcase with a force of 70.0 N directed at an angle of +30.0° to the horizontal. Find the x and y components of this force.
Force in the X-axis
Fx = F(cos θ) =
Force in the Y-axis
Fy = F(sin θ) =
Given:
F = 70 N
Θ = 30.0°
Force in the X-axis
Fx = 70.0 N(cos 30.0°) =
Force in the Y-axis
Fy = 70.0 N(sin 30.0°) =
Sah cah toa
Hyp
Hyp
60.6 N
35.0 N
Y-axis
X-axis
soh cah toa

18. Essential Question
EQ: A fan blows on two balls, a bowling ball and a balloon. Describe what you think will happen.
bowling ball
balloon
Thought Experiment: a Pro Golfer tees up a bowling ball and strikes it with his driver. Describe what you think will happen?

19. Newton’s Laws Inertia: F = ma Action-reaction
An object at rest will stay at rest, and an object in motion will stay in motion at constant velocity, unless acted upon by an unbalanced force.
Force equals mass times acceleration.
Inertia:
F = ma
Action-reaction
For every action there is an equal and opposite reaction.

20. Inertia: Video
Inertia is the tendency of an object to resist being moved or, if the object is moving, to resist a change in speed or direction.
Newton’s first law is often referred to as the law of inertia because it states that in the absence of a net force, a body will preserve its state of motion.
Mass is a measure of inertia.
Why
Why?

21. Newton’s First Law: Inertia
An object in motion remains in motion in a straight line and at a constant speed OR an object at rest remains at rest, UNLESS acted upon by an EXTERNAL (unbalanced) Force.
There are TWO conditions here and one constraint.
Condition #1 – The object CAN move but must be at a CONSTANT SPEED
Condition #2 – The object is at REST
Constraint – As long as the forces are BALANCED!!!!! And if all the forces
are balanced the SUM of all the forces is ZERO.
Pause for a Cause: If acceleration = 0, then the sum of the forces must = ?
The bottom line: There is NO ACCELERATION in this case AND the object
must be at EQILIBRIUM ( All the forces cancel out).
Pause for a Cause: If acceleration = 0, then the sum of the forces must = ?

22. Newton’s 1st Law of MotionV W W Fr Fp Ry Ry
Fr = resistive force
Fp = propulsive force
if v = 0 and SF = 0
DYNAMIC EQUILIBRIUM
if Ry = W
then resultant force = 0
if v = 0 and SF = 0
STATIC EQUILIBRIUM

23. N.F.L and Equilibrium
Since the Fnet = 0, a system moving at a constant speed or at rest MUST be at EQUILIBRIUM.
TIPS for solving problems
Draw a FBD
Resolve anything into COMPONENTS
Write equations of equilibrium
Solve for unknowns

24. Pause for a Cause
A 10-kg box is being pulled across the table to the right at a constant speed with a force of 50N.
Calculate the Force of Friction
Calculate the Force Normal Fn Fa Ff mg

25. Pause for a Cause
Suppose the same box is now pulled at an angle of 30 degrees above the horizontal.
Calculate the Force of Friction
Calculate the Force Normal Fn Fa
Fay Ff 30
Fax mg

26. What if it is NOT at Equilibrium?
If an object is NOT at rest or moving at a constant speed, that means the FORCES are UNBALANCED. One force(s) in a certain direction over power the others. THE OBJECT WILL THEN ACCELERATE.

27. Inertia
The sum of all forces acting on an object is called net force.

28. Equal forces in opposite directions produce no motion
Balanced Force
Equal forces in opposite
directions produce no motion
What is the net force of the truck on bottom?
Unequal opposing forces
produce an unbalanced force
causing motion
Unbalanced Forces

29. Vector Sum of Force
Force 1 (F1) and force 2 (F2) are applied to the ball at the same time with the same magnitude.
Draw a new arrow describing the combined direction of F1 & F2 on the ball. This is called the vector sum.
F1 = F2
Graphing assignment
Graphing assignment

30. Graphing Assignment
Two horizontal ropes are attached to a post that is stuck in the ground. The ropes pull the post producing the following vector forces
Determine the direction and magnitude the support cable must have to prevent pole failure.
F1 = (5, 5) F2 = (-3, 6)
What is the direction and magnitude of the force the pole is acting against the two ropes.
F1 = (5Nx, 5Ny) F2 = (-3Nx, 6Ny)
(X, Y)
(X, Y) (X, Y)
F1 = (5Nx, 5Ny) F2 = (-3Nx, 6Ny)

31. Graphing Assignment Hint: -F3XY = (F1X + F2X) (F1y + F2y)
If F1 + F2 = -F3 diagram the direction and magnitude that the pole pulls back (F3) on the ropes?
Hint: -F3XY = (F1X + F2X) (F1y + F2y)

32. Graphing Assignment What is the angle between F2 & F3?
Hint: tan(θ) = Opposite / Adjacent 
θ = tan -1 (Opposite/Adjacent) or
(Y/X)
(-1) means (1/tan) or inverse.

33. Pause for a Cause: Net Force Level 1
Fnet = ma
Fnet = ma F1
Fnet = F1 + F2
F1 = Fx + Fy
F2 = Fx + Fy
F1xy + F2xy = ma F2

34. Net Force ΣFx = F1x + F2x ΣFy = F1y + F2y Fnet = F1 + F2
Given:
Fnet = ma
a = ?
m = 2.00E3 Kg
Fh = 6.00E2 N
θ1 = 30.0˚
θ2 = 45.0˚ F1
Fnet = F1 + F2
F1 = F1x + F1y
F2 = F2x + F2y F2
ΣFx = F1x + F2x
F1x = F1 cos θ1x =
F2x = F2 cos θ2x =
F1x = 6.00E2 cos 30.0˚ =
F2x = 6.00E2 cos -45.0˚ =
5.02E2 N
4.24E2 N
ΣFx = 9.44E2 N
ΣFy = F1y + F2y
5.02E2 N
F1y = F1 sin θ1y =
F2y = F2 sin θ2y =
F1y = 6.00E2 sin 30.0˚ =
F2y = 6.00E2 sin -45.0˚ =
3.00E2 N
-4.24E2 N
ΣFy = E2 N

35. Magnitude of Acceleration
F = ma
ax = Fx m
ay = Fy
2D: Acceleration
a2 = ax2 = ay2
a = √ax2 = ay2 F2
Given:
a = ?
m = 2.00E3 Kg
Fh = 6.00E2 N
θ1 = 30.0˚
θ2 = 45.0˚
ax = 9.44E2 N =
2.00E3 Kg
0.472 m/s2
m/s2
ay = -1.24E2 N =
2.00E3 Kg
0.472 m/s2
m/s2
a = √ax2 = ay2 
ΣFx = 9.44E2 N
ΣFy = E2 N
a = √(0.472)2 + (-0.062)2 =
0.476 m/s2

36. Direction of Acceleration
The direction will be the ratio of the acceleration in the x-axis & the y-axis
ay  sin θ  tan θ
ax cos θ
= ay ax F2
θ = tan-1 ay ax
Given:
a = m/s2
m = 2.00E3 Kg
Fh = 6.00E2 N
θ1 = 30.0˚
θ2 = 45.0˚
m/s2
0.472 m/s2
θ = tan-1
= ˚
The direction will be the ration of the acceleration in the x-axis & the y-axis
ΣFx = 9.44E2 N
ΣFy = E2 N

37. Newton’s Second Law m m where k1 = constant of proportionality F2 SF a

38. Newton’s Second Law
The acceleration of an object is directly proportional to the NET FORCE and inversely proportional to the mass.
Tips:
Draw an FBD
Resolve vectors into components
Write equations of motion by adding and subtracting vectors to find the NET FORCE. Always write larger force – smaller force.
Solve for any unknowns

39. Pause for a Cause
A 10-kg box is being pulled across a frictionless table to the right by a rope with an applied force of 50N. Calculate the acceleration of the box.
In which direction, is this object accelerating?
The X direction!
So N.S.L. is worked out using the forces in the “x” direction only FN Fa mg

40. Pause for a Cause
A 10-kg box is being pulled across the table to the right by a rope with an applied force of 50N. Calculate the acceleration of the box if a 12 N frictional force acts upon it.
In which direction, is this object accelerating?
The X direction!
So N.S.L. is worked out using the forces in the “x” direction only FN Fa Ff mg

41. Slope Slope = rise / run F F a a Force Acceleration Mass 5 N 2 m/s2

42. Slope
Slope = rise / run = F / a, the slope is equal to the mass. Or, think of y = m x + b, like in algebra class. y corresponds to force, m to mass, x to acceleration, and b (the y-intercept) is zero. F a F a

43. Friction A non-conservative force that opposes motion
Acts parallel to the surfaces in contact.
µ  is the coefficient of friction.
It is unique for each material
It is determined experimentally

44. What about surface area? Which would have a greater friction force?
stress = F a
Friction force does not depend on the area of contact. Same FN and µ!
stress = F/a

45. TWO types of Friction
Static – Friction that keeps an object at rest and prevents it from moving
Kinetic – Friction that acts during motion
A force that opposes motion
Acts parallel to the surfaces
in contact.

46. Force of Friction
The Force of Friction is directly related to the Force Normal.
Mostly due to the fact that BOTH are surface forces
The coefficient of friction is a unitless constant that is specific to the material type and usually less than one.
Note: Friction ONLY depends on the MATERIALS sliding against each other, NOT on surface area.

47. Pause for a Cause
a) What is the coefficient of kinetic friction between the crate and the floor?
A 1500 N crate is being pushed across a level floor at a constant speed by a force F of 600 N at an angle of 20° below the horizontal as shown in the figure. Fa FN
Fay 20
Fax
= 0.33
= 0.33 Ff mg

48. Pause for a Cause = 0.884 m/s2 FN Fa
If the 600 N force is instead pulling the block at an angle of 20° above the horizontal as shown in the figure, what will be the acceleration of the crate. Assume that the coefficient of friction is the same as found in (a)
Fay 20
Fax Ff mg
 FN = mg
= mFN = mg
= m/s2

49. Newton’s Third Law
“For every action there is an EQUAL and OPPOSITE reaction.
This law focuses on action/reaction pairs (forces)
They NEVER cancel out
All you do is SWITCH the wording!
PERSON on WALL
WALL on PERSON

50. N.T.L
This figure shows the force during a collision between a truck and a train. You can clearly see the forces are EQUAL and OPPOSITE. To help you understand the law better, look at this situation from the point of view of Newton’s Second Law.
There is a balance between the mass and acceleration. One object usually has a LARGE MASS and a SMALL ACCELERATION, while the other has a SMALL MASS (comparatively) and a LARGE ACCELERATION.

51. N.T.L Examples Action: HAMMER HITS NAIL Reaction: NAIL HITS HAMMER
Action: Earth pulls on YOU
Reaction:
YOU pull on the earth

52. Tension
In physics, tension is the pulling force exerted by a string, cable, chain, or similar solid object on another object.

53. Pause for a Cause Acceleration = 7.7 m/s2
A mass, m1 = 3.00kg, is resting on a frictionless horizontal table is connected to a cable that passes over a pulley and then is fastened to a hanging mass, m2 = 11.0 kg as shown below. Find the acceleration of each mass and the tension in the cable. FN T T
m1g
7.7 m/s2
m2g
Acceleration = 7.7 m/s2

54. Pause for a Cause

55. T3 =>T3 - Fg T3 – Fg = 0equilibrium T3 = Fg T3 = 100 N
A traffic light weighing 100 N hangs from a vertical cable tied to two other cables that are fastened to a support at the angle illustrated. Find the tension in each of the three cables.
T3 =>T3 - Fg
T3 – Fg = 0equilibrium
T3 =>T3 – Fg equilibrium
T3 = Fg
T3 = 100 N

56. ΣFy = T1y sinθ + T2y sinθ – T3= 0
Hints:
ΣFx there are two forces in the x-axis and they are opposing each other
ΣFy there three forces in the y-axis T1 & T2 are opposing T3
ΣFx = T1x T2x
ΣFx = T1x cosθ - T2x cosθ = 0
Hints:
ΣFx there are two forces in the x-axis and they are opposing each other
ΣFy there three forces in the y-axis T1 & T2 are opposing T3
ΣFy = T1y T2y = T3
ΣFy = T1y sinθ + T2y sinθ – T3= 0

57. 2) ΣFy = T1y sinθ + T2y sinθ – T3= 0
There two equations & two unknows
1) ΣFx = T1x cosθ - T2x cosθ = 0
2) ΣFy = T1y sinθ + T2y sinθ – T3= 0
Solve equation 1 for T2
T1 sinθ + T2 sinθ – T3= 0
T1x cos37˚ - T2x cos53˚ = 0
T2x cos53˚ = T1x cos37˚
T1 sinθ T1 sinθ – T3= 0
T1 sinθ T1 sinθ = T3
T2x = T1x * cos37˚
cos53˚
T1 (sinθ sinθ) = T3
T2x = 1.33T1x
T1 = _____T3_______
(sinθ sinθ)
There two equations & two unknows
T1 = _____100 N_____
(sin37˚ sin53˚)
T1 = N
T2 = N

58. Inclines q q Ff FN q q q Tips Rotate Axis Break weight into components
Write equations of motion or equilibrium
Solve mg q

59. Pause for a Cause
Two packing crates of masses 10.0 kg and 5.0 kg are connected by a light string that passes over a frictionless pulley. The 5.00 kg crate lies on a smooth incline of angle 40.0˚. Find the acceleration of the 5.00 kg crate and the tension in the string. T FN m2
m2gcos40 Ff 40 T
m2g m1 40
m2gsin40
m1g

60. Pause for a Cause Solve 1) for T Sub 1) for T into 2) for T T FN m2 Ff
m2gcos40 Ff 40 T
m2g m1 40
Solve 1) for T
m2gsin40
m1g

61. Pause for a Cause
Masses m1 = 4.00 kg and m2 = 9.00 kg are connected by a light string that passes over a frictionless pulley. As shown in the diagram, m1 is held at rest on the floor and m2 rests on a fixed incline of angle 40 degrees. The masses are released from rest, and m2 slides 1.00 m down the incline in 4 seconds. Determine (a) The acceleration of each mass (b) The coefficient of kinetic friction and (c) the tension in the string. T FN m2
m2gcos40 Ff 40 T
m2g m1 40
m2gsin40
m1g

62. Example

64. The End