1. Chapter 4 The Laws of Motion
Physics 350

2. The Laws of Motion Kinematics Dynamics
Math of HOW things move
Position, velocity, acceleration
Dynamics
WHY do things move?
What causes a body to accelerate?
Forces => Acceleration
The properties of force and the relationships between force and acceleration are given by Newton’s Three Laws of Motion

3. The Laws of Motion
The First Law describes the natural state of motion of a body on which no forces are acting. The other two laws describe the behavior of a body under the influence of forces.
Early 1600’s, theories of object’s tendency to be at rest

4. Laws of Motion Galileo Galilei Sir Isaac Newton
Developed first correct ideas of motion
Gravity and constant acceleration
Forces acting on bodies
Sir Isaac Newton
1687 Principia Mathematica
Laws of Motion
Law of Universal Gravitation
Invented calculus to further describe speed, acceleration

5. Forces Force A central concept in all of physics A vector quantity
Magnitude
Direction
Force is used to describe push or a pull
Forces on objects
springs
rubber bands
ropes
Cables

6. Forces (cont.) Force All examples above are known as “contact forces”
Bouyant Forces
liquids
Friction
Surfaces
All examples above are known as “contact forces”

7. Newton’s First Law Newton’s First Law
“An object moves with a velocity that is constant in magnitude and direction, unless acted on by a nonzero net force.”
The net force on an object is defined as the vector sum of all external forces exerted on the object.
Often called the Law of inertia
Tendency of object in motion to stay in motion

8. ConcepTest 4.2 Cart on Track I
Consider a cart on a horizontal frictionless table. Once the cart has been given a push and released, what will happen to the cart?
1) slowly come to a stop
2) continue with constant acceleration
3) continue with decreasing acceleration
4) continue with constant velocity
5) immediately come to a stop

9. ConcepTest 4.2 Cart on Track I
Consider a cart on a horizontal frictionless table. Once the cart has been given a push and released, what will happen to the cart?
1) slowly come to a stop
2) continue with constant acceleration
3) continue with decreasing acceleration
4) continue with constant velocity
5) immediately come to a stop
After the cart is released, there is no longer a force in the x-direction. This does not mean that the cart stops moving!! It simply means that the cart will continue moving with the same velocity it had at the moment of release. The initial push got the cart moving, but that force is not needed to keep the cart in motion.

10. Newton’s Second Law Newton’s Second Law
“The acceleration a of an object is directly proportional to the net force acting on it and inversely proportional to its mass.”
What it means:
a = ΣF / m
or conversely,
ΣF = ma

11. Newton’s Second Law ΣF = ma A LAW of nature!
Precise definition of FORCE
a and F are in the same direction
after F is completely summed

12. Newton’s Second Law (cont.)
ΣF = ma ΣF
ΣFx = max
ΣFy = may
ΣFz = maz
No net force, means acceleration is zero
Velocity is constant

13. Newton’s Second Law (cont.)
Units
Mass
Kilograms kg
Acceleration
m/s2
Force
Newtons 1N = 1 kg x m/s2 (m x a)
Pounds N = 0.225lb
Pound is defined as F = ma = slug x ft/s2

14. Newton’s Second Law Definition of Mass (physics)
A measure of resistance a body offers to changes in its velocity (acceleration)
Standard is kilogram
Masses can be compared by balances
Mass vs. Weight

15. Newton’s 2nd Law proves that different masses accelerate to the earth at the same rate, but with different forces.
We know that objects with different masses accelerate to the ground at the same rate.
However, because of the 2nd Law we know that they don’t hit the ground with the same force.
F = ma
98 N = 10 kg x 9.8 m/s/s
F = ma
9.8 N = 1 kg x 9.8 m/s/s

16. Newton’s Second Law (cont.)
Example
A mass of 0.2kg slides along the table with a velocity of v = 2.8m/s. It stops in 1.0 m. What force is acting on the mass (neglect friction)?

17. Newton’s Third Law Newton’s Third Law
“If object 1 and object 2 interact, the force F12 exerted by object 1 on object 2 is equal in magnitude but opposite to the force F21 exerted by object 2 on object 1.”
What it means:
“for every action, there is an equal and opposite reaction”

18. Newton’s Third Law F21 = -F12 Action-Reaction Pair
Forces in nature always act in pairs
No single isolated force
The mutual actions of two bodies upon each other are ALWAYS equal and directed contrary to one another

19. Newton’s Third Law Action-Reaction Pairs
Newton’s Law uses the forces acting on an object
n and F are both acting on the object
n is referred as to the normal force and is the force exerted by the TV stand on the TV

20. Applications of Newton’s Laws
An object in equilibrium has no net external force acting on it, and the second law, in component form, implies that  
ΣFx = 0
and  
ΣFy = 0
for such an object. These two equations are useful for solving problems where the object is at rest or moving at constant velocity.

21. Applications of Newton’s Laws
An object under acceleration requires the same two equations, but with the acceleration terms included:  
ΣFx = max
and  
ΣFy = may

22. Applications of Newton’s Laws
Assumptions
Objects behave as particles
can ignore rotational motion (for now)
Masses of strings or ropes are negligible
No stretching – constant length
Interested only in the forces acting on the object
can neglect reaction forces
Pulleys are massless and frictionless
Used to change direction

23. Applications of Newton’s Laws
Free Body Diagram
Diagram representing all the forces applied to an object
Represent the object as a dot
Identify all the forces acting on the object, not exerted
Choose appropriate coordinate system
Incorrect FBD means incorrect solution

24. Applications of Newton’s Laws
The force T is the tension acting on the box
n and F are the forces exerted by the earth and the ground
Only forces acting directly on the object are included in the free body diagram
Reaction forces act on other objects and so are not included

25. Newton’s Third Law n Free Body Diagram Fg ΣFy = N – Fg = n – mg = may
but the TV is not moving, so ay = 0
n – mg = 0
Therefore,
n = mg Fg

26. Solving Newton’s Second Law Problems
Read the problem at least once
Draw a picture of the system
Identify the object of primary interest
Indicate forces with arrows
Label each force
Use labels that bring to mind the physical quantity involved
Draw a free body diagram
If additional objects are involved, draw separate free body diagrams for each object
Choose a convenient coordinate system for each object
Apply Newton’s Second Law
The x- and y-components should be taken from the vector equation and written separately
Solve for the unknown(s)

27. ConcepTest 4.11 On an Incline
Consider two identical blocks, one resting on a flat surface, and the other resting on an incline. For which case is the normal force greater?
1) case A
2) case B
3) both the same (N = mg)
4) both the same (0 < N < mg)
5) both the same (N = 0) A B
[CORRECT 5 ANSWER]

28. ConcepTest 4.11 On an Incline
Consider two identical blocks, one resting on a flat surface, and the other resting on an incline. For which case is the normal force greater?
1) case A
2) case B
3) both the same (N = mg)
4) both the same (0 < N < mg)
5) both the same (N = 0)
In Case A, we know that N = W. In Case B, due to the angle of the incline, N < W. In fact, we can see that N = W cos(q). y x N f A B q Wy W q

29. Newton’s Law review 1st – Law of Inertia 2nd – F = ma
Objects at rest or motion will stay that way unless acted on by a force
Sailboat
Target
2nd – F = ma
The acceleration is proportional to the force and inversely proportional to the mass
3rd – Action-Reaction
if a force is acted on to an object, the object will exert an equal and opposite force
Recoil

30. ConcepTest 4.1c Newton’s First Law
You put your book on the bus seat next to you. When the bus stops suddenly, the book slides forward off the seat. Why?
1) a net force acted on it
2) no net force acted on it
3) it remained at rest
4) it did not move, but only seemed to
5) gravity briefly stopped acting on it 30

31. ConcepTest 4.1c Newton’s First Law
You put your book on the bus seat next to you. When the bus stops suddenly, the book slides forward off the seat. Why?
1) a net force acted on it
2) no net force acted on it
3) it remained at rest
4) it did not move, but only seemed to
5) gravity briefly stopped acting on it
The book was initially moving forward (since it was on a moving bus). When the bus stopped, the book continued moving forward, which was its initial state of motion, and therefore it slid forward off the seat.
Follow-up: What is the force that usually keeps the book on the seat? 31

32. ConcepTest 4.9b Going Up II
A block of mass m rests on the floor of an elevator that is accelerating upward. What is the relationship between the force due to gravity and the normal force on the block?
1) N > mg
2) N = mg
3) N < mg (but not zero)
4) N = 0
5) depends on the size of the elevator m a
[CORRECT 5 ANSWER] 32

33. ConcepTest 4.9b Going Up II
A block of mass m rests on the floor of an elevator that is accelerating upward. What is the relationship between the force due to gravity and the normal force on the block?
1) N > mg
2) N = mg
3) N < mg (but not zero)
4) N = 0
5) depends on the size of the elevator
The block is accelerating upward, so it must have a net upward force. The forces on it are N (up) and mg (down), so N must be greater than mg in order to give the net upward force! m
a > 0 mg N
S F = N – mg = ma > 0
\ N > mg
Follow-up: What is the normal force if the elevator is in free fall downward? 33

34. Applications of Newton’s Laws
Equilibrium
An object either at rest or moving with a constant velocity is said to be in equilibrium
The net force acting on the object is zero (since the acceleration is zero)
ΣF = 0
Should use components
ΣFx = 0 ΣFy = 0

35. Applications of Newton’s Laws
Equilibrium Example - FBD

36. Applications of Newton’s Laws
Equilibrium Example – FBD
Choose the coordinate system with x along the incline and y perpendicular to the incline
Replace the force of gravity with its components

37. Applications of Newton’s Laws
Equilibrium Example?

38. Applications of Newton’s Laws
Equilibrium – Multiple Objects
When you have more than one object, the problem-solving strategy is applied to each object
Draw free body diagrams for each object
Apply Newton’s Laws to each object
Solve the equations

39. Applications of Newton’s Laws
Example
A traffic light weighing 1.00x102 N hangs from a vertical cable tied to two other cables that are fastened to a support, as in the figure to the right. The upper cables makes angles of 37° and 53° with the horizontal. Find the tension in each of the three cables.

40. Applications of Newton’s Laws
Example:
An object with a mass m1 = 5.00kg rests on a frictionless horizontal table and is connected to a cable that passes over a pulley and is then fastened to a hanging object with mass m2 = 10.0 kg, as shown in Figure P Find the acceleration of each object and the tension in the table.

41. Forces of Friction
An object moving on a surface encounters resistance as it interacts through its surroundings. This resistance is called friction.
Friction is a force
The force of friction is opposite of motion
Two types of friction – static and kinetic
Friction is proportional to the normal force
Friction Examples
Car on the road

42. Forces of Friction Static Friction, ƒs
Static friction acts to keep the object from moving
If F increases, so does ƒs
If F decreases, so does ƒs
ƒs  µ n

43. Forces of Friction Kinetic Friction, ƒk
The force of kinetic friction acts when the object is in motion
ƒk = µ n
Variations of the coefficient with speed will be ignored

44. Forces of Friction
Friction Examples

45. Forces of Friction Friction Example Friction acts up the plane
Axes are rotated as usual on an incline
The direction of impending motion would be down the plane
Friction acts up the plane
Opposes the motion
Apply Newton’s Laws and solve equations

46. Static Friction...
We want to know how it acts in fixed or “static” systems: the force provided by friction depends on the forces applied on the system (magnitude: fs ≤ ms N)
Opposes motion that would occur if ms were zero y N
Fapplied x fS mg

47. Static Friction...
If a = 0.
x : Fapplied fS = 0
y: N = mg
While the block is static: fS Fapplied (unlike kinetic friction) y N
Fapplied x fS mg

48. Static Friction...
The maximum possible force that the friction between two objects can provide is fMAX = SN, where s is the “coefficient of static friction”.
So fS  S N.
As one increases F, fS gets bigger until fS = SN and the object “breaks loose” and starts to move. y N F x fS mg

49. Static Friction...
S is discovered by increasing F until the block starts to slide:
x : FMAX SN = 0
y : N = mg
S FMAX / mg
FMAX mg N x y
Smg

50. Additional comments on Friction:
The force of friction does not depend on the area of the surfaces in contact (a relatively good approximation if there is little surface deformation)
Generally S > K for any system

51. Kinetic Friction Dynamics: x-axis max = F  KN
y-axis may = 0 = N – mg or N = mg
so F Kmg = m ax fk v y N F x
max fk
K mg mg

52. Frictionless inclined plane
A block of mass m slides down a frictionless ramp that makes angle  with respect to horizontal. What is its acceleration a ? m a 

53. Angles of the inclined plane
max = mg sin   mg N 
 + f = 90 f 

54. Case 1 - Frictionless inclined plane...
Use a FBD and consider x and y components separately:
Fx max = mg sin  ax = g sin 
Fy may = 0 = N – mgcos  N = mg cos  N
max x y
mg cos  mg 
mg sin  q

55. Case 2 - Inclined plane...static friction
Use a FBD and consider x and y components separately:\
Fx max = 0 = mgsin  - f = g sin  - f
Fy may = 0 = N – mgcos  N = mg cos  N
Friction
Force x y
Special case:
At the breaking point
f = ms N = ms mg cos 
g sin q = f = ms mg cos 
max= 0 mg
mg cos  
mg sin  q

56. Example
A roller coaster reaches the top of the steepest hill with a speed of 6.0 km/hr. It then descends the hill, which is at an average of 45º and is 45.0m. What will its speed be at the bottom? Assume µk = 0.18.

57. Air Resistance and Drag
So far we’ve “neglected air resistance” in physics
Can be difficult to deal with
Affects projectile motion
Friction force opposes velocity through medium
Imposes horizontal force, additional vertical forces
Terminal velocity for falling objects
Dominant energy drain on cars, bicyclists, planes

58. Drag Force Quantified
With a cross sectional area, A (in m2), coefficient of drag of 1.0 (most objects),  sea-level density of air, and velocity, v (m/s), the drag force is:
D = ½ C  A v2  c A v2 in Newtons
c = ¼ kg/m3
Increases as v increases
In falling, when D = mg, then at terminal velocity
Example: Bicycling at 10 m/s (22 m.p.h.), with projected area of 0.5 m2 exerts ~30 Newtons
Requires (F v) of power  300 Watts to maintain speed
Minimizing drag is often important

59. “Free” Fall D = ¼Av2 ≈ mg v ≈ √4mg/A Terminal velocity reached when
Fdrag = Fgrav (= mg)
D = ¼Av2 ≈ mg
v ≈ √4mg/A
For 75 kg person with a frontal area of 0.5 m2,
vterm  50 m/s, or 110 mph
which is reached in about 5 seconds, over 125 m of fall

60. Trajectories with Air Resistance
Baseball launched at 45° with v = 50 m/s:
Without air resistance, reaches about 63 m high, 254 m range
With air resistance, about 31 m high, 122 m range
Vacuum trajectory vs. air trajectory for 45° launch angle.

61. ‘Free’ Fall
Indoor sky-diving