1. Chapter 4
Dynamics: Newton’s Laws
of Motion

2. Units of Chapter 4
• Force
• Newton’s First Law of Motion
• Mass
• Newton’s Second Law of Motion
• Newton’s Third Law of Motion
• Weight – the Force of Gravity; and the
Normal Force

3. Units of Chapter 4
• Solving Problems with Newton’s Laws:
Free-Body Diagrams
• Applications Involving Friction, Inclines
• Problem Solving – A General Approach

4. 4-1 Force
A force is a push or pull. An
object at rest needs a force
to get it moving; a moving
object needs a force to
change its velocity.
The magnitude of a
force can be measured
using a spring scale.

5. 4-2 Newton’s First Law of Motion
Newton’s first law is often called the law of
inertia.
Every object continues in its state of rest, or of
uniform velocity in a straight line, as long as no
net force acts on it.

6. 4-2 Newton’s First Law of Motion
Inertial reference frames:
An inertial reference frame is one in which
Newton’s first law is valid.
This excludes rotating and accelerating frames.

7. 4-3 Mass
Mass is the measure of inertia of an object. In
the SI system, mass is measured in kilograms.
Inertia ≡ The tendency of a body to maintain its
state of rest or motion.
Mass is not weight:
Mass is a property of an object. Weight is the
force exerted on that object by gravity.
If you go to the moon, whose gravitational
acceleration is about 1/6 g, you will weigh much
less. Your mass, however, will be the same.

8. 4-4 Newton’s Second Law of Motion
• 1st Law: If no net force acts, object remains at rest or
in uniform motion in straight line.
• What if a net force acts? Do Experiments.
• Find, if the net force F ≠ 0 ⇒ The velocity v changes
(in magnitude or direction or both).
• A change in the velocity v (Δv)
⇒ There is an acceleration a = (Δv/Δt)
OR
A net force acting on a body produces an
acceleration! F⇒ a

9. 4-4 Newton’s Second Law of Motion
• From experiments: The net force F on a body
and the acceleration a of that body are related.
• HOW? Answer by EXPERIMENTS!
– The outcome of thousands of experiments over
hundreds of years:
a ∝ F/m (proportionality)
• We choose the units of force so that this is not
just a proportionality but an equation:
a ≡ F/m
OR: (total!)→ F = ma

10. 4-4 Newton’s Second Law of Motion
• Newton’s 2nd Law: F = ma
F = the net (TOTAL!) force acting on mass m
m = the mass (inertia) of the object.
a = acceleration of object. Description of the effect of F
F is the cause of a.
• To emphasize that the F in Newton’s 2nd Law is
the TOTAL (net) force on the mass m, text writes:
∑F = ma
∑ = a math symbol meaning sum (capital sigma)

11. 4-4 Newton’s Second Law of Motion
• Newton’s 2nd Law: Based on experiment!
Not derivable
∑F = ma mathematically!!
Force is a vector, so ∑F = ma is true along each coordinate
axis.
∑Fx = max ∑Fy = may ∑Fz = maz
ONE OF THE MOST
FUNDAMENTAL & IMPORTANT
LAWS OF CLASSICAL PHYSICS!!!

12. 4-4 Newton’s Second Law of Motion
Newton’s second law is the relation between
acceleration and force. Acceleration is
proportional to NET force and inversely
proportional to mass.
(4-1)

13. 4-4 Newton’s Second Law of Motion
The unit of force in the SI
system is the newton (N).
Note that the pound is a
unit of force, not of mass,
and can therefore be
equated to newtons but
not to kilograms.

14. 4-5 Newton’s Third Law of Motion
Any time a force is exerted on an object, that
force is caused by another object.
Newton’s third law:
Whenever one object exerts a force on a second
object, the second exerts an equal force in the
opposite direction on the first.
Law of action-reaction: “Every
action has an equal & opposite
reaction”.
(Action-reaction forces act on
DIFFERENT objects!)

15. 4-5 Newton’s Third Law of Motion
A key to the correct
application of the third
law is that the forces
are exerted on different
objects. Make sure you
don’t use them as if
they were acting on the
same object.

16. 4-5 Newton’s Third Law of Motion
Rocket propulsion can also be explained using
Newton’s third law: hot gases from combustion
spew out of the tail of the rocket at high speeds.
The reaction force is what propels the rocket.
Note that the
rocket does not
need anything to
“push” against.

17. 4-5 Newton’s Third Law of Motion
Helpful notation: the first subscript is the object
that the force is being exerted on; the second is
the source.
This need not be
done indefinitely, but
is a good idea until
you get used to
dealing with these (4-2)
forces.

18. Example 4-5

19. 4-6 Weight – the Force of Gravity;
and the Normal Force
Weight is the force exerted on
an object by gravity.
Close to the surface of the
Earth, where the gravitational
force is nearly constant, the
weight is:

20. 4-6 Weight – the Force of Gravity;
and the Normal Force
An object at rest must have no net force on it. If
it is sitting on a table, the force of gravity is still
there; what other force is there?
The force exerted perpendicular to a surface is
called the normal force. It is
exactly as large as needed
to balance the force from
the object (if the required
force gets too big,
something breaks!)

21. Example 4-6
m = 10 kg
The normal
force is NOT
always equal
to the weight!!

22. Example 4-7
m = 10 kg
∑F = ma
FP – mg = ma

23. 4-7 Solving Problems with Newton’s Laws –
Free-Body Diagrams
1. Draw a sketch.
2. For one object, draw a free-body
diagram (force diagram), showing all
the forces acting on the object. Make
the magnitudes and directions as
accurate as you can. Label each force.
If there are multiple objects, draw a
separate diagram for each one.
3. Resolve vectors into components.
4. Apply Newton’s second law to each
component.
5. Solve.

24. 4-7 Solving Problems with Newton’s Laws –
Free-Body Diagrams
When a cord or rope pulls on
an object, it is said to be under
tension, and the force it exerts
is called a tension force.

25. Example 4-11
Fpx = (40.0 N )(cos 30.00 ) = 34.6 N
Fpy = (40.0 N )(sin 30.00 ) = 20.0 N
Fpx
34.6 N
Fpx = ma x ⇒ ax = =
m 10.0 Kg
= 3.46 m / s 2
ΣFy = ma y ⇒ FN + Fpy − mg = ma y
FN + 20.0 − 98.0 = 0 ; a y = 0
FN = 78.0 N

26. Example 4-13
FT − mE g = mE aE = − mE a (1)
FT − mC g = mC aC = + mC a ( 2)
subtract (1) from (2)
(mE − mC ) g = (mE + mC )a
mE − mC
a= g = 0.68 m / s 2
mE + mC
FT = mE g − mE a = mE ( g − a) = 10500 N

27. Example 4-14
The advantage of a pulley
2 FT − mg = ma
const. speed ⇒ a = 0
mg
FT =
2

28. Example 4-15
Σ Fx = FRC cos θ − FRB cos θ = 0 ; a = 0
⇒ FRB = FRC
Σ Fy = Fp − 2 FT sin θ = 0
Fp
FT = = 1700 N
2 sin θ

29. ↑ FT1
Problem 25
Take up as positive!
m1 = m2 = 3.2 kg
↑a m1g = m2g = 31.4 N
m1g ↓ ↓ FT2 Bucket 1: FT1 - FT2 - m1g = m1a
↑ FT2 Bucket 2: FT2 -m2g = m2a
↑a
↓ m2 g
∑F = ma (y direction), for EACH bucket separately!!!

30. Problem 29
Take up positive!
m = 65 kg
mg = 637 N
FT ↑ ↑ FT
FT + FT - mg = ma
2FT -mg = ma
↓ FP FP = - FT (3rd Law!)
↑a
↓ mg
∑F = ma (y direction) on woman + bucket!

31. 4-8 Applications Involving Friction, Inclines
On a microscopic scale,
most surfaces are rough.
The exact details are not
yet known, but the force
can be modeled in a
simple way.
We must account for
Friction to be realistic!

32. 4-8 Applications Involving Friction, Inclines
– Exists between any 2 sliding surfaces.
– Two types of friction:
Static (no motion) friction
Kinetic (motion) friction
– The size of the friction force: Depends on the
microscopic details of 2 sliding surfaces.
• The materials they are made of
• Are the surfaces smooth or rough?
• Are they wet or dry?
• Etc., etc., etc.

33. • Kinetic Friction: Experiments determine
the relation used to compute friction forces.
• Friction force Ffr is proportional to the
magnitude of the normal force FN between
two sliding surfaces.
DIRECTIONS of Ffr & FN are ⊥ each other!!
Ffr ⊥ FN
FN
a
Ffr Fa
mg

34. 4-8 Applications Involving Friction, Inclines
For kinetic (sliding) friction, we write:
µ k is the coefficient of kinetic friction, and is
different for every pair of surfaces.
µ k depends on the surfaces & their conditions
µ k is dimensionless & < 1

35. 4-8 Applications Involving Friction, Inclines
Static friction is the frictional force between two
surfaces that are not moving along each other.
Static friction keeps objects on inclines from
sliding, and keeps objects from moving when a
force is first applied.

36. • Static Friction: Experiments are used again.
• The friction force Ffr exists || two surfaces, even if
there is no motion. Consider the applied force Fa:
FN
The object is not
Ffr Fa moving
mg
∑F = ma = 0 & also v = 0
⇒ There must be a friction force Ffr to oppose Fa
Fa – Ffr = 0 or Ffr = Fa

37. • Experiments find that the maximum static friction
force Ffr (max) is proportional to the magnitude (size)
of the normal force FN between the two surfaces.
DIRECTIONS of Ffr & FN are ⊥ each other!! Ffr ⊥ FN
• Write the relation as Ffr (max) = µ sFN
µ s ≡ Coefficient of static friction
– Depends on the surfaces & their conditions
– Dimensionless & < 1
– Always find µ s > µ k
⇒ Static friction force: Ffr ≤ µ sFN

38. 4-8 Applications Involving Friction, Inclines

39. 4-8 Applications Involving Friction, Inclines
The static frictional force increases as the applied
force increases, until it reaches its maximum.
Then the object starts to move, and the kinetic
frictional force takes over.

40. Example 4-16

41. Example 4-18

42. Example 4-19

43. FN = m A g = 5.0 × 9.8 = 49 N
F fr = µ k FN = 0.20 × 49 = 9.8 N
Σ FBy = mB g − FT = mB a Example
Σ FAx = FT − F fr = m A a 4-20
⇒ FT = F fr + m A a
⇒ mB g − F fr − m A a = mB a
mB g − F fr 19.6 − 9.8
a= = = 1.4 m / s 2
m A + mB 5.0 + 2.0
FT = F fr + mA a = 17 N

44. 4-8 Applications Involving Friction, Inclines
An object sliding down an incline has three forces acting
on it: the normal force, gravity, and the frictional force.
• The normal force is always perpendicular to the surface.
• The friction force is parallel to it.
• The gravitational force points down.
If the object is at rest,
the forces are the same
except that we use the
static frictional force,
and the sum of the
forces is zero.

45. 4-9 Problem Solving – A General Approach
1. Read the problem carefully; then read it again.
2. Draw a sketch, and then a free-body diagram.
3. Choose a convenient coordinate system.
4. List the known and unknown quantities; find
relationships between the knowns and the
unknowns.
5. Estimate the answer.
6. Solve the problem without putting in any numbers
(algebraically); once you are satisfied, put the
numbers in.
7. Keep track of dimensions.
8. Make sure your answer is reasonable.

46. Example 4-21
FGx = mg sin θ
FGy = − mg cos θ
Σ Fy = FN − mg cos θ = 0 ⇒ FN = mg cos θ
Σ Fx = mg sin θ − µ k FN = max ⇒ mg sin θ − µ k (mg cos θ ) = ma x

47. Problem 48
FN
a
Ffr Fp
(m1+m2) g
FN = ( m1 + m2 ) g → Ffr = µ k FN = µ k ( m1 + m2 ) g
∑ Fx = FP − Ffr = ( m1 + m2 ) a → a = 1.9 m/s2
For contact forces analyze forces on the two blocks separately.

48. Summary of Chapter 4
• Newton’s first law: If the net force on an object
is zero, it will remain either at rest or moving in a
straight line at constant speed.
• Newton’s second law:
• Newton’s third law:
• Weight is the gravitational force on an object.
• The frictional force can be written:
(kinetic friction) or (static friction)
• Free-body diagrams are essential for problem-
solving