Newton’s Laws of Motion
• Newton’s Laws
• Forces
• Mass and Weight
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Newton’s First Law (Law of Inertia)
An isolated object, free from external forces, will continue
moving at constant velocity, or remain at rest.
Earlier, Aristotle said objects were “naturally” at rest, and needed a
continuing push to keep moving.
Galileo realized that motion at constant velocity is “natural”, and
only changes in velocity require external causes.
Objects in equilibrium (no net external force) also move
at constant velocity.
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Forces
• A force is a push or pull that tends to cause motion
(more exactly, changes in motion)
• From the Second Law, force should have units of
1 kg  m/s2  1 newton (N)
• Force is a vector
• In Newton’s dynamics, all influences on a particle
from its surroundings are expressed as forces
exerted on that particle
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Newton’s Second Law


Fnet  ma
Fnet (or Ftotal) is the vector sum of all forces acting on the
particle of mass m:



Fnet   Fi   mai
i
i
The acceleration a is parallel to the total force, and proportional
to it. The proportionality constant is the particle’s mass.
Newton defines mass as a measure of an object’s inertia.
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Contact Forces : direct contact is required
examples - normal forces, friction, air resistance,
buoyancy, ...
Non-Contact Forces :
gravity, electromagnetic, weak and strong forces
The gravitational force is also called weight and is
measured in Newtons.
Weight is proportional to mass : Fw = mg, where g is the
gravitational field (and is also the acceleration of an object
in free fall).
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Weight and Mass
Weight is a force; it can be measured using a spring scale
On Earth, a baseball
weighs 2.40 N
On the moon, a
baseball weighs 0.40 N
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•
Mass is a measure of inertia : on the earth or on the moon, a
24.5 N force applied to the baseball will give it an acceleration of
100 m/s2 (its mass is m = F/a = 0.245 kg)
•
We can compare masses with a balance, because of the
remarkable property :
weight  mass
Weights are equal
when masses are
equal

Fg ,1

Fg ,2
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Newton’s Third Law (action and reaction)
If object A exerts a force on object B, object B
exerts an equal, opposite force back on A.

Fg
Block pushes down on table
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Newton’s Third Law : examples
What is the “reaction” to the following forces?

 Fg
You push on a block
Gravity (of block)
pulls earth up
Balloon pushes on
air outside
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Contact Forces
Examples : A heavy block on a table
• The table must push up on the
block to prevent it from falling
Forces on Block

Fg
• The type of contact force is
called a normal force if it is
perpendicular (normal) to the
surfaces in contact.
• The normal force will be as large
as necessary to hold the block
(until the table breaks)

Ftable
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If we look closely, the normal force
arises from the table being bent :
as the table tries to straighten, it
pushes back.
This is really an elastic force; the table behaves like a
spring.
At the atomic level, all contact forces are due to
electromagnetic forces.
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Quiz
A 140-kg wrestler and a 90-kg wrestler try to push each
other backwards out of the ring.
At first they are
motionless as they push; then the large wrestler moves
the other one backwards. Compare the forces they exert
on each other. Which statement is correct?
a) The forces are always equal.
b) The larger wrestler always exerts a larger force.
c) When they are motionless, the forces are equal; they
start to move when the large wrestler exerts a larger
force on his opponent than his opponent exerts back
on him.
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Quiz
You wash your hands, and as you don’t have a towel
handy, you shake them to get rid of the water. You are
able to shake water off of your hands mainly due to:
a) Newton’s First Law
b) Newton’s Second Law
c) Newton’s Third Law
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10 min rest
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Free-Body Diagrams
• Pick one object (the “body”).
• Draw all external forces which act directly on that body
(gravity, contact, electromagnetic).
Imagine cutting around the body to separate it from its surroundings.
Replace each external object with a force applied at the point of
contact.
• Indicate the direction of the acceleration of the object
beside the diagram; but remember, ma is not a force
on the diagram.
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Example of a free-body diagram
m
A block is pulled up a frictionless ramp:
Note :
Forces on Block

a

FT
• title, to indicate the chosen
object (use m or mA etc)
• contact forces, to replace the
rope and the ramp
• gravity doesn’t require contact

mg

n
• a may be indicated for
reference, but is not a force
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Ropes
• A rope attached to something
exerts a force parallel to the rope
• The magnitude of the force is
called the tension in the rope
rope
Force
• Tension is uniform in a rope of negligible mass
• The tension is not changed if the rope passes over
an ideal pulley (assume frictionless and massless)
• Tension has units of force (newtons)
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Moving a Block
!@%&*#$

Fg

FA
Divide the contact force from
the table into two components:

f

n

- normal force : n is perpendicular to the surfaces
in contact

- friction : f is parallel to the surface
friction has a more complex behaviour than
the normal force (next lecture)
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Friction
Friction is the force which resists sliding of two surfaces across
each other.
We distinguish between static and kinetic friction:
Static Friction :
- there is NO relative motion
- fs prevents sliding
Kinetic Friction :
- the block is sliding
- fk is opposite to v
FA

F  0

fs
v
fk
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Friction is complicated. A useful empirical model was
presented by Charles Coulomb in 1781:
1.
The force of static friction has a maximum value; if you
push too hard, the block moves. This maximum value is
proportional to the normal force the surfaces exert on each
other.
2.
Once the object is sliding, kinetic friction is approximately
independent of velocity, and usually smaller than the
maximum static friction force. The force of kinetic friction is
also proportional to the normal force.
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Define two pure numbers (no units):
s (“coefficient of static friction”)
k (“coefficient of kinetic friction”)
(“” is a Greek letter, pronounced “mu”)
Then Coulomb’s rules are:

FA

mg

f

n
f s  s n
f k  k n
Question : would
 itbe correct to write these as vector
equations, f  n ?
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Copper on steel
Aluminum on aluminum
Teflon on Teflon
s
k
0.53
1.5
0.04
0.36
1.1
0.04
• Values depend on smoothness, temperature, etc. and
are approximate
• Usually  < 1, but not always
• Usually, k is less than s , and never larger
• The coefficients depend on the materials, but not on
the surface areas, contact pressure, etc.
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Quiz
A block of mass 10kg is resting on a surface with a
coefficient of static friction μs=0.50. What minimum
force F is needed to move the block?
a) 10 N
b) 49 N
c) 98 N
F
10 kg
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Quiz
A block of mass 10kg is resting on a surface with a
coefficient of static friction μs=0.50.
Once the block is moving, what force is needed to
accelerate it?
a) less than 49 N
b) 49 N
c) more than 49 N
F
10 kg
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Quiz
Each block weighs 100 N, and the coefficient of
static friction between each pair of surfaces is 0.50.
What minimum force F is needed to pull the lower
block out?
a) 50 N
b) 100 N
c) 150 N
F
100 N

100 N
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10 min rest
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Newton’s Laws (III)
• Blocks, ramps, pulleys and other problems
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Equilibrium

• A special case : a  0 (object doesn’t move, or moves
at constant velocity)


• Newton’s second law gives  F  ma  0
The vector sum of forces acting
on a body in equilibrium is zero
• This is equivalent to three independent component
equations:  Fx  0,  Fy  0,  Fz  0
• We can solve for 3 unknowns (or 2, in 2-D problems)
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Remember, when doing problems with “F=ma”
• Draw the free-body diagram carefully.
• You may need to know the direction of a from
kinematics, before considering forces (for friction).
• Any axes will do, but some choices make the algebra
simpler – set up equations for each direction.
• You need one (scalar) equation for each (scalar)
unknown, in general (the mass will often cancel out).
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Block on a ramp
Determine all the forces acting on this block.
Given m, θ and μk, what would the acceleration be:
a) without friction
b) with friction
m
θ
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a)
b)
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Example
A block is in equilibrium on a
frictionless ramp. What is the
tension in the rope?
T
m
f
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Quiz
The block has weight mg and is in equilibrium on
the ramp. If s = 0.9, what is the frictional force?
A)
B)
C)
D)
0.90 mg
0.72 mg
0.60 mg
0.54 mg
37o
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Example
Obtain an expression for the stopping distance for a skier
moving down a slope with friction with an initial speed of v0.
d
θ
Find the distance given that μk=0.18, v=20m/s and θ=5.0º.
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Accelerated motion
Example: A block is pushed with a force FA at an angle to
the horizontal, find the acceleration. Friction is
given by μk.
FA
θ
m
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Question : Can we calculate μs ?
mg
f

n
Increasing  so that    max ,
the block slips, from which we
get:
 s  tan  max
This is an easy method of measuring s
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