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Newton’s Laws of Motion
• Forces
• Force of Gravity (Weight)
• Some WRONG  views about Motion
• Simple Problems
• Galileo’s “Thought Experiments”
• Wagon/Lawnmower-Style Problem
• Inertia
• Newton’s 3rd Law
• Newton’s 1st Law
• The Normal Force
• Newton’s 2nd Law
• Frictional Forces
• Free Body Diagrams (FBDs)
• Block-Pushing-Block Problems
• The 2 Viewpoints
• Train-Style Problems
• View-from-Above Problems
What is a “Force”?
A force is ANYTHING that can make an object accelerate
(speed up, slow down, or change direction)
The SI unit for force is the NEWTON (N). This unit is a
derived unit, meaning it’s made up of base units.
m
kg
sec
m
N = kg 2
s
kg
m
sec
sec
The English unit for force is the POUND (lb).

Contact forces
1 lb = 4.44 N
(pushes & pulls)
Two types of forces
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Field forces
(no contact … gravity
is an example)
What did people long ago believe about motion?
Aristotle, the famous Greek philosopher (~350 BC), studied
the motion of objects and concluded the following:
“It requires a continuous pushing or pulling” to keep
an object such as a rolling stone moving. When the
pushing or pulling is no longer applied, the stone
comes to rest……A FORCE is required to produce a
constant velocity” (Heath, pg. 128)
Sounds good right???
He thought that …
NO!!!!!
Aristotle was wrong 
• With increased force  objects move faster.
• With decreased force  objects move slower.
• With No force  objects will stop.
HOME
Common
sense.
Right?

What did Galileo think about motion?
Galileo Galilei, the great Italian physicist who lived during the
Renaissance, used two different “thought experiments” about
the motion of an object on an inclined plane to explain
motion.
Experiment #1: Galileo imagined a ball rolling down a sloped
plane. He figured that it would speed up. He then imagined
a ball rolling up a slope. He figured that it would slow down.
He reasoned that a ball rolled across a horizontal surface
would neither speed up nor slow down, but rather continue to
move with a constant velocity.
Speed
s up
Slows
Down
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Wants to go at a
constant speed
Experiment #2: Galileo again imagined a ball rolling down a sloped plane.
However, this time, he allowed the ball to roll up a plane afterward. He
reasoned that no matter what the slope of the two planes, the ball would
always attain the same height (the height it was rolled down from equals the
height it rolls up to). Therefore, he concluded that if no“up” plane were
present at the bottom of the “down” plane, the ball would roll on FOREVER,
trying to but never reaching the height from which it were dropped.
Tries to get to
this height again
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Should roll on
forever “trying”
to get back to
original height
Both of Galileo’s thought experiments contradicted his observations of
the same events in real-life. However, he attributed the differences to
“Resistance”, or what we today call FRICTION. Both thought
experiments occurred on a frictionless surface. Galileo argued
that it
was just as ‘natural’ for an object to move
with a constant speed as it is to be at rest. This
contradicted Aristotle’s view. Galileo Published these thoughts about
motion in the early 17th century, and his contemporaries immediately
dropped the Aristotelian view and embraced Galileo’s views of
motion.
AT REST
HOME
MOVING AT A
CONSTANT VELOCITY
The TWO (2) natural
states that objects
want to be in
(according to
Galileo)
Inertia
Galileo noticed that objects had a “natural tendency” to
either stay at rest OR keep moving. This “tendency” to
resist a CHANGE in motion is called INERTIA. To measure
an objects “inertia”, we measure its MASS. Therefore,
like mass, Inertia is a property of an object,
Examples:
• A large rock being difficult to “budge” from rest.
• A heavy grocery cart is difficult to slow down once it is
moving down the grocery aisle.
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• When you are driving and you come to a sudden stop,
your body WANTS to go flying through the windshield.
Mass vs Weight
Mass
• The amount of matter in an
object
• Mass does not change
based on location (moon vs
earth)
• Mass is measured in units of
Kg in the metric system
• Mass is measured with an
instrument called a balance
Weight
• The downward force of
gravity on that mass “Fg”
(weight)
• Weight varies based on
location (moon vs earth)
• Weight is measured in units
of….Newtons in the metric
system (Pounds in English)
• Weights is measured with
an instrument called a scale
What is the difference between a
balance and a scale?
Mass
• A balance has a fulcrum,
and you need the same
amount of “stuff” on each
side to balance
Weight
• A scale has something
inside that stretches or
compresses under the force
of gravity.
• Weight is calculated by
factoring gravity into the
mass of the object.
Multiply mass by gravity.
Practice Questions
• How much does a 6 kg object weigh on the
surface of the earth?
– In the metric system?
– In the English system?
• An object weighs 88N on the earth’s surface,
what is it’s mass?
• An alien with a mass of 45 Kg weighs 700N on
his home planet. What is the gravitational
field strength on the surface of his planet?
• If a person weighs 150 lbs, what is their
weight in the metric system?
Newton’s
st
1
Law
Isaac Newton, an Englishman who lived later in the 17th century,
began his theories of motion by looking at a concept that he called
“Inertia”. It can be thought of as ‘object laziness’. Objects tend to
keep doing what they are doing. It takes force to make an object
start moving or change direction. The more massive an object is,
the larger the force that is required for a given change.” (Holt, teacher’s
addition) Galileo was the first person to formalize this concept.
However, Newton used it to develop his famous “Laws of Motion”,
which he formally published in his book Principia Mathematica.
This book is widely considered to be the greatest scientific work
ever published. Newton’s “First Law” of motion, otherwise known
as “Newton’s Law of Inertia”, informally states that …
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Newton’s
st
1
Law
Sometimes called the
“Law of Inertia”
An object in motion (or at rest) will remain
in motion (or at rest) unless acted upon by
an unbalanced, external force.
Some
Notes “in motion” means “moving at a constant velocity”.
If all the forces acting on an object are BALANCED,
then the object will NOT change its current motion
state.
Only external forces can cause a change in motion.
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Newton’s
st
1
Law … continued
But if this is in fact a “Law” of motion, why don’t we see this happen in
the “real world”. When an object is pushed, it slows down, even when
no other forces act upon it. Or so it would seem. FRICTION is the
“invisible” force that acts on objects to slow them down. However, in
a friction-free case (like in outer space), Newton’s 1st Law will be
“observed” to hold true. And, it holds true in all cases, we simply have
a hard time “observing” it because of friction.
Examples:
• No seatbelt … going through the windshield in a car accident.
• WITH seatbelt … NOT going through the window.
• A magician pulling the tablecloth off but leaving the place-setting.
• Getting a car to stop rolling once it’s started.
• Punching a light “speed bag” vs. punching a “heavy bag”.
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In layman’s terms, please 
Newton’s
st
1
Law
Object’s “naturally” like to be either at rest or
moving with a constant velocity. In order to make
them do anything else (which obviously involves an
acceleration), “something” needs to force them out
of their natural motion state. This “something” is an
unbalanced force that must be applied to the object
from the outside of the object. Forces that occur
inside the object will NOT alter the object’s motion.
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Examples:
Object at rest
The object stays at rest because the
50 N
50 N
Object at rest
50 N
20 N
Object at rest
forces acting on it are BALANCED.
The object will accelerate (to the right)
because the NET force acting on it is not zero
(ie. The forces are UNBALANCED).
The object will accelerate (this way
)
because the NET force acting on it is not zero
(ie. The forces are UNBALANCED).
50 N
50 N
40 N
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Bottom line … whichever
direction the net force is
acting, that is the direction
that the object will
accelerate.
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Examples:
Object moving at a constant
velocity to the right
Object moving at a constant
velocity to the right
50 N
The object will accelerate (to the right) and
speed up because the NET force acting on it is
not zero (ie. The forces are UNBALANCED).
20 N
Object moving at a constant
velocity to the right
30 N
The object will keep doing this forever
unless its forced to change its motion.
The object will accelerate (to the left) and slow
down because the NET force acting on it is not
zero (ie. The forces are UNBALANCED).
50 N
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More Real-Life Examples of
Newton’s 1st Law
A runaway truck on a
highway vs. a runaway
bike. Which one would you
want to have to stop?
I would much rather have to stop the runaway bike. It has
less mass, and therefore less inertia than the truck. Does it
want to stop? No! It is in motion and wants to continue on
in its motion. However, since its inertia is less than the
truck’s, it has less of a tendency to avoid its motion change,
and it will require less force to stop it.
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Real-Life Examples of
Newton’s 1st Law
A car in a collision with a
brick wall (and the person
inside not wearing a
seatbelt)
The car and he person have inertia. When the car hits the
wall, the car has an unbalanced force applied to it, and it
slows down and stops. The person inside, who is in
motion, wants to stay in motion. Without the seatbelt
supplying an unbalanced force, they will continue in their
motion and fly through the windshield.
NEXT
Real-Life Examples of
Newton’s 1st Law
Trying to get a loaded
grocery cart to move from
rest, and then trying to get
it to stop once its moving.
When the cart is at rest, it has inertia. This tends to keep it
at rest, and it requires a pretty large force to get the cart to
move. Once the cart is moving, it wants to keep moving
(since it still has inertia) and it will take a pretty large force
to slow it down.
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Newton’s
nd
2
Law
Let’s Review. Newton’s 1st Law says “An object in motion (or at rest)
tends to stay in motion (or at rest) unless acted upon by an
unbalanced, external force”. Included in this statement is the fact that
objects naturally like to be either at rest or moving at a constant
velocity. Their inertia keeps them in one of these two natural motion
states, and it requires an unbalanced, external force to “knock them
out” of their preferred motion state. Many forces can act on an object
at rest, but unless the forces are unbalanced, the object will not move.
The same can be said for objects moving at a constant velocity.
Now for the one million dollar question. What happens to an object IF
an unbalanced, external force DOES act upon it? The answer is simple:
its motion will change. In other words, it will accelerate. Newton’s 2nd
Law explains what will happen to this object. Stated in the simplest
terms, it says …
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Newton’s
nd
2
Law
F = ma
Do NOT memorize Newton’s 2nd
Law as F = ma!!!!
FNet = ma
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FNet = ma
Newton’s 2nd Law has a vector nature.
Applied
forces
Object
initially at
rest
50 – 30 = 10ax
60 N
10 kg
30 N
FNET,x = max
50 N
ax = 2 m/s2
FNET,y = may
60 – 60 = 10aY
aY = 0 m/s2
60 N
y
x
Since an unbalanced force acts
on the object, it will accelerate
in the +x-direction at 2 m/s2.
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In order to apply Newton’s
2nd Law to a situation, we
often start by making a “freebody” diagram for the object
in question. We will refer to
them as FBD’s, for short.
HOME
Example Free Body Diagrams (FBD)s:
Object sitting at rest
on a surface.
Object falling
through the air.
REAL LIFE
SCENARIO
F
B
D
Force that the surface
pushes back on the
object with (called the
NORMAL FORCE, FN)
Force due to gravity (Fg)
Air Resistance
Force, FA
Fg
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Example FBDs:
Object being pushed
across a surface
Object being pulled
across a surface
FN
FFRICTION
FN
FPULL
Fg
FFRICTION
Notice that friction
always opposes the
motion of the object!
FPUSH
Fg
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Example FBDs:
Object being pushed
at an angle across a
surface
Object being pulled
at an angle across a
surface
FN
FN
FPULL
FP,y
q
FP,y
FP,x
FFRICTION
Fg
Notice that friction
always opposes the
motion of the object!
FPUSH
q
FFRICTION
FP,x
Fg
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There are two “VIEWPOINTS” from
which we can look at a problem.
Example: An object is pulled across a surface
VIEW FROM ABOVE
VIEW FROM THE SIDE
FN
FF
FPULL
Fg points into the page
FN points into the page
FF
FPULL
Fg
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How do you know when to use the
“VIEW FROM ABOVE” view?
Example: A 3 kg box is pulled by three force. The 1st is 70 N [E],
the 2nd is 45 N [NW], and the 3rd force is 15 N [E 20o S].
Find the acceleration of the block.
Notice how this problem is using words like “North, South, East,
and West”, and not “up and down”. When a problem uses
words like “North and East” (which are perpendicular to each
other), then you must use the VIEW FROM ABOVE.
If you see the words “up and down” and “East and West”,
OR “North and South” (but not both together), then use
the VIEW FROM THE SIDE.
HOME
Before we can practice using
Newton’s 2nd Law (along with
Free-Body Diagrams, FBDs) in
order to tackle problems, we
need to learn how to
calculate the FORCE DUE TO
GRAVITY (Fg) acting on an
object.
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Weight and the Force due to Gravity
Force due to gravity: A field force (a vector quantity) that
always is directed towards the center of the earth.
Fg  mag  mg
Weight: The magnitude of the Force due to gravity.
W  Fg  m a g  mg
Mass is an inherent property of an object. It stays the
same no matter where you move. Weight depends on
the acceleration due to gravity (g). It changes
depending on your location.
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Examples:
On the earth your mass is 80 kg. What is
your weight?


W  mg  80kg 9.8 sm2  784 N
If you take a trip to the moon, what will your
new mass and weight be?
Your mass doesn’t change when you change locations.
Therefore, it is still 80 kg. Your weight depends on the
acceleration due to gravity. On the moon, it is 1.6 m/s2 instead of
9.8 m/s2. Therefore, your weight on the moon is …


W  mg  80kg 1.6 sm2  128N
HOME
A 20 kg box is pushed across a frictionless surface by a
force of 500N. Find the box’s acceleration as well as the
normal force acting on the block (from the ground).
FNET,x = max
FN
ax
500 – 0 = 20ax
ax = 25 m/s2
20 kg
500N
y
FNET,y = may
(20)(9.8) = 196N
FN - 196 = 20(0)
FN = 196 N
x
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A 40 kg box is pushed across a rough surface at a
constant velocity by a force of 76 N. Find the force
of friction acting on the block (from the ground).
FN
FNET,x = max
76 – Ff = 40(0)
40 kg
76N
y
Ff
(20)(9.8) = 196N
x
Ff = 76N
From Newton’s 1st Law, constant
velocity means that FNET = 0, or,
in other words, that a = 0.
HOME
A man pushes a 30 kg block with a constant force of 200N at
an angle of 30o above the horizontal. If the surface provides
a constant frictional force of 30N, find the acceleration of
the block and the normal force acting on the block.
FNET,x = max
FN
173.1 – 30 = 30ax
ax = 4.77 m/s2
200N
100N
30o
173.1
30 kg
FNET,y = may
Ff = 30N
ax
y
FN - 100 - 294 = 30(0)
FN = 394 N
(30)(9.8) = 294N
x
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A man pulls a 20 kg block with an unknown force at an
angle of 40o above the horizontal. A constant frictional
force of 60 N acts on the block. If the block moves at a
constant velocity, find the unknown pulling force.
FN
FNET,x = max
F
40o
Ff = 60N
20 kg
Fcos40
Fsin40
Fcos40 - 60 = 20(0)
Fcos40 = 60
F = 60 / cos (40) = 78.3 N
y
Fg
x
HOME
Newton’s
rd
3
Law
For every action force there is
an EQUAL, but OPPOSITE
reaction force.
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Examples:
Action Force
Reaction Force
Pushing against a wall.
The wall pushes back
(normal force).
Pushing against the ground.
The ground pushes back
(normal force).
Feet pushing backward
against the ground.
Ground pushes you forward
(allowing you to move
forward when walking).
Swimmer pulling the water
backward with his hand.
Rocket engines pushing
exhaust gas out into the air at
the back end of the rocket.
Water pushes the swimmer
forward through the water.
Air pushes the rocket
forward through the sky.
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Example: Two astronauts are sitting next to each other
in deep space. One has a mass of 80 kg and the other
has a mass of 120 kg. The more massive astronaut
pushes the les massive one with a force of 200 Newtons.
Find the acceleration of each astronaut.
120 kg
80 kg
200 N
a more massive
200 N
a less massive
200 N   120kga
200 N   80kga
a  1.67 sm2
a  2.5 sm2
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The NORMAL Force
The NORMAL FORCE is a “reactionary” force exerted on one
object by another that is always PERPENDICULAR to the
surface of contact.
It is a direct result of NEWTON’S 3RD LAW
Example:
FN
Object sitting at rest
on a surface.
Fg
Notice how the
normal force is
the ground’s
“reaction” to
the gravity
force.
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Finding the Normal Force
Example: A 50 kg block sits at rest on the ground. Find the
normal force acting on the block.
FN
50 kg
Fg = mg = (50kg)(9.8 m/s2)
= 490N
Since the block does not move UP or DOWN, the
normal force’s job is to provide balance. Therefore,
UPS = DOWNS. And, as a result, in the problem the
normal force is equal to the weight (490N).
NEXT
Finding the Normal Force
Example: A 50 kg block sits at rest on the ground. A man pulls
upward on the block with a force of 200N using a
rope at an angle of 30o above the horizontal. Find
the normal force acting on the block.
FN
200 N
100N
q
50 kg
173N
FFRICTION
490 N
Again, since the block does not move UP or DOWN, UPS
must equal DOWNS. Therefore, FN + 100 = 490. The
Normal Force equals 390 N.
NEXT
Finding the Normal Force
Example: A 50 kg block sits at rest on the ground. A man
pushes downward on the block with a force of
200N (like it’s a lawnmower) at an angle of 30o
below the horizontal. Find the normal force.
FN
200 N
100N
q
50 kg
173N
FFRICTION
490 N
Again, since the block does not move UP or DOWN, UPS
must equal DOWNS. Therefore, FN = 490 + 100. The
Normal Force equals 590 N.
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So, Let’ Review ….
When an object is simply sitting on flat ground, the
normal force is equal to the weight. The ground feels the
full weight, and it reacts by pushing back.
When an object on flat ground is pulled upward, at an
angle, the normal force is reduced. WHY? Because the
process of pulling up relieves some of the weight that the
ground feels … and the thus the block “feels” lighter to
the ground, which has to react less to hold it up.
Vice Versa, when an object on flat ground is pushed
downward at an angle, the normal force is increased,
since the pushing force is “forcing” the block into the NEXT
ground more. The ground reacts accordingly, increasing
the normal force.
Finding the Normal Force … one more time
Example: A 50 kg block is being pushed up against a wall by a
force of 1000 N, at an angle of 30o above the
horizontal. Find the normal force exerted in the
block.
50
kg
Since the block is not
moving INTO the wall
or OUT OF it, INTOs =
OUTOFs
FFRICTION
1000 N
FN
q
Therefore, FN = 865 N
865 N
490 N
HOME
Frictional Forces
Let’s review what we probably already know about friction:
1)Friction USUALLY slows things down, instead of speeding them
up. It USUALLY resists motion.
2)Friction happens when two surfaces rub together. The types of
surfaces determines whether there is a lot or a little friction.
3)The harder you press the surfaces together, the larger the force
of friction. For example, if you are using sandpaper to sand
wood, the harder you push the paper into the wood, the more
“sanding” takes place.
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Frictional Forces
Now, for what you might not know yet:
1)If something is moving at a CONSTANT VELOCITY (a = 0), the
frictional forces must be in balance with any forces pushing or
pulling the block forward.
2)There are two types of friction: The first type acts on an object
when it is at rest (trying to keep it at rest), and the second type
acts on objects when they are moving (trying to slow them
down).
3)When an object sits (at rest) on a flat surface, there is no
friction acting on it. When an object sits at rest on an incline,
friction DOES act on it (to keep it from sliding down).
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The Coefficient of Friction …
… tells you how much friction exists between different surfaces.
alot of
friction
very little friction
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Calculating Friction
The amount of friction between two surfaces can be calculated
using ….
F f  mFN
The coefficient of friction
(which is given the greek
letter “mu”, m)
mu is pronounced “myoo”
The larger the normal force
between the two surfaces, the
larger the friction between the
surfaces.
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Two different kinds of friction
STATIC friction occurs when an
object is trying to be moved from
rest. It is larger than KINETIC
friction. Once the maximum
static frictional force is exceeded,
the object moves.
KINETIC friction occurs when
a there is movement
between the two surfaces
which are in contact.
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“OLD” Friction Problems
(friction is given  simply use)
1) A 50 kg box is pulled across a surface by a force of 100 N. A constant force of
friction of 25 N acts against the object. Find the objects acceleration.
FN
a
25 N
100 N
Fg
FNET  ma
m
a  1.5 2
s

100N  25N  (50kg)a
(constant velocity  find the frictional force)
2) A wagon is pulledat a constant velocity by a force of 100 N at an angle of 30o
above the horizontal. Find the frictional force acting on the block.
FN
a
100 N
50 N
Ff
73 N
Fg
FNET  ma  m(0)  0
Ff  73N
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Example: A 10 kg box is motionless on the floor. If the
coefficient of static friction is 0.4 and the coefficient
of kinetic friction is 0.3 (between the box and the
floor), find the force required to start the block in
motion.
FN
Ff
a
F
Fg
Fg  mg  490N
UPs  DOWNs
 FN  490N
Ff  mFN  (.4)(490)  196N
FNET  ma
F  196N

F 196  m(.00000001)
In order to move, it only needs
to accelerate a little.
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Example: Continuation of the last problem. Once the block
(10 kg) starts moving, find the force needed to keep it
moving. (remember, the coefficient of static friction
is 0.4 and the coefficient of kinetic friction is 0.3
(between the box and the floor).
FN
Ff
a
F
Fg  mg  490N
UPs  DOWNs
 FN  490N
Ff  mFN  (.3)(490)  147N
FNET  ma
F  147N
Fg

F 147  m(.00000001)
Again, in order to keep moving, it
only needs to accelerate a little.

Notice that the force needed to make
something start moving is greater then
the force required to keep it moving!
NEXT
Example: It takes a 50 N horizontal force to pull a 20 kg object
along the ground at a constant velocity. What is the
coefficient of kinetic friction?
FN
Ff
a
50 N
m
Fg  mg  (25kg)(9.8 2 )  245N
s
UPs  DOWNs
 FN  245N
Ff  mFN  m(245)
FNET  ma
50  245m
Fg



50  245 m  (25)(0)
m  .204
Notice that mu doesn’t
have any units.
NEXT
Example: A cart with a mass of 2.0 kg is pulled across a level
desk by a horizontal force of 4.0 N. If the coefficient
of kinetic friction is 0.12, what is the acceleration of
the cart?
FN
Ff
a
4N
m
Fg  mg  (2kg)(9.8 2 )  19.6N
s
UPs  DOWNs
 FN  19.6N
Ff  mFN  (.12)(19.6)  2.352N
FNET  ma
m
a  .824 2
s
Fg

4  2.352  (2)a

NEXT
Example: A small 10 kg cardboard box is thrown across a level
floor. It slides a distance of 6.0 m, stopping in 2.2 s.
Determine the coefficient of friction between the
box and the floor.
FN
a
Ff
Fg
Notice that the acceleration arrow
points backward because the
block is slowing down.
x  6m, t  2.2s, v 2  0

m
v1  5.455
s
m
m
t  2.2s, v 2  0, v1  5.455
 a  2.48 2
s
s
UPs  DOWNs
 FN  Fg  98N
Ff  mFN  98m
FNET  ma
m  .253

98 m  10(2.48)
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Example: A farmer is pushing down a 4 kg shovel with a force of
40 N at an angle of 60 o with the ground. Determine
the acceleration of the shovel if the coefficient of
friction between the shovel and the icy ground is 0.15.
40 N
34.6 N
FN
60
a
20 N
Ff
Fg
UPs  DOWNs

FN  Fg  34.6
FN  39.2  34.6  73.8N
Ff  mFN  (.15)(73.8)  11.07N
FNET  ma
m
a  .223 2
s

20 11.07  4a
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Block-pushing-block problems
200 N
50 kg
150 kg
Step #1: Look at the
entire set of masses at
the same time (MEGA
MASS!!!)
200  200a
m
a 1 2
s
EXAMPLE #1: Assuming a
frictionless surface, find the force
applied by the 50 kg box onto the
150 kg box.
1,960 N
200 kg
a
200 N
1,960 N
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Block-pushing-block problems
200 N
EXAMPLE #1 … CONTINUED
Assuming a frictionless surface,
find the force applied by the 50
kg box onto the 150 kg box.
150 kg
50 kg
Step #2: Look at the masses
individually (MINIs!!!)
490 N
200 N
1,470 N
a
150
R
50
R
a
1,470 N
490 N
200  R  50a
R 150a 150(1) 150N
NEXT
Block-pushing-block problems
F
40 kg
EXAMPLE #2: If m = 0.4 and if the
force between the blocks is 50N,
find the force pushing the blocks.
10 kg
490 N
Step #1: Look at the
MEGA MASS!!!
50 kg
F
F 196  50a
a  ???
490 N
a
F f  mFN
 (.4)(490)
 196N
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Block-pushing-block problems
F
40 kg
10 kg
Step #2: Look at the MINIs!!!
392 N
F
40
50 N
F f  (.4)(392)
a
392 N
 156.8N
F  50 156.8  40a
a
 ???
EXAMPLE #2: If m = 0.4 and if the
force between the blocks is 50N,
find the force pushing the blocks.
98 N
50 N
10
F f  (.4)(98)
a
98 N
 39.2N
50  39.2  10a
m

a  1.08 2
s
NEXT
Block-pushing-block problems
F
40 kg
10 kg
EXAMPLE #2: If m = 0.4 and if the
force between the blocks is 50N,
find the force pushing the blocks.
Step #3: Plug “a” back into the other equations to find
the missing force.
m
a  1.08 2
s

F  50 156.8  40a
F  50 156.8  40(1.08)
F  250N
NEXT
RECAP of Block-pushing-block problems
#1) Look at the entire set of masses at once (the MEGA MASS).
Draw an accurate FBD, making sure to account for all forces.
Use F = ma. If you can solve for an unknown, great! If not, we
can always come back to this equation later.
#2) Look at each individual part separately (the MINIs). Draw an
accurate FBD for each, accounting for ALL the forces. Use F = ma
for each one.
#3) Whenever one block pushes another, the pushing force and
the “push back” force are the same (Newton’s 3rd Law).
#4) If you got information using the MEGA MASS, plug it into the
MINIs (since both the MEGA and the MINIs accelerate at the
same rate). Otherwise, solve one of the MINIs equations for an
unknown and plug it into the MEGA equation.
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Train-Style problems
400 N
250 kg
350 kg
Step #1: Look at the
entire set of masses at
the same time (MEGA
MASS!!!)
400  600a
m
a  0.67 2
s
EXAMPLE #1: Assuming a frictionless
surface, find the acceleration of the
train as well as the tension in the
rope between the blocks.
a
400 N
5,880 N
600 kg
5,880 N
NEXT
Train-Style problems
EXAMPLE #1: Assuming a frictionless
surface, find the acceleration of the
train as well as the tension in the
rope between the blocks.
400 N
250 kg
350 kg
Step #2: Look at the masses
individually (MINIs!!!)
2,450 N
400 N
3,430 N
a
350
T
250
a
2,450 N
T
400  T  250a
3,430 N
T  350a  350(0.67)  234.5N
NEXT
Train-Style problems
F
30 kg
40 kg
EXAMPLE #2: The train shown starts
from rest and covers 50 meters in 8
seconds, all the while accelerating at a
constant rate on a frictionless surface.
Find the force pulling the train as well as
the tension between the blocks.
Step #1: Use an equation of motion to get “a”.
1 2
x  at  v1t
2
1
50  a(8) 2  (0)(8)
2
m
a  1.5625 2
s
NEXT
Train-Style problems
F
30 kg
40 kg
Step #2: Look at the
entire set of masses at
the same time (MEGA
MASS!!!)
EXAMPLE #2: The train shown starts
from rest and covers 50 meters in 8
seconds, all the while accelerating at a
constant rate on a frictionless surface.
Find the force pulling the train as well as
the tension between the blocks.
a
F
F  70a
F  70(1.5625)  109.375N
686 N
70 kg
686 N
NEXT
Train-Style problems
F
30 kg
40 kg
EXAMPLE #2: The train shown starts
from rest and covers 50 meters in 8
seconds, all the while accelerating at a
constant rate on a frictionless surface.
Find the force pulling the train as well as
the tension between the blocks.
Step #3: Look at a MINI
a
T
T  40a
T  40(1.5625)  62.5 N
392 N
40 kg
392 N
NEXT
Train-Style problems
800 N
30 kg
40 kg
EXAMPLE #3: If m = 0.3 for the train at
the right, how far will it travel, starting
from rest, in 4 seconds. Then, find the
tension in the rope.
Step #1: Use the MEGA mass.
800  205.8  70a
m
a  8.489 2
s
686 N
a
800 N
70 kg
F f  (.3)(686)
 205.8N
686 N

NEXT
Train-Style problems
800 N
30 kg
40 kg
EXAMPLE #3: If m = 0.3 for the train at
the right, how far will it travel, starting
from rest, in 4 seconds. Then, find the
tension in the rope.
Step #2: Use an equation of motion.
1 2
1
x  at  v1t  (8.489)(4) 2  (0)(4)  67.9m
2
2

NEXT
Train-Style problems
800 N
30 kg
40 kg
EXAMPLE #3: If m = 0.3 for the train at
the right, how far will it travel, starting
from rest, in 4 seconds. Then, find the
tension in the rope.
Step #3: Use a MINI.
T 117.6  40a
T 117.6  40(8.489)
392 N
a
T
40 kg
F f  (.3)(392)
 117.6N
392 N
T  457.2N

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