Newton's Three Laws of Motion
Newton’s First Law – Inertia
-Isaac Newton was a 17th century physicist who
formulated three laws that can help us describe the
motion of any object. His first law, sometimes called
the law of inertia, says:
-Objects in motion tend to stay in motion and
objects at rest tend to stay at rest unless acted
upon by an unbalanced force.
-essentially this is saying that objects will tend to
remain doing what they are doing.
-inertia – the property of matter that causes objects to
resist changes in motion.
-the tendency of an object to resist changes in its state
of motion is dependent upon its mass.
-the base unit for mass is the kilogram (kg)
-the terms mass and weight are often used in the
same context, but it is incorrect to do so. The base unit
of weight is the Newton (N).
-mass – the amount of matter in an object (eg. 20 kg)
-weight – the force on an object due to gravity (eg. 15
N)
Newton’s Second Law – Acceleration
-Newton’s first law of motion indicated to us how a
body acts when there is no net applied force
(unbalanced force). In other words when the forces
1
were in equilibrium. What happens when there is a
net force?
-Newton’s second law states that:
-the acceleration produced by a net force on an
object is directly proportional to the magnitude
of the net force, is in the same direction as the
net force, and is inversely proportional to the
mass of the object.
-if the object is at rest when the force is applied, it will
begin to move in the direction of the force and will
move faster and faster as long as the force continues.
-Newton’s second law can be written in formula form
as:
Fnet = ma
where,
Fnet = net force in Newtons (N)
m = mass in kilograms (kg)
a = acceleration (m/s2, or N/kg)
-the unit of force, the Newton (N), is a derived unit,
that is, a unit created from more basic units. So,
1 N = 1 kg x m/s2
-therefore one Newton is the force required to
accelerate one kilogram of mass at one meter per
second each second.
-a lot of the physics that we study involves that which
we observe here on Earth. Since gravity is the force on
2
an object that causes it to accelerate, we can have the
formula:
Fnet = mg
-or more commonly:
Fw = mg
-it will however be useful for us to actually build this
equation from scratch so as to see some of the useful
working equations that can arise.
-from Newton’s second law we see that the
acceleration acquired by an object is directly
proportional to the amount of force applied. So let’s
compare two objects of equal mass and write them in
the form of Newton’s Second Law. So that we don’t
mix up the two objects, we will denote one of the
objects with the ‘prime’ symbol. So we’ll have:
-let’s solve both of these for the mass:
-since both masses are identical, it follows that:
-since we're on Earth, one of the forces is known. It
equals the weight, which we will denote as Fw.
-also due to being on Earth, one acceleration is known
as well, g, or 10 m/s2.
3
-so let’s put Fw into the
spot.
spot, and a g into the
-Sidenote: when talking about weight, it is sufficient to
use the absolute value of g, which would be +10 m/s2.
-substituting these in we can have:
-finally, just cross multiply the up with the Fw and
you have arrived at your working equation:
-with this equation we could calculate the force
needed to give any desired acceleration to an object of
known weight here on Earth.
-continuing on, since F = ma, we can substitute the ma
into the above equation and have:
-then by getting rid of the two a’s and solving for Fw,
we come full circle from where we started and we
arrive back at:
-from this equation we can now calculate the weight
of an object if we know it’s mass. (recall the difference
between weight and mass)
-Aside: Fw = mg is the same formula as Fgrav = mg
4
Example
What is the weight of a 4.0 kg object?
Solution
= (4.0 kg)(10 m/s2)
= 40 N or 4.0 x 101 N
Example
A car weighs 1.50 x 104 N. What force is required to
accelerate the car at 4 m/s2? (Neglect friction or air
resistance)
Solution
F = (1.50 x 104 N)(4 m/s2)
10 m/s2
F = 0.6 x 104 N or 6.0 x103 N
-there is one more equation that involves Newton’s
second law, and this is if we start to consider any
frictional forces in a given situation. The formula for
this is:
Ffrict = µFnorm
Where,
Ffrict
5
= frictional force (Newtons)
µ
Fnorm
= coefficient of friction (essentially just a
multiplier indicating the relative amount
of friction that a surface has)
= normal force (Newtons)
-the frictional force is the force that would be required
to get an object moving that is initially at rest.
Example
A book is lying on a table, whose normal force is 10 N.
If the table has a frictional coefficient of 0.2, what
frictional force must be overcome to get the book
moving?
Solution
Ffrict = µFnorm
Ffrict = (0.2)(10 N)
Ffrict = 2 N
-free-body diagrams are useful in figuring out what
forces are at work in any situation.
-a free-body diagram is a drawing that represents the
object and the forces acting on that object.
Example
There is no net force in the following situation. What
must be force A and force B for this to be true?
6
Solution
A = 50 N, the horizontal forces must be balanced.
B = 200 N, the vertical forces must be balanced.
Example
A rightward force of 25 N is applied to a 4-kg object to
move it across a rough surface with a rightward
acceleration of 2.5 m/s2. Use the diagram to
determine the gravitational force, normal force,
frictional force, net force, and the coefficient of
friction between the object and the surface.
Solution
Fnet can be found using Fnet = ma
Fnet = (4 kg)(2.5 m/s2)
Fnet = 10 N, right.
-Since the mass is known, Fgrav can be found by using
the equation Fgrav = mg
Fgrav = (4 kg)(10 m/s2) = 40 N.
7
-Since there is no vertical acceleration, there is no net
vertical force, thus Fnorm = Fgrav = 40 N.
-Since Fnet = 10 N, right, the rightward force (Fapp) must
be 10 N more than the leftward force (Ffrict); thus, Ffrict
must be 15 N.
-Finally, to find the coefficient of friction, use
Ffrict = µFnorm
µ = Ffrict /Fnorm
µ = (15 N)/(40 N) = 0.375
Newton’s Third Law - Action and Reaction
-a force is a push or a pull upon an object which results
from its interaction with another object.
-some forces result from contact interactions (normal,
frictional, tensional, and applied forces) and other
forces result from action-at-a-distance interactions
(gravitational, electrical, and magnetic forces).
-according to Newton, whenever objects A and B
interact with each other, they exert forces upon each
other.
-there are two forces resulting from this interaction,
and these two forces are called action and reaction
forces.
-Newton's third law is:
-for every action, there is an equal and opposite
reaction
-the statement means that in every interaction, there
is a pair of forces acting on the two interacting objects.
8
-the size of the force on the first object equals the size
of the force on the second object.
-the direction of the force on the first object is
opposite to the direction of the force on the second
object.
Dynamics
-this is the study of interacting forces on objects.
-the forces can be all be at equilibrium, or can be
happening at angles.
-in physics, one way to look at equilibrium is to say
that the velocity of the object in question is not
changing, ie. that the object is not accelerating.
-this could mean that the object is at rest, or that the
object is moving at a constant velocity.
-therefore an object is in equilibrium when it has zero
acceleration.
-for the object to have zero acceleration, the sum of
the forces acting on the object must be zero.
-if this is the case, we say that the net force on the
object is zero.
-recall from Newton's Second Law the equation
Fnet = ma, and if a = 0, then Fnet = 0 as well.
-if we consider a two-dimensional situation, ie. forces
occurring along x and y planes, then the sum of all the
9
x components of the forces must be zero, and the sum
of all of the y components must likewise be zero.
-this can be expressed mathematically as:
∑Fx = 0 and ∑Fy = 0
-the forces acting on an object in equilibrium must
balance.
-if an object is accelerating, then the forces acting on
an object do not balance, and the object is not in a
state of equilibrium.
Strategies for Analyzing Equilibrium Problems
1. Select the object to be studied. This object is often
called a "system". In some problems, two or more
objects may be connected by a rope or cable. It may
then be necessary to treat each object separately and
the follow the rest of the steps below.
2. Draw a free-body diagram for each object chosen.
3. Choose a set of x and y axes for each of the objects
being analyzed and resolve the free body diagram into
components that point along these axes. The axes may
not necessarily point vertically and horizontally or up
and down the way we are normally accustomed to
seeing them on a graph. The axes should be selected
in such a way that as many forces as possible point
directly along the x axis and y axis.
4. For an equilibrium situation, we must set up the
equations in such a way that the sum of the x
components of the forces is zero, and the sum of the y
components is also equal to zero.
5. Solve the equations for the unknown quantities.
10
Equilibrium in One Dimension
-one common example of equilibrium in one
dimension is a hanging sign suspended by vertical
cords.
Example
Suppose you had a sign with a mass of 20 kg
suspended by two ropes, one on each side of the sign.
What would be the tension in each of the ropes?
Solution
Follow the five steps outlined above.
Step 1. Select the object to be studied. In this case the
object is the sign.
Step 2. Draw a free-body diagram. In our diagram,
there are three forces acting on this object. Gravity is
acting down and this force is balanced by the two
upward forces exerted by the cords. The cords are the
same distance from the centre.
11
Step 3. Choose a set of x and y axes for each of the
objects being analyzed and resolve the free body
diagram into components that point along these axes.
In this case, there is only one axis, the vertical one.
There are only y components of the forces. There are
no x components.
Step 4. Set up the equations in such a way that the
sum of the x components of the forces is zero, and the
sum of the y components is also equal to zero. There
are two tension forces (T) pointing upwards and one
force of gravity (Fg) pointing downwards. The sum of
these forces is zero. Our equation will be:
2T -Fg = 0
-We modify this equation according to what
information we have, and since Fg = mg, our equation
becomes:
2T - mg = 0
Step 5. Solve the equations for the unknown
quantitie(s).
2T - mg = 0
2T = mg
T = mg/2
T = (20 kg)(10 N/kg)/2
T = 200 N/2
T = 100 N in each rope
Equilibrium in Two Dimensions
-The example of a hanging sign could also be applied in
two dimensions.
12
Example
Suppose a sign is hanging from two wires as per the
diagram below. The mass of the sign is 30 kg. What is
the tension in each of the wires?
Solution
Apply the five steps:
Step 1. Select the object to be studied. The object we
are studying is the sign, but in this case, when we draw
the free body diagram, it would be better to choose
the point at which the three wires join, because it is
here that all three forces are acting. The sign would
not be a good choice for the object because, out of the
three forces indicated by the diagram, only two forces
act on the sign; the force of gravity downwards and
the equal and opposite force of the wires upward.
Step 2. Draw a free-body diagram. The free-body
diagram would contain the three tensions in the wires
at the angle shown in the diagram.
13
Step 3. Choose a set of x and y axes for each of the
objects being analyzed and resolve the free-body
diagram into components that point along these axes.
-So to not get confused as to which component is
which, we can use numbers and letters as subscripts.
Notice that T1 and T2 are not exactly along either the x
or y axes. This means that T1 and T2, because they are
at angles other than 0 and 90, will each have a
horizontal and a vertical component. If you like, you
could redraw the free-body diagram to look like:
-The x and y components of each of the vectors T1 and
T2 are as follows:
T1x = -T1cos45o (note the negative direction because
the force is to the left)
T1y = T1sin45o (positive, because the force is upwards)
T2x = T2sin37o (positive, because the force is to the
right)
T2y = T2sin37o (positive, because the force is upwards)
14
Step 4. Set up the equations in such a way that the
sum of the x components of the forces is zero, and the
sum of the y components is also equal to zero.
-There are two forces in x (horizontal) direction, and
their sum must be zero. Therefore the basic equation
is:
T1x + T2x = 0
-Rewrite this as:
-T1cos45o + T2cos37o = 0
-There are three forces in the y (vertical) direction, and
their sum must also be zero. Therefore the basic
equation is:
T1y + T2y - T3 = 0
-Rewrite this as:
T1sin45o + T2sin37o - mg = 0
Step 5. To solve a set of equations with two unknowns,
solve one of the equations for an unknown of your
choice, and put this answer into the second equation.
-From the equation in the horizontal direction, we
could solve it for T2 as follows:
-T1cos45o + T2cos37o = 0
T2cos37o = T1cos45o
T2 = T1cos45o
cos37o
15
-Now put this answer into the equation for the vertical
direction as follows:
T1sin45o + T2sin37o - mg = 0
T1sin45o + T1cos45o sin37o - mg = 0
cos37o
T1(0.707) + T1[0.707/0.798](0.601) - (30 kg)(10 N/kg) =
0
T1(0.707) + T1(0.532) - 300 N = 0
T1(1.239) = 300
T1 = 300/1.239
T1 = 242 N
-Now that we know T1, put this answer back into the T2
equation and solve:
T2 = T1cos45o
cos37o
T2 = (242 N )cos45o
cos37o
T2 = (242 N )(0.885)
T2 = 214 N
-Finally, T3 is just the weight:
T3 = mg
T3 = (30 kg)(10 N/kg)
T3 = 300 N
16
Pulling at an Angle: No Friction
-In this example, we will examine pulling a rectangular
box at an angle assuming there is no friction present.
-in the drawing below, a 10 kg box is being pulled to
the right with a force of 40 N at an angle of 30o.
-find the normal force, the x component of the applied
force (the pull), and the rate of acceleration of the
box.
-when drawing the free-body diagram for this
situation, it is useful to draw the force vectors from a
dot at the centre of the box:
-in the y direction, the sum of the forces is equal to
zero because the block doesn't rise off the table or fall
into it. Therefore the net force in the y direction is
zero, and our formula for these forces becomes:
FN + Fasinθ - Fg = 0
-isolate FN, fill out the formula so it's a bit more
complete, and plug in values to find the normal force:
17
FN + Fasinθ - Fg = 0
FN = -Fasinθ + Fg
FN = -Fasinθ + mg
FN = -(40 N)(sin30) + (10 kg)(10 N/kg)
FN = -20 N + 100 N
FN = 80 N
-in this scenario, there is only one force along the x
direction, and it is the x component of the applied
force. We could label this x component of the applied
force as Fax and then our formula to find it would be:
Fax = Facosθ
Fax = (40 N)(cos30)
Fax = 34.64 N
-this is the force that makes the block accelerate, so
we can use Fax as the net force. To find the
acceleration, use:
Fnet = ma
a = Fnet/m
a = 34.64 N/10 kg
a = 3.46 m/s2
18
Pulling at an Angle: With Friction
-use all the same numbers as the above example,
except introduce a coefficient of friction on the surface
of 0.3. Find the acceleration and the net force.
-tweak your free-body diagram to include the force of
friction:
-nothing changes for the y components, but the x
components do change.
-the sum of the x components of the forces is equal to
the mass of the block times the acceleration. So our
formula becomes:
Fax - Ff = ma
-now fill it out a bit more, because remember that
Ffrict = μFnorm, Ff = μFN
Facosθ - Ff = ma
Facosθ - μFN = ma
*aside: how would you know when you have to fill out
your formula a bit more like this? It all depends on
what information you are given, and in this scenario
we are given the coefficient (μ) and we are given the
normal force (FN).
19
-plug in values, recalling that we already know the
normal force from the previous non-friction example:
Facosθ - μFN = ma
(40 N)(cos30) - (0.3)(80N) = (10 kg)a
34.64 N- 24 N= (10 kg)a
10.64 N = (10 kg)a
a = 10.64 N/10 kg
a = 1.06 N/kg or m/s2
-to find the net force:
Fnet = ma
Fnet = (10 kg)(1.06 N/kg)
Fnet = 10.64 N
Inclines: No Friction
-a block with a mass of 50 kg is sliding down a slope of
250. Find the normal force and the rate of acceleration.
-the force of gravity still acts straight downwards, but
the normal force will be perpendicular to the surface
of the slope. Therefore the normal force and the force
of gravity are not straight up and down, but rather:
20
-to draw the free-body diagram in this situation, it is
good to set the coordinates parallel to the surface of
the incline and perpendicular to the ground. It is also
good to set the direction down and to the right as the
positive x direction and to set the direction of the
normal force vector to be the positive y direction.
What this does for us is it will now show the force of
gravity vector broken up into its two components. So
the free-body diagram could look like:
-the normal force is not equal to the gravitational
force because these two forces do not point in the
same direction.
-but the y component of the gravitational force, Fgy, is
equal to the normal force, and these two forces must
add up to zero. So our formula becomes:
FN - Fgy = 0
-solve for the normal force:
21
FN - Fgy = 0
FN = Fgy
FN = mgcosθ
FN = (50 kg)(10 N/kg)(cos25o)
FN = 453.15 N
-the force down the slope is just the x component of
the force of gravity. It is also the net force causing the
object to accelerate down the slope, and therefore
Fnet = Fgx
-so to find acceleration, use:
Fnet = ma
Fgx = ma
mgsinθ = ma
gsinθ = a
(10 N/kg)(sin25o) = a
a = 4.22 N/kg or m/s2
Inclines: With Friction
--use all the same numbers as the above example,
except introduce a coefficient of friction on the surface
of 0.3. Find the acceleration.
- tweak your free-body diagram to include the force of
friction:
22
-in adding friction, the components of the forces in the
y direction have not changed, but the components of
the forces along the x direction have changed.
-the sum of the forces along the x direction is equal to
the net force or the mass times acceleration, and it is
this force that causes the object to accelerate down
the incline.
-so our formula becomes:
Fgx - Ff = ma
-modify it to include what we know about the friction:
Fgx - μFN = ma
-but since the FN is not equal to Fg, but rather a
component of Fg, FN = mgcosθ. So our formula
becomes:
Fgx - μmgcosθ = ma
-almost done! Since Fgx from the previous example was
equal to mgsinθ, put this into your formula as well:
mgsinθ - μmgcosθ = ma
-simplify to:
23
gsinθ - μgcosθ = a
-and solve... whew!
(10 N/kg)(sin25o) - (0.3)(10 N/kg)(cos25o) = a
a = 1.50 N/kg or m/s2
24