### Video Transcript

In this video, we’ll learn how to
solve problems using Newton’s third law of motion. By this stage, you should’ve
encountered and applied Newton’s first and second laws of motion. Newton’s first law states that an
object will remain at rest or in uniform motion in a straight line unless acted upon
by an external force. And Newton’s second law says that
the force acting on an object is equal to the mass of that object multiplied by its
acceleration, 𝐹 equals 𝑚𝑎. So what does Newton’s third law
tell us?

Newton’s third law is most commonly
stated as for every action, there’s an equal and opposite reaction. But we’re not talking about the
type of reaction you have when, say, your sibling steals your favorite item of
clothing. It actually says that if object A
exerts a force on object B, then object B must exert a force of equal magnitude and
opposite direction on object A. As with Newton’s other laws, this
is quite intuitive. For example, during a rocket
launch, burning fuel releases gas at high velocity. The rocket exerts a downward force
on the gas in the combustion chamber. And so the gas exerts an equal and
opposite reaction force upward on the rocket. The special name for this is
thrust.

Another very important example
applies to you at this very second. Whether you’re sitting on a chair
or standing up, you are exerting a downward force on the chair or the floor. The chair or the floor is therefore
exerting an upwards reaction force on you. This is the very thing that’s
stopping you from falling straight through the chair or the ground, like in some
sci-fi film. And so we have Newton’s third law
of motion and some context behind it. Note that in classical mechanics,
this law very rarely stands alone, and for this reason, you must make sure that
you’re confident in applying the first two.

Let’s begin then by applying the
law in a very simple example.

An elevator is moving vertically
upward at a constant speed. A man of mass 150 kilograms is
standing inside. Determine the reaction force of the
floor on the man. Take 𝑔 to be equal to 9.8 meters
per square second.

With questions like these, it can
be really helpful to begin by drawing a very simple diagram. Here’s the lift moving upward at a
constant speed. Now, if the speed is constant,
acceleration must be equal to zero meters per square second. A man of mass 150 kilograms stands
inside the lift. And so this means that the man must
exert a downward force on the floor of the lift. Now, in fact, to find the value of
that downward force, we can use Newton’s second law, force is equal to mass times
acceleration. We call the downward force due to
the mass “weight”, 𝑤. And we say it’s equal to the mass
of the object multiplied by acceleration due to gravity, so 𝑚𝑔. This must mean that since the man
has a mass of 150 kilograms, the downwards force of weight is 150𝑔 on the floor of
the lift.

Then we recall Newton’s third law
of motion. And this says that if object A
exerts a force on object B, then object B must exert a force of equal magnitude and
opposite direction on object A. And so since the man exerts a
downwards force on the floor of the elevator, the floor of the elevator must exert
an upward force on the man itself. And that’s all the forces we’re
interested in for now. We aren’t interested in, for
example, tension forces in the cable for the lift. And so we’re going to go back to
Newton’s second law of motion, force is equal to mass times acceleration. We’re going to define upwards in
this case as being positive, and we’re going to find the net sum of the forces.

In the upwards direction, we have
𝑅, and then downwards, we have 150𝑔. This is acting in the negative
direction. And so the net force on our diagram
is 𝑅 minus 150𝑔. This is equal to mass times
acceleration, 150 times zero. 150 times zero is zero. So our equation becomes 𝑅 minus
150𝑔 equals zero. We can solve for 𝑅 by adding 150𝑔
to both sides, and we find 𝑅 is 150𝑔. But of course we were told 𝑔 is
9.8, so we substitute 𝑔 equals 9.8 into this expression. And we get 𝑅 equals 150 times 9.8,
and that’s 1470. We’ve worked with kilograms and
meters per square second, so the units for our force are newtons. The reaction force of the floor on
the man is 1470 newtons.

Now it’s worth noting that we could
have gone straight to this step. We didn’t actually need the first
line of working. It’s essentially the reverse of the
first law. We’re told that an object will
remain at rest or in uniform motion unless acted upon by an external force. In other words, if the sum of the
forces acting on the body is zero, then the object will remain at rest or in uniform
motion. Now we know that our object remains
in uniform motion, and so the sum of the forces did have to be equal to zero.

We’ll see why, though, in our next
example, it can be really helpful to use Newton’s second law most of the time.

An elevator was accelerating
vertically downward at 1.7 meters per square second. Given that the acceleration due to
gravity is 𝑔 equals 9.8 meters per square second, find the reaction force of the
floor to a passenger of mass 103 kilograms.

Let’s begin by sketching a
diagram. We have an elevator accelerating
vertically downward at 1.7 meters per square second. We know that a passenger in the
lift has a mass of 103 kilograms. This passenger must exert a
downward force on the lift, and we call that downward force weight, where weight is
mass times 𝑔 gravity. In this case then, since the
passenger has a mass of 103 kilograms, his weight is 103𝑔. Newton’s third law then tells us
that if an object exerts a force on a second object, that second object must exert a
force of equal magnitude and opposite direction on the first. And so the floor of the lift exerts
a reaction force on the passenger itself.

We now have all of the forces that
we’re interested in, and so we recall Newton’s second law of motion, force is equal
to mass times acceleration. We’re going to find the net sum of
the forces in our diagram, and we’re going to set that equal to the mass of the
passenger multiplied by acceleration. And so we’re going to need to
define a positive direction. Now, this is very much personal
preference. I prefer to take the positive
direction as the direction at which the object is moving. So in this case, I’m going to
assume that downwards is the positive direction. As long as we’re consistent,
though, it doesn’t matter which direction we choose.

So forces acting in the positive
direction, we have 103𝑔. In the negative direction, we have
𝑅, and so the net sum of the forces in this diagram is 103𝑔 minus 𝑅. That’s equal to the mass of the
passenger multiplied by acceleration, so 103 times 1.7. Remember, we’re trying to find the
reaction force of the floor to the passenger, so we need to make 𝑅 the subject. Let’s begin by adding 𝑅 to both
sides. We could, of course, have
multiplied by negative one instead, but we do need 𝑅 to be a positive. Then we’re going to subtract 103
times 1.7 from both sides, and so we get 103𝑔 minus 103 times 1.7 equals 𝑅.

It can be sensible to factor where
possible. This makes any mental calculation a
little easier. And the left-hand side then becomes
103 times 𝑔 minus 1.7. But, of course, 𝑔 is equal to 9.8,
and so since 9.8 minus 1.7 is 8.1, we find 𝑅 must be equal to 103 times 8.1. That’s 834.3. We have been working with kilograms
and meters per square second, and so the units for our force are newtons. The reaction force of the floor to
a passenger of mass 103 kilograms is 834.3 newtons.

Now, comparing this reaction force
to the weight force of the man might remind you of a time where you’ve been in a
lift yourself. The weight force 103𝑔 is 103 times
9.8, and that’s 1009.4. 1009.4 is greater than the reaction
force, and this actually makes a lot of sense. If you’ve ever been in a lift which
accelerates downwards, you almost feel a little bit lighter than normal. Similarly, if you’re in an elevator
accelerating upwards, you might feel a little bit heavier than normal. And this is because the reaction
force will be greater than the weight force.

In our next example, we’re going to
consider how to work with a spring balance.

A body of mass 32 kilograms was
suspended from a spring balance fixed to the ceiling of an elevator. Given that the elevator was
accelerating upward at 405 centimeters per square second, find the apparent weight
of the body. Take 𝑔 to be equal to 9.8 meters
per square second.

Let’s begin by sketching a
diagram. We have a body of mass 32 kilograms
suspended from a spring balance. That spring balance is fixed to the
ceiling of an elevator. And so there are a couple of forces
at play here. Firstly, there’s the downwards
force of the body. That force is the weight, and it’s
equal to mass times gravity, so 32 times 𝑔. Now, since the body itself is
exerting a downward force on the spring balance, Newton’s third law tells us that
the spring balance must also exert an upward force on the body itself. Let’s call that force tension, and
actually that tension will give us the reading and the apparent weight of the
body.

We know that the elevator itself is
accelerating upward at 405 centimeters per square second. But actually, since there are 100
centimeters in a meter, we’re going to divide this value by 100 to get 4.05 meters
per square second. And that’s because we’re currently
working in kilograms. We’ve got gravity in meters per
second, and so our final answer is going to be in newtons. And so we need to be consistent
with our units throughout. Now, actually, the spring balance
itself will exert a force on the ceiling of the elevator. But we’re not interested in that
system of forces. And so we move on, and we use
Newton’s second law of motion, force is equal to mass times acceleration.

We’re going to work out the net sum
of the forces in our system, and then we’re going to set that equal to the mass of
the body times the acceleration. And so we need to choose a positive
direction. Let’s take upwards to be positive,
since that’s the direction in which the body is moving. And since we have 𝑇 tension acting
upward and 32𝑔 acting in the negative direction, the net sum on our system is 𝑇
minus 32𝑔. That’s equal to the mass of the
body times acceleration. So that’s 32 times 4.05. We’re trying to find 𝑇 since that
will tell us the apparent weight of the body. And so we’re going to add 32𝑔 to
both sides. 𝑇 is 32 times 4.05 plus 32𝑔,
which is equal to 32 times 4.05 plus 32 times 9.8. That gives us a value of 443.2. The apparent weight of the body
then is 443.2 newtons.

In our final example, we’ll look at
how we can find acceleration by using a system of simultaneous equations.

A body is suspended from a spring
balance fixed to the ceiling of an elevator. The reading of the balance was 50
kilogram-weight when the elevator was accelerating upward at 𝑎 meters per square
second, and the reading was 10 kilogram-weight when the elevator was accelerating
downward at five-thirds 𝑎 meters per square second. Given that the acceleration due to
gravity is 𝑔 equals 9.8 meters per square second, determine the value of 𝑎.

Let’s begin by sketching a diagram
of each part of this scenario. Initially, the elevator is
accelerating upward at 𝑎 meters per square second. The reading of the balance at this
point was 50 kilogram-weight. Try not to worry too much about the
units here. Kilogram-weight is just another way
of measuring force, and so we simply write 50 as the reading on our balance, and
that’s essentially the upward force. That upward force is a reaction to
the downward force of the weight of the body. Now we don’t know the mass, so
we’ll let that to be equal to 𝑚. And the downward force is mass
times gravity; it’s 𝑚𝑔. We then have a reading of 10
kilogram-weight when the elevator is accelerating downward at five-thirds 𝑎 meters
per square second.

The downwards force of the weight
𝑚𝑔 remains unchanged. And so now we have the relevant
forces. Let’s use Newton’s second law,
force is equal to mass times acceleration, taking upwards to be positive in our
first diagram and the net force is 50 minus 𝑚𝑔. That’s equal to mass times
acceleration 𝑚𝑎. Then taking the downwards direction
to be positive in our second diagram, the sum of the forces is 𝑚𝑔 minus 10. That’s equal to mass times the new
acceleration, 𝑚 times five-thirds 𝑎. Note that we could have chosen
upwards to be positive. It really doesn’t matter. We’ll get the same result.

We rewrite the right-hand side as
five-thirds 𝑚𝑎, and the question tells us that we’re trying to find the value of
𝑎. So we’re going to make 𝑚 the
subject in each of our equations, and then we can set them equal to one another. In our first equation, we add 𝑚𝑔
to both sides, and then we factor. So we get 50 equals 𝑚 times 𝑎
plus 𝑔. Finally, we divide by 𝑎 plus 𝑔,
and so 𝑚 is 50 over 𝑎 plus 𝑔. In our second equation, we subtract
five-thirds 𝑚𝑎 and add 10. We factor once again, and we’re
going to divide by the expression inside our parentheses. 𝑚 is 10 over 𝑔 minus five-thirds
𝑎.

Now that we have two equations for
𝑚, we’re going to set them equal to one another. 50 over 𝑎 plus 𝑔 equals 10 over
𝑔 minus five-thirds 𝑎, and then we divide through by 10. We’re going to multiply by both
denominators. That gives us five times 𝑔 minus
five-thirds 𝑎 equals 𝑎 plus 𝑔. Let’s distribute the parentheses
and then make 𝑎 the subject. Now, of course, 𝑔 itself is a
number; it’s 9.8. Adding 25 over three 𝑎 to both
sides and subtracting 𝑔, and we get four 𝑔 equals twenty-eight thirds 𝑎. But of course, four 𝑔 is four
times 9.8, which is 39.2. Our final step is to divide both
sides of this equation by twenty-eight thirds. 39.2 divided by twenty-eight thirds
is 4.2, and so the acceleration 𝑎 is 4.2 meters per square second.

Let’s now recap the key points from
this lesson. In this lesson, we learned that
Newton’s third law of motion says that if object A exerts a force on object B, then
object B must exert a force of equal magnitude and opposite direction on object
A. We also saw that this law is very
rarely used alone. It will be used quite commonly
alongside the first and second laws for motion.