Video Transcript
In this video, we will learn about
Isaac Newton’s famous second law of motion and apply it to find the accelerations
caused by forces that act in various directions. So let’s start by first
understanding what a force actually is. Often intuitively thought of as a
push or a pull, a force is an interaction that causes a change in the motion of an
object. Of course, this assumes that the
force in question is not balanced or cancelled out by another force. But we’ll get to that
momentarily.
So let’s now see how a force
specifically causes a change in the motion of an object. This is when Newton’s second law of
motion comes into play. Let’s first consider a block of,
say, wood. Let’s say that this block of wood
is sitting on the floor and that this block can slide about freely on the floor. Let’s also arbitrarily say that the
mass of the block of wood is 𝑚.
Now this block of wood with mass 𝑚
is currently just sitting on the floor basically doing nothing. However, let’s imagine that we now
push the block of wood from the left. In other words, we exert a force —
let’s call this force 𝐹 — on the block of wood. Well, that force exerted on the
block of wood is going to result in an acceleration of the block of wood. In other words, the block is going
to stop being stationary and it’s going to start moving in the direction of the
force. This is very important. And let’s say that the block of
wood will have an acceleration, which we will call 𝑎.
Well, in this situation, Newton’s
second law of motion will tell us the relationship between the force 𝐹, the mass
𝑚, and the acceleration 𝑎. The relationship is that the
acceleration experienced by the block of wood, 𝑎, multiplied by the mass of that
block of wood is equal to the force exerted on the block of wood. In other words, 𝐹 is equal to
𝑚𝑎. And this is Newton’s famous second
law of motion. In other words, it gives us the
relationship between the force exerted on an object, the mass of the object, and the
acceleration experienced by the object due to the force.
However, there is one thing that we
need to be extremely careful of here. The force in question is actually
the net or overall force on the object. In this scenario where we
considered our block of wood sitting on the floor, we were only exerting one force
on the object. That force was force 𝐹. And so the net or overall or
resultant force on the object was just the force 𝐹.
However, if we now say that the
force we exert on the block of wood is called 𝐹 subscript one and we also introduce
the concept of friction, in other words, the fact that the floor will exert a force
on our wooden block in order to try and oppose the motion of the block. So let’s say that the fact that the
block of wood and the floor are in contact with each other and the block of wood is
trying to move this way to the right as we’ve drawn it. Then the floor is going to exert
another force on the base of the block of wood to try and oppose this motion. Let’s call that force 𝐹 two. And this is a friction force.
Well, in this case, we can see that
now we’ve got a large force to the right on the block, which is 𝐹 one, and we’ve
got a slightly smaller force, 𝐹 two, to the left. So in this case, what’s the
relationship between the mass of the wooden block, the acceleration of the wooden
block, and the forces acting on the wooden block, 𝐹 one and 𝐹 two?
Well, Newton’s second law of motion
tells us that 𝐹 is equal to 𝑚𝑎. But like we said earlier, 𝐹 has to
be the net or overall or resultant force on the object. In other words, let’s say we were
trying to calculate the acceleration of the block. What we’d have to do is to
rearrange the Newton’s second law equation by dividing both sides by the mass 𝑚,
which would give us 𝐹 divided by 𝑚 is equal to 𝑎. In other words, the force — we’ll
come back to that in a second — divided by the mass of the block is equal to the
acceleration experienced by the block.
So what exactly is the force that
we’re talking about? Is it the force 𝐹 one? Or is it the force 𝐹 two? Well, actually, it’s a combination
of the two. Like we said earlier, we need to
consider the net force on the block. And the way to do this is to add
the forces together as vectors. In other words, the force 𝐹 one we
can say is acting towards the right on our object, whereas 𝐹 two is acting in the
opposite direction.
Well, if this is the case, then the
resultant force on the object is going to be something like this, where the size or
magnitude of that force is 𝐹 one minus 𝐹 two. This is because 𝐹 two is acting to
oppose the force 𝐹 one. And so, overall, the object is
going to experience a force 𝐹 one minus 𝐹 two to the right. And that is the force 𝐹 that we’re
talking about in this equation. It’s the net or resultant
force. And so we’re actually best off
placing a subscript net in this equation. This way, we’ll remember that
Newton’s second law deals with the net force on an object and that that net force is
equal to the mass of the object multiplied by the acceleration experienced by the
object.
Of course, if there’s only one
force acting on the object in question, then the net force simply is going to be
that force. However, if we’re dealing with
multiple forces, then we have to add all of those forces acting on the object as
vectors, in other words making sure we account for the direction in which they’re
acting. And hence, we find the net force on
the object, which allows us to use Newton’s second law of motion.
It’s also worth quickly noting by
the way that the forces acting on an object don’t necessarily have to be acting on
the same line of motion. Because the situation that we’ve
drawn here we’ve got a force acting to the right and a force acting to the left.
It is, however, entirely possible
that we could place another force acting upwards, for example, if somebody tries to
lift the block of wood off the floor. Let’s call that force 𝐹 three. Well, in that case, what we’d have
to do is to add all three of the forces as vectors. Let’s simplify our wooden block
just to a little blob and think about the force 𝐹 one that’s acting to the right,
the force 𝐹 two that’s acting to the left, and the force 𝐹 three that’s acting
upwards.
We can start by vectorially adding
𝐹 one and 𝐹 two so that the resultant force is something like this, which is 𝐹
one minus 𝐹 two, to the right. And then we can combine this force,
𝐹 one minus 𝐹 two, with 𝐹 three, the resultant of which is going to look
something like this. And to find out the size of that
blue vector, we would have to use Pythagoras’s theorem, with 𝐹 one minus 𝐹 two as
one of the shorter sides and 𝐹 three as the other shorter side. But the point is that we need to
consider all of the forces and all of the directions in which all of those forces
are acting in order to find the net force on an object. This is because Newton’s second law
of motion deals with the net force on an object.
So now that we’ve looked at
Newton’s second law of motion in some detail, let’s get a little bit of practice
using this law by looking at a couple of example questions.
How much force is applied to an
object of mass five kilograms that is accelerated by that force at a rate of two
metres per second squared?
So in this question, we’ve got
an object which has a mass — let’s say 𝑚 — of five kilograms and is accelerated
— let’s say it’s accelerated to the right — because we can randomly choose a
direction. And we’ve been told that this
acceleration, which we’ll call 𝑎, is two metres per second squared. We’ve been asked to find the
force applied to this object.
Now because the acceleration
we’ve said is to the right, this must mean that the force exerted is also to the
right, since the force and acceleration must be in the same direction. And we’ll call this force
𝐹.
Now to find the value of this
force, we need to recall Newton’s second law of motion. This law tells us that the net
force on an object, 𝐹, is equal to the mass of that object multiplied by the
acceleration it experiences. So with this equation, we’ll be
able to find the net force on the object. And this is exactly what we’ve
been asked to find. Even though the question
doesn’t necessarily say find the net force, we’ve been asked to find how much
force is applied to an object.
Now this may imply that there’s
only one force acting on the object, in which case that force is the net force
anyway. Or there could be multiple
forces acting on the object, but the resultant force or the net force is going
to be a combination of all of those forces, whilst accounting for the direction
in which they act. And that overall combination
can also be thought of as the amount of force applied to an object. And so we can use Newton’s
second law of motion to find our answer to this question.
We can therefore say that the
force on the object, the net force, is equal to the mass, which is five
kilograms, multiplied by the acceleration of the object, which is two metres per
second squared. Now taking a quick look at the
units, we see that we’ve got kilograms and metres per second squared. And those are the base units of
mass and acceleration, respectively. Therefore, when we find our
final answer for the force, it’s going to be in its own base unit as well.
Now the base unit of force is
the newton, which is equivalent to a kilogram multiplied by metres per second
squared. And so when we evaluate our
answer — that’s five kilograms multiplied by two metres per second squared — we
find that the force is 10 kilograms metres per second squared or 10 newtons. And hence, our answer to this
question is that the force applied to this object is 10 newtons.
Okay, so let’s look at another
question where, this time, we have to consider multiple forces acting on an
object.
A swimmer of mass 45 kilograms
uses her legs to push herself away from a swimming pool wall, applying a force
of 280 newtons. The water that the swimmer
accelerates through applies 160 newtons of force in the opposite direction to
that in which she accelerates. What is the swimmer’s
acceleration through the water?
Okay, so in this question, we’ve
got a swimmer moving through some water. So here’s our swimmer. And we’ve been told that she uses
her legs to push herself away from a swimming pool wall. So let’s say that this is the wall
that she’s pushing off from. And just for fun, we can draw in
the surface of the water.
Anyway, so we’ve been told that the
swimmer firstly has a mass of 45 kilograms. Let’s label her mass 𝑚 and say
it’s 45 kilograms. We’ve also been told that the
swimmer pushes herself off the wall, applying a force of 280 newtons. In other words, the swimmer is
exerting a 280-newton force on the wall. But what this means is that we can
recall Newton’s third law of motion.
What Newton’s third law of motion
tells us is that if object A exerts a force on object B, then object B exerts an
equal and opposite force on object A. Now in our situation here, the
swimmer is object A. It’s exerting a 280-newton force on
object B, which is the wall. Therefore, Newton’s third law of
motion tells us that the wall, object B, will exert an equal, 280 newtons, and
opposite, to the right as we’ve drawn it, force on the swimmer, 280 newtons.
Now this is important because, as
well as this 280-newton force on the swimmer to the right as we’ve drawn it, we’ve
been told that the water exerts a 160-newton force on the swimmer in the opposite
direction to that in which she accelerates. Well, the swimmer is accelerating
to the right because she’s trying to push away from the wall. And of course, once again, we’ve
just arbitrarily drawn the wall to the left of the swimmer. We could’ve equally drawn the
swimmer going in the opposite direction and the wall to the right. But the point is that the swimmer
is trying to accelerate away from the wall.
And so we can say that her
acceleration is to the right as we’ve drawn it. And we’ll call this acceleration
𝑎. But then the water we’ve been told
exerts a 160-newton force on the swimmer in the opposite direction to which she’s
accelerating. This means that there are now two
forces acting on the swimmer: the 280-newton force to the right because of the wall
and 160-newton force to the left because of the water.
We can simplify all of this by
drawing the swimmer just as a blob and considering that there’s a 280-newton force
to the right and a 160-newton force to the left. Then we can account for these two
forces and find the resultant or overall force on the swimmer. This overall force is also called
the net force. And if we consider arbitrarily that
the direction to the right is positive and the direction to the left is negative,
then we can add these two forces together.
And the net force, which we’ll call
𝐹 subscript net, is equal to the positive 280 newtons, cause that force is to the
right, minus the 160 newtons, which is to the left. And that net force, therefore, ends
up being 120 newtons. Therefore, it’s a 120-newton force
to the right. And this kind of makes sense. If there’s a 280-newton force to
the right and a 160-newton force to the left, then 160 of those 280 newtons to the
right are being cancelled by this force to the left. And whatever remains — that’s the
120 newtons — is going to be the resultant force to the right.
If, however, we had arbitrarily
chosen the left to be positive and the right to be negative, then we’d have found
the net Force, 𝐹 subscript net, to be the positive 160 newtons minus the 280
newtons. And that would’ve given us a net
force of negative 120 newtons. So what gives?
Well, the reason for this is
because we chose this way, to the left, to be positive. And hence, what we found is that
the net force is negative 120 newtons to the left. Or in other words, it’s 120 newtons
to the right, exactly as what we found here. So even though we do need to be
careful with signs, regardless of which convention we chose, we’d find the net force
to be 120 newtons to the right.
So now that we found the net force
on our swimmer, we can recall Newton’s second law of motion. This law tells us that the net
force on an object is equal to the mass of that object multiplied by the
acceleration it experiences. And since we’ve already found the
net force and we’ve been given the mass of the swimmer, we can find out the
acceleration experienced by the swimmer.
We can do this by rearranging the
equation by dividing both sides of the equation by the mass, which results in the
mass cancelling out on the right. This way, what we’re left with is
that the net force on the swimmer divided by the mass of the swimmer is equal to the
acceleration experienced by the swimmer. So we can say that the
acceleration, 𝑎, is equal to the 120-newton net force to the right divided by the
mass of the swimmer, which is 45 kilograms.
We can also see that we’re working
in base units because the base unit of force is the newton and the base unit of mass
is the kilogram. And so our answer for the
acceleration is going to be in its own base unit, which is metres per second
squared. And so when we evaluate this
fraction on the right-hand side, we find that the acceleration is equal to 2.6
recurring. And the unit is metres per second
squared.
Now because the smallest number of
significant figures that we’ve been given in the question is two significant
figures, we can give our final answer to two significant figures as well. Now 2.6 recurring is basically two
point and then six occurring many times. And the reason we write it like
this is so that we can work out that this is the first significant figure and this
is the second.
Well, the second significant figure
is either going to stay the same or it’s going to round up, depending on what the
next significant figure is. The next one is a six. Since six is larger than five, this
second significant figure is going to round up. And so our value becomes 2.7 metres
per second squared. So at this point, we have our final
answer. The swimmer’s acceleration is 2.7
metres per second squared.
All right, so now that we’ve had a
look at a couple of examples, let’s summarise what we’ve learnt this lesson. We’ve learnt that the net force on
an object is given by multiplying the mass of that object by the acceleration
experienced by that object. It’s important that the force we’re
discussing is the net force. And it’s also important to remember
that because the net force and the acceleration are both vectors, we need to account
for the directions. The direction in which the net
force on an object acts is the same as the direction in which the object will
accelerate. And that is Newton’s second law of
motion.
