### Video Transcript

In this video, we will be looking
at how to distinguish between the weight of an object and other forces acting on the
object. The reason that we would want to do
this is because it is often very easy to confuse the weight of an object with an
overall or net downward force acting on the object. So let’s clear up some
misconceptions.

Let us start by recalling what we
mean when we talk about the weight of an object. Whenever we place an object, such
as this tennis ball, into a gravitational field, such as this gravitational field of
the Earth, that object experiences a force. And that force acts to pull the
object towards the centre of mass of whatever is generating the gravitational
field. In this case, the tennis ball
placed in the gravitational field of the Earth experiences this gravitational force
that pulls the tennis ball towards the centre of the Earth. This gravitational force is known
as the weight of the ball.

Now in a constant gravitational
field, this weight force is a constant as well. We can recall how to calculate the
weight of an object by remembering that the weight of the object is defined as the
mass of the object multiplied by the strength of the gravitational field. Which by the way is also sometimes
known as the acceleration due to gravity.

Now this weight force, 𝑊, as we’ve
called it, will continue to act on the object that’s placed in the gravitational
field for as long as that object is in the gravitational field. Closer to the surface of the Earth,
the gravitational field becomes almost a constant. And so we essentially treat the
Earth’s gravitational field close to the surface as a constant field.

And we can also recall that the
value of this gravitational field strength or acceleration due to gravity near the
surface of the Earth is 9.8 metres per second squared. So in Earth’s gravitational field,
near the surface of the Earth, the weight of any object is given by multiplying the
mass of that object by the gravitational field strength, which is 9.8 metres per
second squared.

Now there are a few misconceptions
about the notion of an object’s weight that we can clear up in this video. The first of these misconceptions
is the assumption that any downward net force on an object is the same thing as its
weight. This is definitely incorrect. The weight of an object is very
specifically the gravitational force exerted on that object. And that force is given by this
equation. 𝑊 is equal to 𝑚𝑔, where 𝑊 is
the weight, 𝑚 is the mass of the object, and 𝑔 is the gravitational field
strength, as we saw earlier.

To illustrate this point, let’s
think about a rocket. But let’s imagine that this rocket
is flying in a downward direction, that is, pointing towards the surface of the
Earth. And not only is it just falling
downwards, but its thrusters are actually firing and providing an extra downward
force onto the rocket. So what are the forces acting on
the rocket?

Well, first of all, we’ve got the
weight of the rocket, which is the gravitational force exerted on the rocket simply
because it’s in the gravitational field of the Earth. And we can call this force 𝑊. Secondly, there’s an additional
downward force on the rocket that is provided by the firing thrusters. And that force is also pushing the
rocket in a downward direction. So we’ll call this second force 𝑇
for the thrust from the thrusters.

And then, of course, if we wanted
to, we could also consider the air resistance on the rocket because this rocket is
flying through the atmosphere. And so it’s going to experience a
resistive force that acts in the opposite direction to its motion. Let’s call this resistive force
𝑅. And let’s assume that this
resistive force is relatively small.

Now in this situation, it is very
easy to assume that the weight of the rocket is the same thing as the overall or net
or resultant force acting on the rocket. And to find out the resultant force
acting on the rocket, we can first of all define arbitrarily that the rocket moving
in a downward direction means that it’s moving in the positive direction. And therefore, anything acting in
the upward direction is acting in the negative direction.

Then we can say that the net force
on the rocket, which we’ll call 𝐹 subscript net, is equal to firstly the weight of
the rocket. And that’s a positive force because
it’s acting in the downward direction. Plus the thrust on the rocket from
the boosters, once again a positive force. Minus the resistive force acting on
the rocket, because it’s acting in the negative direction. So this is an expression for the
overall force acting on the rocket.

And what this is equivalent to is
just one force acting on the rocket that’s equal to the net force acting on the
rocket. So we could replace these labels
for 𝑅, 𝑇, and 𝑊 and say that this entire situation is equivalent to the rocket
having just one force with a magnitude of 𝑊 plus 𝑇 minus 𝑅 acting on it. Because, to any of observer, that’s
exactly what this looks like.

The rocket’s acceleration downwards
will be at a rate given by Newton’s second law of motion, which tells us that the
net force on an object, 𝐹 subscript net, is equal to the mass of that object
multiplied by its acceleration. And so this rocket will not be
accelerating downwards at the same rate as the acceleration due to gravity.

Remember, we assumed that gravity’s
acceleration is 𝑔 is equal to 9.8 metres per second squared. This rocket will not be
experiencing that specific acceleration. And this is because the rocket has
other forces acting on it, that is, the thrust from the boosters and the air
resistance. But that does not mean that the net
force acting on the rocket is the same thing as its weight.

It’s very easy to imagine that,
just because the rock is accelerating downwards at a certain rate 𝑎, that the force
that causes this, which is actually the net force on the rocket, is equivalent to
the weight of the rocket. But as we’ve seen already, this is
not true. The weight of an object has a very
specific magnitude given by multiplying the mass of the object by the strength of
the gravitational field that the object is placed in. And therefore, any downward net
force on an object is not the same thing as its weight. So that’s one misconception about
weight dealt with. Let’s now look at another one.

This second misconception is that
objects that are stationary on Earth’s surface do not have weight. Now this statement once again is
most definitely not true. And this statement kind of links to
the previous one as well. It’s related to a misunderstanding
of the net force acting on an object as compared to the weight of an object.

Let’s think about an object such as
this block of wood, for example, sitting stationary on the Earth’s surface. Well, if this block of wood is
stationary, then we know that it has a velocity of zero. In other words, the block is not
moving. And if the block is not moving,
then there’s no way that it can be accelerating. So we can also say that the
acceleration of the block is zero.

But then if we recall Newton’s
second law of motion, which tells us that the net force on an object, 𝐹 subscript
net, is equal to the mass of that object multiplied by the acceleration it
experiences. And then we substitute in a value
of zero for the acceleration. Then we see that the entire
right-hand side becomes zero because we’ve got the mass of the block multiplied by
zero.

And then if the right-hand side is
zero, then the left-hand side must be zero as well. Therefore, the net force on this
object, our wooden block, is zero. Now this might lead us to believe
that the weight of the wooden block is zero because the block is just sitting
there. We don’t see the effects of the
weight of the block.

But remember, Newton’s second law
of motion refers to the net force on the object. And actually, for however long this
block is in the gravitational field of the Earth, there will be a gravitational
force acting on the block. The only difference here is that
because the block is sitting on the floor, the floor exerts an upward force onto the
block, which is normal to the surface of the Earth. And this force, which we’ll call
𝐹, is also known as a contact force. This force only occurs because of
the contact between the block and the surface of the Earth.

Now because the block is stationary
and therefore not accelerating, we know that this means that the net force on the
block is zero. Therefore, this force, 𝐹, which is
the normal force on the block, must exactly cancel out the weight of the block. And the fact that these two forces
cancel each other out means that the net force on the block is zero. And therefore, the block is not
moving.

So the idea is that the
gravitational force on the block is still acting. But in the situation where the
block is placed on the floor, this force is counteracted by the normal force. So the misconception is that
because the block is sitting on the floor, it’s stationary. We don’t see it falling through the
Earth’s surface or anything like that. And so we don’t see the effects of
the weight of the block. Hence, we assume that the weight of
the block doesn’t exist.

But the fact of the matter is that,
for however long any object with mass is placed into a gravitational field, that
object will experience a gravitational force, otherwise known as the weight of the
object. If we want our object which has
mass to have zero weight, then we need to take it away from any gravitational
fields. In other words, we need to take it
deep into outer space away from any stars or planets or galaxies that could produce
gravitational fields. And only then will our block of
wood have a weight of zero.

So now that we’ve cleared out these
misconceptions, let’s have a look at an example question.

A rock with a mass of 2.5 kilograms
is thrown vertically upward by an applied force of 60 newtons. What is the rock’s weight? What is the net vertically upward
force applied to the rock? What is the rock’s rate of upward
vertical acceleration just before it is released? What is the rock’s rate of upward
vertical acceleration just after it is released?

Okay, so in this question, we’ve
been told that we have a rock that is thrown vertically upward. We’ve been told that the rock is
thrown vertically upward by an applied force of 60 newtons. So we can say that the hand, for
example, doing the throwing of the rock is applying an upward force of 60
newtons. But as well as this, we need to
recall that the rock is going to have a gravitational force acting on it because the
rock is found in the gravitational field of the Earth.

Therefore, the rock is going to
have a weight force, which we’ll call 𝑊, acting on it. And the first part of the question
actually asks us to find the rock’s weight. So to do this, we can recall that
the weight of an object is found by multiplying the mass of that object by the
gravitational field strength of the Earth in this case. And from the question, we already
know the mass of the rock. And we can recall that the
gravitational field strength of the Earth is 9.8 metres per second squared.

Therefore, to find the weight of
the rock 𝑊, we’ll say that this is equal to 2.5 kilograms — that’s the mass of the
rock — multiplied by 9.8 metres per second squared, the gravitational field
strength. Now before we evaluate this, we can
see that we’re working in base units, kilograms for mass and metres per second
square for the gravitational field strength, otherwise known as the acceleration due
to gravity. Therefore, the weight of the rock
is going to be in its own base unit, which is the newton. And this is because weight is a
force.

And so the right-hand side of the
equation becomes 24.5 newtons. Hence, we can say that the weight
of the rock, the answer to the first part of the question, is 24.5 newtons.

Moving on then, we need to find the
net vertically upward force applied to the rock. In other words, we need to find the
overall force acting on the rock when the two individual forces acting on the rock
are the 60-newton upward force applied by the hand and the downward gravitational
force, otherwise known as the weight. And because in the question we’ve
been asked to find the net vertically upward force, let’s arbitrarily choose the
upward direction to be positive and the downward direction therefore will be
negative.

Now it’s of course worth noting
that we’re trying to find the net vertically upward force applied to the rock just
before the rock has been let go from the hand. Because once the hand lets go of
the rock, this 60-newton upward force will no longer be acting on the rock. However, because we’ve been asked
to find the net force, we therefore need to consider the two forces acting on the
rock. And this is just before the rock
has been released.

Hence, we can say that the net
force on the rock before it’s been released, which we’ll call 𝐹 subscript net, is
equal to the upward 60-newton force, which is positive because it’s pointing in the
upward direction, minus its weight, which we’ve seen is 24.5 newtons, because the
weight is acting in the downward direction. And so the overall or resultant or
net upward force is 35.5 newtons. And this is our answer to the
second part of the question. So let’s move on to the third
part.

What is the rock’s rate of upward
vertical acceleration just before it is released?

Well, like we said already, just
before the rock is released from the hand, it has two forces acting on it: the
upward 60-newton force and the downward weight force. And as we’ve already seen, the net
force on the rock because of this is 35.5 newtons upward. And that’s because this net force
is positive and we said the positive direction was the upward direction.

Then we can recall Newton’s second
law of motion, which tells us that the net force on an object is equal to the
object’s mass multiplied by the acceleration it experiences. Now in this scenario, just before
the rock is released from the hand, we already know the net force and we also know
the mass of the rock. Therefore, we can solve for the
acceleration of the rock. And coincidentally, because the net
force on the rock is in the vertically upward direction, this means that using our
equation will give us the rock’s rate of upward vertical acceleration, just as we
need.

So to solve for the acceleration,
let’s divide both sides of this equation by the mass of the rock 𝑚. Doing this means that the 𝑚 on the
right-hand side cancels. And what we’re left with is that
the net force on the rock divided by the mass of the rock is equal to the
acceleration experienced by the rock.

Then when we plug in our values, we
see that the acceleration experienced by the rock is equal to the net force, 35.5
newtons, divided by the mass, 2.5 kilograms. This evaluates to an acceleration
of 14.2 metres per second squared. And therefore, we found the answer
to the third part of our question. So we can move on to the final
part.

What is the rock’s rate of upward
vertical acceleration just after it is released?

Well, just after the rock is
released, the hand is no longer exerting that upward 60-newton force on the
rock. Therefore, the only force now
acting on the rock is the weight of the rock. And so the new net force acting on
the rock is simply the weight of the rock.

Writing this mathematically, we can
say that the net force on the rock, 𝐹 subscript net, comma after the rock has been
released from the hand is now simply given by the weight of the rock. But remember, the weight of the
rock is acting in the downward or negative direction. Therefore, we can say that the net
force on the rock is equal to negative 𝑊.

But then we know that the weight of
an object is defined as the mass of that object multiplied by the gravitational
field strength. Therefore, using that equation, we
can say that negative 𝑊 is equal to negative the mass of the rock multiplied by the
gravitational field strength of Earth. And we also know that the net force
on an object at any time is given by multiplying the mass of that object by the
acceleration experienced by the object. And the effect of all of this is to
realise the mass of the rock multiplied by its acceleration is the same thing as
negative the mass of the rock multiplied by the gravitational field strength of
Earth.

Therefore, we can get rid of
anything else in the middle of this equation. And we can say that 𝑚𝑎 is equal
to negative 𝑚𝑔. Then we can divide both sides of
the equation by the mass of the rock, 𝑚. And this results in 𝑚 cancelling
on both sides. And so what we’re left with is that
the acceleration experienced by the rock just after the rock is released from the
hand is equal to negative 𝑔 or negative 9.8 metres per second squared. And that is our answer to the very
final part of this question.

Okay, so now that we’ve had a look
at an example question, let’s summarise what we’ve talked about in this lesson. Firstly, we’ve seen that an
object’s weight is given by finding the product between its mass and the
acceleration due to the gravitational field it is placed in. In other words, the weight of the
object is equal to the mass of the object multiplied by the gravitational field
strength. 𝑊 is equal to 𝑚𝑔.

Secondly, we’ve seen that the
gravitational force acts on an object for however long it is in the gravitational
field. In other words, the weight of the
object doesn’t stop existing just because the object is not moving, for example,
when the object is stationary on the surface of the Earth.

And finally, we’ve seen that the
weight of an object is not necessarily the same as the net vertically downward force
on the object. Now the weight of the object can be
the same thing as the net vertically downward force on the object. But that’s only true if the weight
is the only force acting in the upward or downward direction. However, in general, the net force
downwards on the object is not the same thing as its weight.