Video Transcript
In this video, we will learn how to
use Newton’s second and third laws of motion together in order to analyse systems of
unbalanced forces. So let’s start by recalling
Newton’s second and third laws.
So let’s start by looking at
Newton’s second law. This law tells us that the net
force on an object is equal to the mass of the object multiplied by the acceleration
experienced by the object. In symbols, we often write this
equation as 𝐹 subscript net, the net force on an object, being equal to the mass of
the object, 𝑚, multiplied by the acceleration of the object, 𝑎. And it’s worth noting that Newton’s
second law doesn’t need refer to the net force on the object, not just any forces
acting on the object.
For example, let’s imagine that
we’ve got a box here sitting on the floor. Now in this situation, there’s
bound to be a few forces acting on the box. Firstly, we know that the box’s
weight is going to act in a downward direction. And secondly, we know that there’s
going to be an upward contact force from the floor acting onto the box. Now these two vertical forces
cancel each other out. So if these are the only two forces
acting on the box, then the net force on the box is zero. Even though there are individual
forces acting on the box, they are all completely cancelled out by other forces. And hence, because the net force on
the box is zero, the acceleration of the box is also going to be zero, regardless of
what the mass of the box is.
However, if we were to now add a
rightward-acting force, in other words if somebody was to push the box towards the
right, then this force we can see is not balanced by another force acting towards
the left. And so, in this case, there is a
resultant force or unbalanced force acting on the box. And this would even be true if we
had a small leftward-acting force on the box. When we say small, we mean smaller
than the magnitude of the rightward-acting force. Because now even though there is a
leftward-acting force trying to counter the rightward-acting force, the magnitude or
size of the leftward-acting force is not large enough. And so, there is a resultant force
acting towards the right. And hence, there is a net force
acting on the box. And therefore, the box will
accelerate in the direction of the net force, which in this case is towards the
right. So that is Newton’s second law of
motion.
What about Newton’s third law
then? Well, Newton’s third law says that
when an object — let’s say object 𝐴 — exerts a force on another object — let’s say
object 𝐵 — object 𝐵 exerts an equal and opposite force on object 𝐴. Now when we say equal and opposite,
we mean equal in magnitude or size but opposite in direction. In other words then, if we think
about a box sitting on the floor once again and this time we have a person pushing
our box towards the right with let’s say a force which has a magnitude 𝐹. Then because of Newton’s third law
of motion, the box will exert an equal in magnitude but opposite in direction force
on the person doing the pushing. In other words, on the person’s
hands, the box will exert a force of magnitude 𝐹 towards the left in the opposite
direction. So that in a nutshell is Newton’s
third law of motion.
Now Newton’s second law and
Newton’s third law can be used in conjunction with each other. A good example is when we consider
the collision between two objects. So let’s say here’s our first
object and here’s our second object. Let’s also say that the first
object is moving towards the second object, whereas the second object we see to be
stationary. Now eventually, the first object
arrives at the second object and they collide. And in this situation, let’s say
that the second object exerts a force to the left on the first object. And this is the force due to the
collision. Let’s call this force 𝐹.
Well then by Newton’s third law of
motion, we can say that because this object exerts a force on this object, that
means that the first object will exert a force on the second object with the same
magnitude but in the opposite direction. In other words, the first object
will exert a force towards the right with magnitude 𝐹 on the second object. And since we’re ignoring all other
forces, for example, we’re not even considering gravity here, we know that the only
force acting on the second object is the force 𝐹. And this means that the force 𝐹
towards the right on the second object is equal to the net force acting on the
second object because there are no other forces. And so that overall resultant or
net force on the second object is simply the force 𝐹 towards the right.
Then, we can use Newton’s second
law of motion which tells us that the net force on an object is equal to the mass of
the object multiplied by its acceleration. And in this particular case for the
orange object, we know that the net force is 𝐹. And we know that this is equal to
the mass of the orange object which we will call 𝑚 subscript 𝑜 multiplied by the
acceleration experienced by the orange object which we will call 𝑎 subscript
𝑜. And so now we know that because the
blue object collides with the orange object, the orange object is going to
accelerate towards the right with acceleration 𝑎 subscript 𝑜. In other words, the acceleration is
going to be in the same direction as the net force on the object. And so, that is how we use Newton’s
second and third laws in conjunction to analyse objects and in general systems with
unbalanced forces acting. Now the best way to learn about
this is to attempt a few example questions. So let’s go about doing that.
Three demolition workers push on a
wall. The workers push with forces of 80
newtons, 50 newtons, and 60 newtons on the same side of the wall parallel to each
other. What is the total force acting on
the wall?
Okay, so in this question, we are
considering a wall. So let’s say that this is the wall
that we’re talking about and we’re looking at it from above. In other words, we’ve got a bird’s
eye perspective. Now, we’ve been told that three
demolition workers are pushing on this wall. We’ve been told that the first
demolition worker — let’s say they are pushing at this point here — pushes with a
force of 80 newtons. The second demolition worker we’ve
been told pushes with 50 newtons of force. And the third one we’ve been told
pushes with 60 newtons. So those are the three forces
acting on the wall.
Now the reason that we can draw
them like this is because we’ve been told that the forces are exerted on the same
side of the wall and parallel to each other. And because they’re all pushing
forces, this means that they must all be pointing in the same direction. What direction that is doesn’t
matter as long as we draw them pointing all the same way. In other words, we could have drawn
the wall like this for example and said that the workers were pushing in this
direction. But as long as we drew all three
forces acting in the same direction, then it would have been fine. So anyway, we’ve been asked to find
the total force acting on the wall. In other words, we need to combine
these three forces acting on the wall to find one overall or resultant or net force
acting on the wall.
To do this, we simply add up all of
the forces, whilst also remembering that forces are vector quantities. We can recall that a vector
quantity is one which has magnitude or size and direction. And as we can see from our three
labels, each one of the forces has a certain magnitude: 80 newtons, 50 newtons, and
60 newtons. And each one also has a specific
direction. In this particular case, they’re
all pointing in the same direction which actually makes life easier for us. Because if they’re all acting in
the same direction, then to find the total force acting on the wall, we simply add
them up.
If, however, one of the forces was
acting in the opposite direction, then we would need to make sure that we label
these forces correctly. In other words, if we said that
forces acting to the right were positive, then any forces acting towards the left
would have to be labelled as negative. But in this case, we don’t have to
do that. Instead, we can just say that the
total force acting on the wall which we will call 𝐹 subscript total is equal to 80
newtons plus 50 newtons plus 60 newtons. And that ends up being 190
newtons. In other words then, this wall
having three demolition workers pushing on it with magnitudes of 80 newtons, 50
newtons, and 60 Newtons, respectively, all in the same direction is equivalent to
the wall having one net force of 190 newtons acting in the same direction.
Now in this question, we haven’t
been told what that direction is, just that the three forces acting on the wall are
acting in the same direction. And so, in order to answer our
question, we only need to give the magnitude of the total force. And as we’ve seen already, this is
190 newtons. So we’ve just answered the question
where we find the net force or total force acting on an object.
Let’s now look at another example
question.
A piano of mass 300 kilograms is
moved along a horizontal surface by a person pushing from one side and another
person pulling from the other. The piano accelerates at 0.25
meters per second squared. The pushing force is 120
newtons. And a friction force of 60 newtons
acts in the opposite direction to the piano’s velocity. What is the pulling force? Consider the direction in which the
piano moves to be the positive direction.
So let’s say that this is the piano
that’s being moved and this is the floor. Let’s also say that the piano is
moving in this direction, towards the right. And the way that this is happening
is that one person is pushing from one side and another person is pulling from the
other side. So in the question, we’ve been told
that the force exerted by the person who’s pushing — in other words the force
exerted on this side — is equal to 120 newtons. And we’ve also been told that
there’s a 60-newton friction force acting in the opposite direction to the piano’s
velocity.
Based on this information as well
as the fact that the piano has a mass of 300 kilograms and the fact that it
accelerates at 0.25 meters per second squared, we need to find the pulling
force. In other words, we can say that the
force exerted by the person pulling is 𝑥 and that is what we’re trying to find
here. Now to do this, we can start by
recalling Newton’s second law of motion. Newton’s second law tells us that
the net force on an object is equal to the mass of the object multiplied by the
acceleration experienced by the object.
In this case, the object in
question is our piano. And luckily, we already know the
mass of the piano, 300 kilograms. And we know its acceleration, 0.25
meters per second squared. So using this equation for Newton’s
second law, we can calculate the net force on the piano. In other words then, we can say
that the net force on the piano is equal to the mass, 300 kilograms, multiplied by
the acceleration, 0.25 meters per second squared. And because we’re working in base
units for mass and base units for acceleration, our answer will have the base units
of force which is newtons. So when we evaluate the right-hand
side of this equation, we find that the net force on the piano is 75 newtons. And that net force must be acting
towards the right.
Now the reason for this is that
we’ve been told that the direction in which the piano moves is the positive
direction. In other words then, the direction
towards the right is positive. And therefore, anything towards the
left is negative. And in our calculation, we took the
acceleration of the piano to be positive. We said it was 0.25 metres per
second squared and not negative 0.25 meters per second squared. And based on that calculation, we
got a net force of 75 newtons, which is also a positive value. And so, the net force 𝐹 subscript
net will also be acting towards the right as we’ve drawn this diagram.
Now the net force on the piano is
the overall force acting on the piano when we account for all of the individual
forces acting on the piano. In other words then, we can say at
𝐹 subscript net, the net force on the piano is equal to firstly the pushing force,
which we will call 𝐹 subscript push, plus the pulling force, which we will call 𝐹
subscript pull, plus the frictional force, which we will call 𝐹 subscript
friction. But we need to be very careful here
when we substitute in all our values because remember the 60-newton friction force
acts in the opposite direction to the piano’s velocity.
And so, the value of 𝐹 subscript
friction, the frictional force, is actually negative 60 newtons. And hence, we can say that the net
force on the piano is equal to 𝐹 subscript push, the pushing force which is 120
newtons, plus 𝐹 subscript pull, the pulling force which is 𝑥 and that’s what we’re
trying to find, plus the frictional force, which we’ve said is negative 60
newtons. And so, even though we’re adding
the frictional force, we’re actually adding a negative value. And so, we’ve accounted for the
directionality of the frictional force. And so, what we find on the
right-hand side is 120 newtons minus 60 newtons plus 𝑥 which altogether ends up
being 60 newtons plus 𝑥.
But then, this expression on the
right-hand side is equal to the net force on the piano which we’ve already found is
75 newtons. And so, we can say that 60 newtons
plus 𝑥 is equal to 75 newtons. And when we equate the two of the
equation, it looks something like this. Then, we can rearrange simply by
subtracting 60 newtons from both sides of the equation because this way the positive
60 newtons cancel with the negative 60 newtons. And we’re left with just 𝑥 on the
right-hand side. And on the left, we’re left with 75
newtons minus 60 newtons. At which point, we arrive at our
answer. We find that the pulling force, the
force exerted by the person pulling on the piano, is 15 newtons. And that is the answer to our
question.
So now that we’ve had a look at a
couple of examples, let’s summarise what we’ve talked about in this lesson.
Firstly, we’ve seen that an
unbalanced force acting on an object means that there is a net force on the
object. Well, of course, an unbalanced
force means that the overall force acting on the object is not zero. And this is when we’ve accounted
for all of the forces acting on an object. Now, if we have an unbalanced force
acting on an object or a nonzero net force acting on the object, then the object
itself will accelerate according to Newton’s second law of motion. This law tells us that the net
force on the object is equal to the mass of the object multiplied by its
acceleration. And secondly, we saw that we can
use Newton’s second and third laws of motion as necessary. So in other words, we don’t always
have to use both. Sometimes, we’ll use one, sometimes
the other, and sometimes both. And we can use these laws as
necessary to analyse systems of unbalanced forces.
