Video Transcript
In this video, we’re going to talk
about tension forces. These are forces that are
transmitted to objects through things like ropes or cables or chains. We’re going to learn about tension
forces under two different conditions: first, when objects are in equilibrium and,
second, when they’re accelerating.
Now, we mentioned that tension
forces are always transmitted forces. What we mean by that is they’re
always communicated by objects such as, for example, this rope here. Right now, this rope is fastened on
one end and free on the other. But if we were to pick up the free
end and pull on it, we would establish tension in this rope. We can better understand this
tension by considering the horizontal forces involved here.
First, there’s the force we’re
exerting. We can call that 𝐹 pull, and we
know that’s acting to the left. But we know since our rope isn’t
moving that this isn’t the only horizontal force acting on it. The wall the rope is tied to is
also exerting a force equal in magnitude, but opposite in direction to our pull. So our rope is in equilibrium, even
though forces are acting on it to stretch it out, and therefore it’s under
tension.
Tension can be a bit tricky to
understand because, as in the case of this rope, it acts at every point on the
rope. At any location along the rope
then, say this one here, the force of our pull and the force of the wall would be
transmitted along our rope, meaning that these two forces would effectively be
acting at that point or any other point along the rope. This is what we mean when we say
that tension involves the transmitting of forces. All of this may raise the question
though: then, what is the tension, we often represent it using a capital 𝑇, in this
rope? Is it equal to 𝐹 pull plus 𝐹 wall
or something else? This question points us to
something we’ve learned earlier related to Newton’s second law of motion.
Recall that this law tells us that
the net force acting on some object with a mass 𝑚 is equal to that object’s mass
multiplied by its acceleration. From this equation, we can arrive
at a couple of important implications. The first is that if an object’s
acceleration is zero, that is, it’s in equilibrium, then the net force on it must
also be zero. And then it’s equally true that if
an object has a nonzero acceleration, it also has a nonzero net force. These are basically the two
options. Either a system is or is not in
equilibrium. All of this is helpful as we study
tension because, for a given scenario, we can classify it either as equilibrium or
nonequilibrium. That will tell us about the forces
that may be involved.
Getting back to our example of this
rope, we can see that, in this case, we’re working with an equilibrium scenario. The rope is not accelerating; and
therefore, Newton’s second law tells us that the net force acting on it is zero. To see this a bit more clearly,
let’s draw a free-body diagram of the forces acting at this point in our sketch. If we model this location as a
point, we know that there is the force of the wall acting on that point to the
right. And because our system is in
equilibrium, we know that that’s not the only force. Along with this, there is the force
of the rope pulling to the left. That’s the tension force. And we see now that it must be
equal in magnitude but opposite in direction to the force of the wall on the
rope.
So in answer to our question, what
is the tension in this rope, the magnitude of that tension is equal to the magnitude
of the force of the wall on the rope, which is also equal to the magnitude of our
pulling force. Now, this partly answers our
question of what is the tension in the rope, but not completely. That’s because tension, like any
force, is a vector with both magnitude and direction. We’ve answered what the magnitude
of the tension is. But now, which way does it act?
Interestingly, the answer to that
question depends on where along the rope we’re looking. At the point where the rope is
fastened to the wall, the tension force acts to the left, opposing the force of the
wall on the rope. But if we created a free-body
diagram here, where our hands are grasping onto the rope, then in that case, we have
𝐹 sub pull to the left and tension to the right. So to answer the question of in
what direction the tension force acts, we need to be considering a specific point
along the rope. On the other hand, if we only wanna
solve for the magnitude of the tension force, then we can solve for that anywhere
and the answer is the same.
Now that we’ve looked at an
equilibrium scenario, let’s consider a different situation where the objects are not
in equilibrium. Imagine that we have two unequal
masses 𝑚 one and 𝑚 two and that they’re attached by a cord. And then say that another cord
attached to the other side of 𝑚 one is being pulled with a force 𝐹. Given all this and assuming that
the two cords have negligible mass and that the surface the masses move across is
perfectly smooth, we want to solve for the two tensions 𝑇 two and 𝑇 one. If this were an exercise question,
the first thing we would want to figure out is whether our objects are accelerating
or not. And that has to do with whether the
forces on an object balance out.
So what about 𝑚 one and 𝑚
two? We know that 𝑚 one experiences
this force 𝐹 which is mediated by this cord here. So then we ask ourselves, is there
any force acting on 𝑚 one to the left? We know that our surface is
perfectly smooth, so there’s no friction force involved. We do see, though, this cord here
with a tension we’ve called 𝑇 two. This cord will supply a
leftward-acting tension force on the mass 𝑚 one because it’s effectively dragging
along the mass 𝑚 two.
There’s something very important,
though, we must realize about this tension force 𝑇 two. This force, which is nonzero, is
what we can call an internal force. That is, if we think of our system
as these two masses and the two cords, then 𝑇 two pulls on one part of our system
in one direction and another part of the system in an opposite direction. And these pull magnitudes, the
strength of these forces, are equal. For the system then, there is only
one external force involved. And therefore, the net force on the
system is nonzero, and so it must accelerate. To figure out what the acceleration
of our system overall is, we must consider its overall mass and the overall external
forces acting on it.
Starting with the forces, if we
think only in the horizontal direction, that is, ignoring the force of gravity which
is vertical, then the only external force on this system, as we’ve said, is the
force 𝐹. But then how does that force act on
the system? Well, we see it’s communicated by
this cord, which is under a tension we’ve called 𝑇 one. The mechanism by which the force 𝐹
is actually exerted on the system then is via this cord, which is under this
tension. So we can say that tension is equal
to 𝐹, which, by Newton’s second law, is equal to the mass of our system, which is
𝑚 one plus 𝑚 two, multiplied by the overall system’s acceleration.
So, we now have an expression for
the tension force 𝑇 one. But recall we also wanted to solve
for the tension force 𝑇 two. We might be tempted to think that
𝑇 two is equal in magnitude to 𝑇 one. But if that were the case, then if
we were to draw a free-body diagram of the forces acting on 𝑚 one, those horizontal
forces would cancel out. It wouldn’t accelerate. We know, though, that 𝑚 one does
accelerate off to the right, as does 𝑚 two which is attached to it. To solve for 𝑇 two, which we
called an internal force when we considered our system to be the two masses and the
two cords, what we’re going to do is redefine for a moment what our system is.
For the moment, we’ll just consider
the mass 𝑚 two and the tension 𝑇 two. This, we can say, is our new
system. And we’ll apply Newton’s second law
to it. Thinking this way, the only
horizontal force involved is the tension force 𝑇 two. So that must be equal to the mass
of our system, which is just the mass 𝑚 two, multiplied by its acceleration. Now, notice that we’ve said that
the acceleration of the mass 𝑚 two is the same as that of the mass 𝑚 one. This is true because as 𝑚 one
moves and it’s connected by a cord to 𝑚 two, 𝑚 two must accelerate at the same
rate. So, considering these two
expressions for the two different tensions, notice that they’re not equal, but that
for masses connected by cords or ropes like this, the accelerations of the masses
are equal.
So far, we’ve only considered
tension forces that act in the horizontal direction. But we can also look
vertically. We can clear some space on screen
and then say we draw a line here which represents a ceiling surface. From that surface, we use a rope
with a negligibly small mass to suspend a mass 𝑚. If we call the tension in this rope
𝑇 one, what, we wonder, is that tension? Once again, we can start out by
considering whether this is an equilibrium or nonequilibrium scenario. Since our mass is just hanging
there stationary, this is an equilibrium scenario, which means the net force
involved on this mass must be zero. That’s helpful to know because
currently the only force we see acting on it is the tension force from this
rope.
If we model this mass as a dot, we
could see that tension force acting upward. So there must be some
downward-acting force that balances it. And indeed there is. It’s the weight force of this mass,
𝑚 times 𝑔. So here the tension force balances
out the force of gravity. But now, imagine that we add two
more identical masses to hang from the first one by two more massless ropes. If we do this, it’s no longer the
case that 𝑇 one is equal and opposite to 𝑚𝑔. In this updated situation, each of
the ropes involved has its own tension force. We’ve called them 𝑇 one, 𝑇 two,
and 𝑇 three.
To solve for these three tension
forces, one place to start is by considering only the lowest mass and the rope
attached to it. If we were to draw a free-body
diagram of just this mass, it would actually look very much like the one we drew
earlier. There’s the vertically downward
force of 𝑚 times 𝑔, the weight force acting on this mass. And then because this system is an
equilibrium, there’s an equal and opposite force, in this case, the tension force 𝑇
three. This means we can write that the
magnitude of the tension force 𝑇 three is equal to 𝑚 times 𝑔.
Knowing this, let’s know expand our
definition of the system to include the two lower masses. If we were to draw a free-body
diagram of the forces acting on this second-lowest mass, then we would have number
one the weight force of the second second-lowest mass 𝑚 times 𝑔. But then, along with that, we would
have the weight force of the very-lowest mass, which is also 𝑚 times 𝑔. And again, since our system is in
equilibrium, there must be an equal and opposite force acting upward. This, we see from our diagram, is
the tension force 𝑇 two. So, we can then write that the
magnitude of 𝑇 two is equal to two times 𝑚 times 𝑔.
And then finally, we can think of
the system of all three masses together. In this case, our free-body diagram
of the forces acting on the top-most mass here would include three vectors pointing
downward, each of magnitude 𝑚 times 𝑔. Here, the equal-in-magnitude but
opposite-in-direction force that balances them all out is 𝑇 one. The magnitude of this tension then
is three times 𝑚 times 𝑔. We see then that in studying
tension forces, two helpful tools are Newton’s second law of motion and free-body
diagrams. Knowing all this, let’s now get
some practice of these ideas through an example.
Two identical objects are connected
to each other by a rope as shown in the diagram. A second rope is connected to one
of the objects. The masses of the ropes are
negligible. A short time after a constant force
𝐹 is applied to the end of the second rope, both objects uniformly accelerate in
the direction of 𝐹 across a smooth surface. Tension 𝑇 one is produced in the
rope that the force is applied to, and tension 𝑇 two is produced in the rope that
connects the objects. Which of the following statements
correctly represents the relationship between 𝑇 one and 𝑇 two?
Before we consider these
statements, let’s take a look at our diagram. We see here two masses, which we’re
told are identical, connected by this rope. Then there’s another rope here,
which has a constant force 𝐹 applied to the end of it, pointing to the right. Under this influence, we’re told
that both objects accelerate uniformly in that direction. And note that there’s no friction
force opposing this acceleration because we’re told that the movement is across a
smooth surface. So we have two ropes, one under
tension 𝑇 one and the other under tension 𝑇 two. And we want to pick out what
relationship correctly describes them.
Here are our options. (A) 𝑇 one equals 𝑇 two, (B) 𝑇
one equals 𝑇 two divided by two, (C) 𝑇 one plus 𝑇 two equals zero, (D) 𝑇 one
equals two times 𝑇 two.
Now, here’s one way we can think
about this scenario. We essentially have a system where
that system consists of these two identical masses in the two ropes. We have an external force 𝐹 being
applied to the system and causing it to accelerate. This can remind us of Newton’s
second law of motion, which tells us that the net force acting on an object of mass
𝑚 is equal to that mass multiplied by the object’s acceleration. Now, in our case, if we think only
of forces acting in a horizontal direction, we can say that there’s one external
force acting on our system. That’s the force 𝐹. That force is transmitted through
the first rope and then pulls on the first mass, then transmitted through the second
and pulls on the second mass.
Effectively then, this force is
pulling our whole system, both masses and both ropes. Therefore, 𝐹 is equal to our
system’s mass times its acceleration. Now, we’re told that the two ropes
in our scenario are massless, but we’re not told the masses of these two
objects. We do know, though, that they’re
identical. So, just to give them a name, let’s
say that they each have a mass 𝑚. This means that the total mass of
our system is two times 𝑚. Again, the mass of our ropes is
considered zero. So, if we call the acceleration of
our two objects 𝑎, then we can say that 𝐹 is equal to two times 𝑚 times 𝑎.
But then, looking at our diagram,
we see that this force 𝐹 is being applied to the end of our first rope. And therefore, the tension in this
rope is equal to that applied force. This means we can write that two
times 𝑚 times 𝑎 is also equal to 𝑇 one. Because we don’t know 𝑚 or 𝑎, we
can’t go about calculating a numerical value for 𝑇 one. But all we want to do is compare it
to the other tension force 𝑇 two to arrive at an expression for that variable. Instead of considering our two
masses and the two ropes, let’s just consider the second mass and the rope under
tension 𝑇 two. Focusing in here, we can say that
𝑇 two is the only horizontal force acting on this second mass. And therefore, by Newton’s second
law, it’s equal to the mass of this object, which is 𝑚, times its acceleration
𝑎.
And note that this object’s
acceleration is equal to the acceleration of the system overall. This is because both of our masses
move together and accelerate equally. We see then that the tension force
𝑇 two is equal to this unknown quantity 𝑚 times 𝑎 and that the tension force 𝑇
one is equal to two times that same quantity. So, if we replace 𝑚 times 𝑎 here
with 𝑇 two, which is equal to that product, then we find that two times 𝑇 two is
equal to 𝑇 one. And we see that that corresponds to
answer choice (D). The correct relationship between
these two tension forces is that 𝑇 one is equal to two times 𝑇 two.
Let’s summarize what we’ve learned
about tension forces. In this lesson, we learned that
tension is a force transmitted along ropes, cords, strings, etcetera. Tension forces are a response to
other forces such as gravity or pulling on the end of a rope. Tension can arise in either
equilibrium — that is, acceleration equals zero — or nonequilibrium — that is,
nonzero acceleration — scenarios. And lastly, we saw that two helpful
tools in solving for tension forces are Newton’s second law of motion and free-body
diagrams.
