Main content

# Moments

Introduction to moments. Created by Sal Khan.

## Want to join the conversation?

- why is F capitalized and D is not?(30 votes)
- Its just the symbols for force it is 'F' and for displacement it is 's' or 'd'(7 votes)

- what is the difference between moment and torque? are the both same ?(32 votes)
- Moment refers to inertia, ie. rotational mass. Depending on the axis of rotation, you have different moments for the same mass. Thus inertia is a tensor quantity.(21 votes)

- At09:55, he said counter-clockwise, drawing an arrow depicting clockwise. Shouldn't it point the other way?(22 votes)
- It's also ok because if he consistently gets all of them backwards, he'll get the right answer again. :)(21 votes)

- Can somebody tell me what exactly is Axis of Rotation?

Thanks in advance.(6 votes)- It is an
**imaginary**line around which the object is rotating. For instance in a door, the axis of rotation is the line that passes through the hinges. When we speak about the rotation of the Earth, is the line that passes through the Poles.(26 votes)

- ok, this is kind of a stupid question, but he said that F1d1=F2d2. But that's only when it is in equilibrium, right?(5 votes)
- That's really the same thing as saying that they're still in equilibrium with all three forces. If the system isn't in equilibrium, you can solve for the net torque if you know the angular acceleration it causes as well as the rotational inertia of the object (T=Ia (not the same "a")). Think of this as the direct equivalent of typical translational forces, where F=ma.(2 votes)

- how to calculate the reaction force if you have many forces acting on one object(4 votes)
- You have to use vector analysis to figure out what the net force is.(5 votes)

- A SPHERE cannot roll on a SMOOTH INCLINED SURFACE . Why ?

Please reply soon .

Thank you .(8 votes) - What is a Pseudo-vector?It is mentioned by Sal Sir at4:16.(4 votes)
- A vector-like object which is invariant under inversion is called a pseudovector, also called an axial vector (as a result of such vectors frequently arising as vectors describing rotation. Hope This Works!(2 votes)

- what if we assume that F, the unknown force is from a clockwise direction.... will our answer be wrong then??(3 votes)
- It doesn't really matter whether it is clockwise or anti-clockwise (counter clockwise) but it is typical to do moments in the anti-clockwise direction. You simply need to decide in which direction you want to solve for and then evaluate your result.

Let's say you wanted to solve for anti-clockwise direction, and the answer was 5N. If instead you for the clockwise direction then your answer would be -5N (negative) indicating that the force is actually in the opposite direction.(3 votes)

- In general, when you are balancing torque you are balancing forces acting in OPPOSITE DIRECTIONS and not on OPPOITES SIDES OF THE PIVOT POINT , right?(2 votes)
- You have to consider both the direction of the force and position with respect to the pivot. A force with the same direction but on opposite sides of the pivot will produce torques of different directions.(4 votes)

## Video transcript

Welcome to the presentation
on moments. So just if you were
wondering, I have already covered moments. You just may not have recognized
it, because I covered it in mechanical
advantage and torque. But I do realize that when I
covered it in mechanical advantage and torque, I think
I maybe over-complicated it. And if anything, I didn't cover
some of the most basic moment to force problems that
you see in your standards physics class, especially
physics classes that aren't focused on calculus or going
to make you a mechanical engineer the very next year. So we did that with-- why did
I write down the word "mechanical?" Oh yeah,
mechanical advantage. If you do a search for
mechanical advantage, I cover some things on moments
and also on torque. So what is moment of force? Well, it essentially is the
same thing as torque. It's just another word for it. And it's essentially force times
the distance to your axis of rotation. What do I mean by that? Let me take a simple example. Let's say that I have
a pivot point here. Let's say I have some type
of seesaw or whatever. There's a seesaw. And let's say that I were to
apply some force here and the forces that we care about-- this
was the exact same case with torque, because there's
essentially the same thing. The forces we care about
are the forces that are perpendicular to the distance
from our axis of rotation. So, in this case, if we're here,
the distance from our axis of rotation is this. That's our distance from
our axis of rotation. So we care about a perpendicular
force, either a force going up like that or a
force going down like that. Let's say I have a force
going up like that. Let's call that F, F1, d1. So essentially, the moment of
force created by this force is equal to F1 times d1, or the
perpendicular force times the moment arm distance. This is the moment
arm distance. That's also often called the
lever arm, if you're talking about a simple machine, and I
think that's the term I used when I did a video on
torque: moment arm. And why is this interesting? Well, first of all, this force
times distance, or this moment of force, or this torque, if it
has nothing balancing it or no offsetting moment or torque,
it's going to cause this seesaw in this example to
rotate clockwise, right? This whole thing, since it's
pivoting here, is going to rotate clockwise. The only way that it's not going
to rotate clockwise is if I have something keep-- so
right now, this end is going to want go down like that, and
the only way that I can keep it from happening is if I exert
some upward force here. So let's say that I exert some
upward force here that perfectly counterbalances, that
keeps this whole seesaw from rotating. F2, and it is a distance d2
away from our axis of rotation, but it's going in a
counterclockwise direction, so it wants to go like that. So the Law of Moments
essentially tells us, and we learned this when we talked
about the net torque, that this force times this distance
is equal to this force times this distance. So F1 d1 is equal to F2 d2, or
if you subtract this from both sides, you could get F2 d2 minus
F1 d1 is equal to 0. And actually, this is how we
dealt with it when we talked about torque. Because just the convention with
torque is if we have a counterclockwise rotation, it's
positive, and this is a counterclockwise rotation
in the example that I've drawn here. And if we have a clockwise
rotation, it has a negative torque, and that's just the
convention we did, and that's because torque is a
pseudovector, but I don't want to confuse right now. What you'll see is that these
moment problems are actually quite, quite straightforward. So let's do a couple. It always becomes a lot easier
when you do a problem, except when you try to erase
things with green. So let's say that --
let me plug in real numbers for these values. Let me erase all of this. Let me just erase everything. There you go. All right, let me draw
a lever arm again. So what we learned when we
learned about torque is that an object won't rotate if the
net torque, the sum of all the torques around it, are zero,
and we're going to apply essentially that same
principle here. So let's do it with masses,
because I think that helps explain a lot of things and
makes this seesaw example a little bit more tangible. Let's say I have a 5-kilogram
mass here, and let's say that gravity is 10 meters
per second squared. So what is the downward
force here? What is the downward force? It's going to be the mass times
acceleration, so it's going to be 50 Newtons. And let's say that the distance,
the moment arm distance or the lever arm
distance here, let's say that this distance right
here is 10 meters. Let's say that I have
another mass. Let's say it's a 25 kilogram--
no, that's too much. Let's say it's 10 kilograms.
Let's say I have a 10-kilogram mass. And I want to place it some
distance d from the fulcrum or from the axis of rotation so
that it completely balances this 5-kilogram mass. So how far from the axis of
rotation do I put this 10-kilogram mass? This is the distance, right? Because we actually carry
the distance to the center of the mass. Well, how much force
is this 10-kilogram mass exerting downwards? Well, it's 10 kilograms times 10
meters per second squared, it's 100 Newtons. This is acting what? This is acting clockwise,
right? This one's acting clockwise
and this one's acting counterclockwise, right? So they are offsetting
each other. So we could do it a
couple of ways. We could say that 50 Newtons,
the moment in the counterclockwise direction, 50
Newtons times 10 meters, in order for this thing to not
rotate has to be equal to the moment in the clockwise
direction. And so the moment in the
clockwise direction is equal to 100 Newtons times some
distance, let's call that d, 100 Newtons times d, and
then we could just solve for d, right? We get 50 times 10 is 500. 500 Newton-meters is equal
to 100 Newtons times d. That's 100. Divide both sides by 100, you
get 5 meters is equal to d. So d is equal to 5. That's interesting. And I think this kind of
confirms your intuition from playing at the playground that
you can put a heavier weight closer to the axis of rotation
to offset a light weight that's further away. Or the other way to put it is
you could put a light weight further away and you kind of get
a mechanical advantage in terms of offsetting the
heavier weight. So let's do a more difficult
problem. I think the more problems we
do here, the more sense everything will make. So let's say that we have
a bunch of masses. Actually, let's not
do it with masses. Let's just do it with forces
because I want to complicate the issue. So this is the pivot. And let's say I have a force
here that's 10 Newtons going in the clockwise direction, and
let's say it is at-- let's say if this is 0, let's say that
this is at minus 8, so this distance is 8, right? Let's say that I have another
force going down at 5 Newtons. And let's say that its
x-coordinate is minus 6. Let's say I have another force
that's going up here, and let's say that it
is 50 Newtons. This might get complicated. 50 Newtons, and it's at minus
2, so this distance right here is 2. Let's say that I need to figure
out-- and I'm making this up on the fly. Let's say that I have another
force here that is 5 Newtons. No, let's make it a weird
number, 6 Newtons, and this distance right here
is 3 meters. And let's say that I need to
figure out what force I need to apply here upwards or
downwards-- I actually don't know, because I'm doing this on
the fly-- to make sure that this whole thing
doesn't rotate. So to make sure this whole
thing doesn't rotate, essentially what we have to
say is that all of the counterclockwise moments or all
of the counter clockwise torques have to offset all
of the clockwise torques. And notice, they're not
all on the same side. So what are all of the things
that are acting in the counterclockwise direction? So counter clockwise
is that way, right? So this is acting
counterclockwise, this is acting counterclockwise,
and that's it, right? So the other ones
are clockwise. And we don't know this one. Let's assume for a second. We could assume either way. And if we get a negative, that
means it's the opposite. So let's assume that this is a--
all of the clockwise ones I'll do in this dark brown. Let's assume this is clockwise,
let's assume that this is clockwise, and let's
assume that our mystery force is also clockwise. All of the counterclockwise
moments have to offset all the clockwise moments. So what are the counterclockwise
moments? Well, this one's
counterclockwise, so it's 10 Newtons, 10 times its distance
from its moment arm. We said it's 8, because it's
at the x-coordinate minus 8 from 0, so it's 10
times 8, plus 50. This is also counterclockwise
times 6, 50 times 6, and those are all of our counterclockwise
moments and that has to equal the
clockwise moments. So clockwise moments,
let's see. We have 5 Newtons that's going
clockwise times 6. 5 Newtons. Actually, was this 6? No, if this is 6, I must have
written some other number here that I can't read now. How far did I say this was? Let's say that this is 2. So that 50, let's say this is
2, it's negative 2, because that's what it looks like. I apologize for confusing you. So what were all the
counterclockwise moments? This 10 Newtons times its
distance 8, the 50 Newtons times this distance, 2. Don't get confused
by the negative. I just kind of said we're in
the x-coordinate axis or at minus 8 if this is 0, but it's
8 units away, right? And this 50, its moment arm
distance is 2 units. So that has to equal all of
the clockwise moments. So the clockwise moments
is 5 Newtons times 6. Its distance is 6 and it's
5 Newtons going in the clockwise direction. And then we have plus
6 Newtons times 3, plus 6 times 3. And then we're just assuming,
we don't know for sure. Let's say we're applying
the force. I should have told you
ahead of time so you could do this problem. Let's say that we're applying
the force at 10 meters away from our fulcrum arm. So force times 10. So now let's just solve
for the force. We get 80 plus 100 is equal
to 30 plus 18 plus 10F. We get 180 is equal
to 48 plus 10F. What's 180 minus 48? It's 132 is equal to
10F, or we get F is equal to 13.2 Newtons. So we guessed correctly that
this is going to be a-- sorry, this is going to be a-- I keep
mixing up all of the clockwise and counterclockwise. This is going to be
a clockwise force. These were all of the-- sorry,
this is going to be a counterclockwise force, right? A clock, this is
counterclockwise. Let me label that because I
think I said it wrong several times in the video. So these go clockwise. And it's this one
and this one. And what were the
counterclockwise? These go counterclockwise. So we have to apply a 13.10
Newton force 10 meters away, which will generate 132
Newton-meters moment in the counterclockwise direction,
which will perfectly offset all of the other moments, and
our lever will not move. Anyway, I might have confused
you with all the counterclockwise/clockwise. But just keep in mind that all
the moments in one rotational direction have to offset all
the moments in the other rotational direction. All a moment is is the force
times the distance from the fulcrum arm, so force times
distance from fulcrum arm. I'll see you in the
next video.