Main content

## Physics library

### Course: Physics library > Unit 7

Lesson 2: Torque, moments, and angular momentum- Introduction to torque
- Moments
- Moments (part 2)
- Finding torque for angled forces
- Torque
- Rotational version of Newton's second law
- More on moment of inertia
- Rotational inertia
- Rotational kinetic energy
- Rolling without slipping problems
- Angular momentum
- Constant angular momentum when no net torque
- Angular momentum of an extended object
- Ball hits rod angular momentum example
- Cross product and torque

© 2023 Khan AcademyTerms of usePrivacy PolicyCookie Notice

# Moments (part 2)

2 more moment problems. Created by Sal Khan.

## Want to join the conversation?

- Eh, but, why didn't he take the center of mass as the center of the table? The mass isn't concentrated on the left, so by what we've learned, we can't start the "moment arms" from there, they should go from the center....did I miss something?(26 votes)
- It doesn't matter where you place your point of reference as your rotation point.

Let me choose the center of the table as the pivot point. Let's call FA and FB the reactive forces exerted by the left and the right leg respectively. Let's call WB and WT to the weight of the books and the table respectively.

M(FA) - M(WL) - M(FB) = 0 (for there to be no rotation)

<=> 2.FA - 2FB = 100

FA + FB = WB + WT (for there to be no translation)

<=> FA + FB = 120

If you work out that system of equations, you'll get:

FA = 85 N , FB = 35 N(34 votes)

- Why aren't the clockwise forces marked with a negative sign when Sal does the algebra? As always, thanks a bunch.(6 votes)
- The reason why there is no negative sign in the equation is because He is setting it as in equilibrium, the rightwise is equal to the counter-clockwise. This is like up is equal to down, it is implied that the forces are already acting in opposite directions so the negative sign is not needed. Another way to look at this would be The net movement of the table equals zero; 0 = clockwise force minus counter-clockwise force. Add counter-clockwise force on each side and you get. clockwise force = counter-clockwise force.(22 votes)

- I'm not sure whether or not I may have missed a title, but is there a video that addresses the parallel axis theorem? I have a textbook that briefly addresses the theorem, but it gives a poor example and moves right along.(10 votes)
- I don't think there is one specifically for that topic, but the parallel axis theorem as I know states that the moment of inertia of a body about any axis is equal to the sum of [the moment of inertia of the body about an axis (parallel to the axis taken) at its center of mass] and [the product of the mass of the body and the square of the distance between the two parallel axes].

That is : I = Icm + (m).(d^2)

{where I is the moment of inertia about the axis taken; Icm is the moment of inertia about the center of mass and parallel to the axis taken; And m is the mass of the body, d is the distance between the two parallel lines}

I hope the statement wasn't so confusing...(11 votes)

- please can give me answer of this ques , 10N weight can be lifted in one end of a thin rod by applying a 5N force on the

other end. If the length of the rod is 1.5 meter, where is the fulcrum situated? and please tell me how to measure(4 votes)- The fulcrum is located somewhere between the two weights. The distance from the fulcrum for the 5N weight is twice the distance from the fulcrum (on the opposite side) for the 10N weight.(1 vote)

- when calculating the counterclockwise forces, why isn't the weight of the right leg counted? only the force up is counted not the one going down? why is that ?(3 votes)
- Wouldn't the axis of rotation be at the base of the left leg? If you removed the right leg, the table would rotate on the base of the left leg. If the axis was at the top, that left leg would slip off the floor and rotate clockwise. That would be weird.(2 votes)
- Yes you are right...the axis should be at the base of the left leg...but when calculating the torque it wont matter as perpendicular distance of the 100 N force from the axis will be same in both cases(you know, since force is a vector you can translate the force vector or extrapolate its line of action downwards without changing its value) ....

Hope this helps :)(2 votes)

- just with the table in general, i was wondering why you didn't have to calculate the effect of the weight of the table on the outside of the pivots?(2 votes)
- I didn't get how the forces of 20 N and and the force exerted by the book are clockwise ?(2 votes)
- is it because he takes the leg as the fulcrum that the moment is 1*100, why would the moment of the block not be anticlockwise and be 3*100??(2 votes)
- New to this process. Dont know if i should be asking this here or els where. When considering a rolling ball down an incline with no slipping , how does moments of forces come into that? would very helpful if you could do a problem on that, because its stummping me.(2 votes)

## Video transcript

I'm going to do a couple more
moment and force problems, especially because I think
I might have bungled the terminology in the previous
video because I kept confusing clockwise with counterclockwise. This time I'll try to
be more consistent. Let me draw my lever again. My seesaw. So that's my seesaw, and that is
my axis of rotation, or my fulcrum, or my pivot point,
whatever you want to call it. And let me throw a bunch
of forces on there. So let's say that I have a
10-Newton force and it is at a distance of 10, so distance
is equal to 10. The moment arm distance is 10. Let's say that I have a
50-Newton force and its moment arm distance is equal to 8. Let's say that I have a 5-Newton
force, and its moment arm distance is 4. The distance is equal to 4. That's enough for that side. And let's say I have a I'm
going to switch colors. Actually, no, I'm going to keep
it all the same color and then we'll use colors to
differentiate between clockwise and counterclockwise
so I don't bungle everything up again. So let's say I have a 10-Newton
force here. And, of course, these vectors
aren't proportional to actually what I drew. 50 Newtons would be huge if
these were the actual vectors. And let's say that that moment
arm distance is 3. Let me do a couple more. And let's say I have a moment
arm distance of 8. I have a clockwise force of 20
Newtons, And let's say at a distance of 10 again, so
distance is equal to 10, I have my mystery force. It's going to act in a
counterclockwise direction and I want to know what
it needs to be. So whenever you do any of
these moment of force problems, and you say, well,
what does the force need to be in order for this see
saw to not rotate? You just say, well, all the
clockwise moments have to equal all of the
counterclockwise moments, So clockwise moments equal
counterclockwise. I'll do them in different
colors. So what are all the
clockwise moments? Well, clockwise is this
direction, right? That's the way a clock goes. So this is clockwise,
that is clockwise. I want to go in this
direction. And so this is clockwise. What are all the clockwise
moments? It's 10 Newtons times its
moment arm distance 10. So 10 times 10 plus 5 Newtons
times this moment arm distance 4, plus 5 times 4, plus 20
Newtons times its moment arm distance of 8, plus 20 times 8,
and that's going to equal the counterclockwise moments,
and so the leftover ones are counterclockwise. So we have 50 Newtons acting
downward here, and that's counterclockwise, and it's at
a distance of 8 from the moment arm, so 50 times 8. Let's see, we don't
have any other counterclockwise on that side. This is counterclockwise,
right? We have 10 Newtons acting
in the counterclockwise direction, and its moment arm
distance is 3, plus 10 times 3, and we're assuming our
mystery force, which is at a distance of 10, is also
counterclockwise, plus force times 10. And now we simplify. And I'll just go to a neutral
color because this is just math now. 100 plus 20 plus 160 is equal
to-- what's 50 times 8? That's 400 plus 30 plus 10F. What is this? 2, 50 times 8. Right, that's 400. OK, this is 120 plus
a 160 is 280. 280 is equal to 430-- this is
a good example-- plus 10F, I just realized. Subtract 430 from both sides. So what's 430 minus 280? It's 150. So it's minus 150
is equal to 10F. So F is equal to minus
15 Newtons in the counterclockwise direction. So F is minus 15 Newtons
in the counterclockwise direction, or it means that
it is 15 Newtons. We assumed that it was in the
counterclockwise direction, but when we did the math,
we got a minus number. [SNEEZE] Excuse me. I apologize if I blew out your
speakers with that sneeze. But anyway, we assume it was
going in the counterclockwise direction, but when we did the
math, we got a negative number, so that means it's
actually operating in the clockwise direction at 15
Newtons at a distance of 10 from the moment arm. Hopefully, that one was less
confusing than the last one. So let me do another problem,
and these actually used to confuse me when I first learned
about moments, but in some ways, they're the
most useful ones. So let's say that I have
some type of table. I'll draw it in wood. It's a wood table. That's my table. And I have a leg here,
I have a leg here. Let's say that the center
of mass of the top of the table is here. It's at the center. And let's say that
it has a weight. It has a weight going down. What's a reasonable weight? Let's say 20 Newtons. It has a weight of 20 Newtons. Let's say that I place some
textbooks on top of this table, or box, just to make
the drawing simpler. Let's say I place a box there. Let's say the box weighs 10
kilograms, which would be about 100 Newtons. So let's say it weighs
about 100 Newtons. So what I want to figure out,
what I need to figure out, is how much weight is being
put onto each of the legs of the table? And this might not have even
been obviously a moment problem, but you'll see in
a second it really is. So how do we know that? Well, both of these legs are
supporting the table, right? Whatever the table is exerting
downwards, the leg is exerting upwards, so that's the amount of
force that each of the legs are holding. So what we do is we pick-- so
let's just pick this leg, just because I'm picking
it arbitrarily. Let's pick this leg,
and let's pick an arbitrary axis of rotation. Well, let's pick this is as
our axis of rotation. Why do I pick that as the
axis of rotation? Because think of it this way. If this leg started pushing more
than it needed to, the whole table would rotate in the counterclockwise direction. Or the other way, if this leg
started to weaken and started to buckle and couldn't hold
its force, the table would rotate down this way, and it
would rotate around the other leg, assuming that the other
leg doesn't fail. We're assuming that this leg
is just going to do its job and it's not going to move
one way or the other. But this leg, that's why we're
thinking about it that way. If it was too weak, the whole
table would rotate in the clockwise direction, and if it
was somehow exerting extra force, which we know a leg
can't, but let's say if it was a spring or something like that,
then the whole table would rotate in the
counterclockwise direction. So once we set that up, we can
actually set this up as a moment problem. So what is the force
of the leg? So the whole table is exerting
some type of-- if this leg wasn't here, the whole table
would have a net clockwise moment, right? The whole table would tilt down
and fall down like that. So the leg must be exerting a
counterclockwise moment in order to keep it stationary. So the leg must be exerting
a force upward right here. The force of the leg, right? We know that. We know that from
basic physics. There's some force coming down
here and the leg is doing an equal opposite force upwards. So what is that force
of that leg? And one thing I should
have told you is all of the distances. Let's say that this distance
between this leg and the book is 1 meter-- or the box. Let's say that this distance
between the leg and the center of mass is 2 meters, and so
this is also 2 meters. OK, so we can now set this
up as a moment problem. So remember, all of the
clockwise moments have to equal all of the
counterclockwise moments. So what are all of the
clockwise moments? What are all of the things that
want to make the table rotate this way or this way? Well, the leg is the
only thing keeping it from doing that. So everything else is essentially a clockwise moment. So we have this 100 Newtons,
and it is 1 meter away. Its moment arm distance is 1. So these are all the clockwise
moments, 100 times 1, right? It's 100 Newtons acting
downwards in the clockwise direction, clockwise moment, and
it's 1 meter away, plus we have the center of mass at the
top of the table, which is 20 Newtons, plus 20 Newtons, and
that is 2 meters away from our designated axis,
so 20 times 2. And you might say, well, isn't
this leg exerting some force? Well, sure it is, but its
distance from our designated axis is zero, so its moment
of force is zero. Even if it is exerting a million
pounds or a million Newtons, its moment of force,
or its torque, would be zero because its moment arm distance
is zero, so we can ignore it, which makes
things simple. So those were the only
clockwise moments. And what's the counterclockwise
moment? Well, that's going to be the
force exerted by this leg. That's what's keeping the whole
thing from rotating. So it's the force of
the leg times its distance from our axis. Well, this is a total of 4
meters, which we've said here, times 4 meters. And so we can just solve. We get 100 plus 40, so we get
140 is equal to the force of the leg times 4. So what's 140-- 4 goes
into 140 35 times? My math is not so good. Is that right? 4 times 30 is 120. 120 plus 20. So the force of the leg
is 35 Newtons upwards. And since this isn't moving,
we know that the downward force right here must
be 35 Newtons. And so there's a couple of ways
we can think about it. If this leg is supporting 35
Newtons and we have a total weight here of 120 Newtons, our
total weight, the weight at the top of the
table plus the bookshelf, that's 120 Newtons. So the balance of this
must be supported by something or someone. So the balance of this
is going to be supported by this leg. So it's 120 minus 35 is what? [PHONE RINGS] Oh, my phone is ringing. 120 minus 35 is what? 120 minus 30 is 90. And then 90 minus
5 is 85 Newtons. It's so disconcerting
when my phone rings. I have trouble focusing. Anyway, it's probably because
my phone sounds like a freight train. Anyway, so there you go. This type of problem is actually
key to, as you can imagine, bridge builders, or
furniture manufacturers, or civil engineers who are bridge
builders, or architects, because you actually have to
figure out, well, if I design something a certain way, I have
to figure out how much weight each of the supporting
structures will have to support. And as you can imagine,
why is this one supporting more weight? Why is this leg supporting more
weight than that leg? Well, because this book, which
is 100 Newtons, which is a significant amount of the total
weight, is much closer to this leg than it
is to this leg. If we put it to the center, they
would balance, and then if we push it further to the
right, then this leg would start bearing more
of the weight. Anyway, hopefully you found
that interesting, and hopefully, I didn't
confuse you. And I will see you
in future videos.