Welcome back. We’ll now use a little bit of

what we’ve learned about work and energy and the conservation

of energy and apply it to simple machines. And we’ll learn a little bit

about mechanical advantage. So I’ve drawn a simple

lever here. And you’ve probably

been exposed to simple levers before. They’re really just kind

of like a seesaw. This place where the

lever pivots. This is called a fulcrum. Just really the pivot point. And you can kind of view this

as either a seesaw or a big plank of wood on top of a

triangle, which essentially is what I’ve drawn. So in this example, I have

the big plank of wood. At one end I have this 10

newton weight, and I’ve written 10 in there. And what we’re going to figure

out is one, how much force– well, we could figure out

a couple of things. How much force do I have

to apply here to just keep this level? Because this weight’s going

to be pushing downwards. So it would naturally

want this whole lever to rotate clockwise. So what I want to figure out is,

how much force do I have to apply to either keep the

lever level or to actually rotate this lever

counterclockwise? And when I rotate the lever counterclockwise, what’s happening? I’m pushing down on this

left-hand side, and I’m lifting this 10 newton block. So let’s do a little thought

experiment and see what happens after I rotate this

lever a little bit. So let’s say, what I’ve drawn

here in mauve, that’s our starting position. And in yellow, I’m going to draw

the finishing position. So the finishing position

is going to look something like this. I’ll try my best to draw it. The finishing position is

something like this. And also, one thing I want to

figure out, that I wanted to write, is let’s say that the

distance, that this distance right here, from where I’m

applying the force to the fulcrum, let’s say that

that distance is 2. And from the fulcrum to the

weight that I’m lifting, that distance is 1. Let’s just say that, just for

the sake of argument. Let’s say it’s 2 meters and 1

meter, although it could be 2 kilometers and 1 kilometer,

we’ll soon see. And what I did is I pressed down

with some force, and I rotated it through

an angle theta. So that’s theta and this

is also theta. So my question to you, and

we’ll have to take out a little bit of our trigonometry

skills, is how much did this object move up? So essentially, what

was this distance? What’s its distance in the

vertical direction? How much did it go up? And also, for what distance did

I have to apply the force downwards here– so that’s this

distance– in order for this weight to move up this

distance over here? So let’s figure out

either one. So this distance is what? Well, we have theta. This is the opposite. This is a 90 degree

angle, because we started off at level. So this is opposite. And this is what? This is the adjacent angle. So what do we have there? Opposite over adjacent. Soh Cah Toa. Opposite over adjacent. Opposite over adjacent. That’s Toa, or tangent. So in this situation, we know

that the tangent of theta is equal to– let’s call

this the distance that we move the weight. soon. So that equals opposite over

adjacent, the distance that we moved the weight over 1. And then if we go on

to this side, we can do the same thing. Tangent is opposite

over adjacent. So let’s call this the distance

of the force. So here the opposite of the

distance of the force and the adjacent is this 2 meters. Because this is the hypotenuse

right here. So we also have the tangent of

theta– now you’re using this triangle– is equal to

the opposite side. The distance of the force

over 2 meters. So this is interesting. They’re both equal to

tangent of theta. We don’t even have to

figure out what the tangent of theta is. We know that this quantity is

equal to this quantity. And we can write it here. We could write the distance of

the force, that’s the distance that we had to push down on

the side of the lever downwards, over 2, is equal to

the distance of the weight. The distance the weight traveled

upwards is equal to the distance, the weight,

divided by 1. Or we could say– this

1 we can ignore. Something divided

by 1 is just 1. Or we could say that the

distance of the force is equal to 2 times the distance

of the weight. And this is interesting, because

now we can apply what we just learned here to figure

out what the force was. And how do I do that? Well, when I’m applying a

force here, over some distance, I’m putting energy

into the system. I’m doing work. Work is just a transfer of

energy into this machine. And when I do that, that

machine is actually transferring that energy

to this block. It’s actually doing work on the

block by lifting it up. So we know the law of

conservation of energy, and we’re assuming that this is a

frictionless system, and that nothing is being lost to

heat or whatever else. So the work in has to be

equal to the work out. And so what’s the work in? Well, it’s the force that I’m

applying downward times the distance of the force. So this is the work in. Force times the distance

of the force. I’m going to switch colors

just to keep things interesting. And that has to be the same

thing as the work out. Well, what’s the work out? It’s the force of the weight

pulling downwards. So we have to– it’s essentially

the lifting force of the lever. It has to counteract the force

of the weight pulling downwards actually. Sorry I mis-said it

a little bit. But this lever is essentially

going to be pushing up on this weight. The weight ends up here. So it pushes up with the

force equal to the weight of the object. So that’s the weight of the

object, which is — I said it’s a 10 newton object — So

it’s equal to 10 newtons. That’s the force. The upward force here. And it does that for

a distance of what? We figured out this object, this

weight, moves up with a distance d sub w. And we know what the distance

of the force is in terms of the distance of w. So we could rewrite this as

force times, substitute here, 2 d w is equal to 10 d w. Divide both sides by 2 you d w

and you get force is equal to 10 d w 2 two d w, which is

equaled to, d w’s cancel out, and you’re just left with 5. So this is interesting. And I think you’ll see where

this is going, and we did it little complicated this time. But hopefully you’ll realize

a general theme. This was a 10 newton weight. And I only had to press down

with 5 newtons in order to lift it up. But at the same time, I pressed

down with 5 newtons, but I had to push down

for twice as long. So my force was half as much,

but my distance that I had to push was twice as much. And here the force is twice as

much but the distance it traveled is half as much. So what essentially just

happened here is, I multiplied my force. And because I multiplied

my force, I essentially lost some distance. But I multiplied my

force, because I inputted a 5 newton force. And I got a 10 newton force out,

although the 10 newton force traveled for

less distance. Because the work was constant. And this is called mechanical

advantage. If I have an input force of 5,

and I get an output force of 10, the mechanical

advantage is 2. So mechanical advantage is equal

to output force over input force, and that should

hopefully make a little intuitive sense to you. And another thing that maybe

you’re starting to realize now, is that proportion of the

mechanical advantage was actually the ratio of this

length to this length. And we figured that out by

taking the tangent and doing these ratios. But in general, it makes sense,

because this force times this distance has

to be equal to this force times this distance. And we know that the distance

this goes up is proportional to the length of from the

fulcrum to the weight. And we know on this side the

distance that you’re pushing down, is proportional to the

length from where you’re applying the weight

to the fulcrum. And now I’ll introduce you

to a concept of moments. In just a moment. So in general, if I have, and

this is really all you have to learn, that last thought

exercise was just to show it to you. If I have a fulcrum here, and

if we call this distance d 1 and we called this

distance d 2. And if I want to apply

an upward force here, let’s call this f 1. And I have a downward force,

f 2, in this machine. f 2 times d 2 is equal

to d 1 times f 1. And this is really all

you need to know. And this just all falls

out of the work in is equal to the work out. Now, this quantity isn’t

exactly the work in. The work in was this force–

sorry, F2– is this force times this distance. But this distance is

proportional to this distance, and that’s what you

need to realize. And this quantity right here is

actually called the moment. In the next video, which I’ll

start very soon because this video is about to end. I’m running out of time. I will use these quantities to

solve a bunch of mechanical advantage problems. See

it`s easy to just apply the formula, but it`s hard to understand where the formula comes from.

thx this help a bit for a very confusing test on mechanical advantage

your learning physical science not physics

Isn't the force times distance equal to the torque?

great…great…you are my master………..

Mr. Khan, even if you're lifting the object upwards, how can u have a 90 degrees triangle. Because i mean, the lenth of the lever didnt extend or anything, it's the same lever. So, even if the lever turns upwards, it has to move in a circular motion.

Another way u can look at it is, it's like a clock. 1 m = hour hand 2 m = minute hand,

it's just turning around like a circle. having 90 degrees thr is actually impossible.

@marcellohro that would be the inductive method…you gather the data, insert it in a graph, and you use simple 9th grade mathematics to solve the problem.

Too complex – Very complicated all round. Use similar triangles instead of all that trig., same angle twice as large. Very complicated all round. Also your argument is circular. You assume the conservation of energy is the same at the conservation of work done and that is what you prove.

@willraytaylor yes too complex, I didnt even take triginometry. but all of your other vids are very use ful

Yeah I would have gone with similar triangles.

I think the instruction is wrong. There are no way that degree is 90 degree. The length of is 1m and 2m and the ending length is still 1m and 2m so there are no way that angle is 90 and therefore it cannot use tan(theta) to calculate.

im in grade 7.. and its pretty confusing but i understand a little… well done though

you messed up on your angles, the hypotenuses would be 1m and 2m. they stay the same length but shift an angle theta

@parklinkin52 mathemaphysicist? haha

Hello Salman first of all i want to thank you for all your efforts.Salman when you draw this picture andwhen you applied Tangent on both of the triangles you said that both of the angles were equal to theta well i want to ask that if we separate these theta as theta1 and theta2.will they will be also equal to each other after considering them as theta1 and 2 🙂

my fav subject

where the fuck is more moments?

Guys I suggest you watch trignometry videos before watching this in order to understand how all this is possible and watch some kinematics video in order to develop your logic more.

if the result is 5N … the fulcrum will not move … :))

Why did the lever start out at equilibrium?

Hi Sal, a question for me here is that it would seem that 5N would only BALANCE the 10N in equilibrium and even assuming a frictionaless system it would take some extra force to actually move the weight. Furthermore since the system is rotating the "opposite" side would be curved and the weight tipped to the left, though over such a short distance I might agree it doesn't matter very much.

61 vids done.. 100 to go:)

Great vids:)

this is what young future scientists really need an intuition not cramming formulas. thanks Mr sal

Very good videos, very helpful

I think you have a mistake. The hyp is 2m and 1m in the triangles because when you apllyed the force the length of this platue/stick is not changing

It would be correct if you use sin instead of tan

He did NO MISTAKE

thanks

wait waa?

This is hardly an "introduction"

Is it true to say that

The greater the mechanical advantage the more efficient the machine is

And

The greater the velocity ratio the less efficient the machine is

Oh man this makes so much sense. THX KHAN!

this is on the Mcat ugh

Shouldn't the force of the weight be 98N? Which was 10 N × 9.8 m/s^2. Making the answer 50N

Shouldn't the force of the weight be 98N? Which was 10 N × 9.8 m/s^2. Making the answer 50N

… i didn't get he was talking about maybe i'll have to watch it again later

This appears to have a reasonably significant flaw. In the diagram, the lever appears to just become longer, which makes no sense. It would make more sense that it goes in a circular path but by similar circles the ratio of the distances is still 2.

not helpful

bring him back, this is much better than the people reading from a script

thanks for the video

wow ! im like reading the comments and everbody is saying stuff like " but dosnt 3=5%50/200=300+50= 3$00(**80)" and im just sitting here like whaaaaa????

"leave er"

Sal you are best

but with 5N wouldnt it be in equilibrium?, and therefore it wouldnt move?

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great explanation thank you

This is confusing ._.

Or is just cause i'm just dum .-.

Very good

this is so confuuuuuusiiiiiiiiinggggggg

set speed to 0.75 everything will turn better

Good job

I saw this video, I’m interested in, can we make a more efficient drive train connector, other than a Clutch or a torque converter, that can be instantaneous. This will improve acceleration. When accelerating, there is always delay because torque multiplication is always too slow, so automatics are ok, but the torque converter needs to do a better job at 2 things; speed and multiplying torque

umm…yeah..i….understand…

Man, I prefer biology!

Forget all the formulae that we use for energy/work/moments of force' that explain levers. Now describe to me physically how a larger force can materialise from a smaller force via the medium of the lever? Further , explain to me why there is a dependency on the position of the fulcrum? Is it some dependency/relationship relating to 'elastic restoring forces' between the atoms/molecules of the medium? That somehow the position of the fulcrum relative to the forces applied on either side of it will cause a difference in the elastic restoring forces in the medium (and a mechanical advantage)? If you search the internet for an answer you will get opinions but no-one seems to have a confirmed explanation.