Welcome back.
Now that we've hopefully, learned a little bit about
Newton's law, let's apply them to solve some problems.
Let's say I have a-- I don't know-- some kind of vehicle, a
car, a motorcycle or something.
And let's say its mass is 500 grams. And let's say that I
can accelerate this vehicle at an acceleration of-- I don't
know-- 3 centimeters per second squared.
What's the force that I need to apply to this mass to
accelerate it at this speed?
And we want the answer in Newton's.
So what's the force?
So we're just going to use Newton's second law and
Newton's second law tells us force is equal to mass times
acceleration.
So you might want to, or you might be tempted just to
multiply mass times acceleration and you'd get
force is equal to-- let's see, 500 grams times 3 centimeters
per second squared.
And you would get force is equal to 1,500 gram
centimeters per second squared.
And if you did this, you would be right.
Although your answer would not be in Newtons and now you
would have to somehow, try to convert this
set of units to Newtons.
And what are Newtons?
Well we learned when we did Newton's laws that a Newton,
so 1 Newton is equal to 1 kilogram
meter per second squared.
So somehow we have to convert the gram to kilograms and we
have to convert the centimeters to meters.
We could do it after the fact here, or what I find it easier
to do is actually to convert the mass and the acceleration
units first and then just do the F equals ma.
So what's 500 grams in kilograms?
Well 500 grams is half-- well, a kilogram as a thousand
grams. So 500 is going to be half a kilogram.
1 kilogram is a thousand grams, so 0.5 kilograms is 500
grams.
Similarly, 3 centimeters is how many meters?
Well, 1 meter is 300-- sorry.
I think I'm dehydrated.
1 meter is a hundred centimeters, right?
So 3 centimeters is 0.03 meters per second squared.
Hopefully this make sense to you.
3 centimeters is 0.03 meters.
And now we already.
We have our mass in kilograms and we have our acceleration
in meters per second squared.
And if this is confusing, you should watch the unit videos
because this is all I'm doing; I'm just doing unit
conversion.
So let's go back to force equals mass times
acceleration.
So the force was equal to 0.5 kilograms times the
acceleration, which is 0.03 meters per second squared.
And this equals--
What's 0.5 times 0.03?
I'll do it down here just because multiplying decimals
seems to be a problem for a lot of people, including
myself, many times.
So what you do, you just multiply the numbers.
5 times 03 or 5 times 3 is 15.
And then, how many points do we have behind the decimal?
How many digits behind the decimal?
Let's see.
We have 1, 2, 3.
So 1, 2, 3.
We have to add the 0 because we need three spaces behind
the decimal point.
So we get the force is equal to 0.015 kilogram meters per
second squared.
And this is a Newton.
So the force is equal to 0.015 Newtons.
Let's do another problem.
And this one's going to be-- and actually, I think you'll
find most of the difficult Newton's laws problems or
force problems, they're just some combination of making
sure you get the units right when we're talking about in
one dimension.
The difficult part is usually getting the units right or
just the math, just the algebra.
So if you have trouble with this it's usually because you
have to just brush up a little bit on the algebra.
The physics itself is just force equals mass times
acceleration as we will see in this problem.
So let's say that when I apply some force, some particular
force, I use this little 0 here.
So I call that force F sub 0.
So this means a particular force.
This is some value.
When I apply that force to some mass, let's call that m
sub 0, I get some acceleration.
I get acceleration a sub 0.
We could've put numbers here.
We could've said, well, if I apply a force of 10 Newtons to
a mass of-- I don't know-- to a mass of let's say 2
kilograms, I have an acceleration of 5 meters per
second squared.
But I'm just doing this because this could be any
relationship.
And let's say the problem tells us that when I put
another mass with this first mass, so let's say that, you
know-- let me draw this diagram.
So here's my mass, m sub 0.
When I apply a force of f sub 0 to it, I get an acceleration
of a sub 0.
Now the problem tell us when I add another mass-- so let's
says I stack it up and we're like in an ice
skating ring or something.
And I stack another mass up here, and let's
call this mass m1.
When I stack another mass on here-- so let me redraw it
actually down here.
Because it's a different case.
And I apply the same force, and now I have
this new mass on here.
I'll do it in red.
This is m1.
The problem tells us that my new acceleration is 1/5 of the
original acceleration.
So it's 1/5 of whatever this was.
So it's 1/5 a sub 0.
So the question is, what is the ratio of m
sub 0 to m sub 1?
So m sub 0 to m sub 1 is equal to what?
And I'm going to keep it in abstract variables just to
confuse you.
And then I'll show you that you can actually substitute
numbers and the problem becomes a little easier.
And you might want to pause it and try it for yourself.
So let's work it through.
So we know we have this relationship
to start off with.
And just for simplicity, let's write what m sub 0 is in terms
of F and a.
So we just divide both sides by a sub 0 and you get F sub 0
divided by a sub 0 is equal to m sub 0.
Good.
So let's just put that aside for a second.
And let's do that same relationship here with this.
So here, this relationship tells us that F sub 0 is equal
to m1 plus m0 times this new acceleration, which
is 1/5 a sub 0.
And so, if we divide both sides by this term right here,
we get-- dividing by 1/5 is the same thing as
multiplying by 5.
So you get 5 F sub 0 over a sub 0 is equal to m1 plus m0.
I just divided both sides by this term right here.
Well, what's F sub 0?
What's this?
Let me switch colors again.
What's F sub 0 divided by a sub 0?
Well it's here's.
It's what we solved for in the beginning.
We just got it from this relationship.
So we could substitute.
5 times F sub 0 divided by a sub 0 is the same
thing as 5 times m0.
Draw a line here.
So we have a new relationship.
5m0 is equal to m1 plus m0.
All I did is I substituted this for
this, or this for that.
And I used this relationship, which we got in the beginning
to do that.
And now what do we have?
We have 5 m sub 0 is equal to m1 plus m0.
We could subtract m0 from both sides.
You get 4 m0 is equal to m1.
You could divide both sides by m0 and you got 4 is equal to
m1 over m0.
And you could invert this relationship and you can get
m0 over m sub 1 is equal to 1/4.
So what we learned is the ratio of the old mass to the
new mass is 1 to 4.
And that's a problem.
And actually, I will leave it for you as an exercise to
figure out-- to just do the same
problem using the numbers.
I will do that actually, in the next video just to show
you that that actually would've been a
simpler way to do it.
But it's good to get used to this just so you can solve
things in general terms. I'll see you in the next video.