Newton’s Laws Examples (part 2)

Welcome back. And I will now do that same
problem in a much easier way. Let me clear it a little bit. So the original problem we said,
you know, we apply some force to m sub 0 and
that gives some acceleration, a sub 0. And then we said when we apply
the same force to a combination of m sub 0 and
m sub 1, we get 1/5 the acceleration. And we worked it through
with all the variables. What I’ll show you is you can
actually do this type of problem just by substituting
numbers. This is kind of quick and dirty,
but it’s good to do a reality check. And oftentimes, you can solve
the problem without having to go through all the
variable mess. So I can just pick some
numbers here. So I could say, well what if F
sub 0 is equal to 10 Newtons, m sub 0 is equal to– I don’t
know– 2 kilograms. Than a sub 0 is equal to– well, it would
be 10 divided by 2. Because force is equal to
mass times acceleration. So it’d be 5 meters per
second squared. And then in this case, this
would be 10 Newtons. 1/5 a sub 0? Well that will be 1/5 this. So it would be 1 meters
per second squared. And then we could solve for
what the new mass is. How would we do that? Well we have force, which is
10 Newtons, is equal to the sum of the masses. So m1 plus m sub 0. But m sub 0 we already learned
is 2 kilograms. Times the acceleration. Times 1/5 a sub 0, which is 1
meter per second squared. So then we have– this 1, we
could ignore it essentially. So then we essentially have that
10– and since all of our units are right, we can kind of
drop the units because we know they work out. 10 is equal to m1 plus 2. So you get m1 is equal to 8. And then once again, if we want
to know the ratio of m sub 0 to m sub 1, we can just
substitute the numbers. m sub 0 is 2 kilograms. m
sub 1 is 8 kilograms. So the ratio is 1:4. You probably find that
a little bit easier. Let’s do another problem. Whoops. Invert colors. OK. This next problem I think
you’ll find interesting. So let’s say I have a sky diver
and he’s in his sky diver position, falling
towards the ground. And let’s say he weighs 70
kilograms. So his mass is equal to 70 kilograms. Let’s say
that the terminal velocity is 120 miles per hour. So he’s moving downward at 120
miles per hour, which is actually accurate. I’ve gone sky diving. And if we convert that– you
could convert that for fun into the metric system. But I’ll do that for you. But it’s good to know just so
you have a sense of how fast you fall when you sky dive
before the parachute opens. This translates to about
53.6 meters per second. I’m reading this from a problem
from a website at the University of Oregon. But anyway, they are asking
us, what force does air resistance exert on
the sky diver? So let’s be clear
a couple things. This 120 miles per hour,
this is the sky diver’s terminal velocity. And if you’re not familiar with
what terminal velocity is, I will now explain
it to you. So when you fall from a plane,
you have a bunch of wind pushing on you. You have a lot of
wind resistance. It causes friction; it slows you
down as you can imagine. I mean that’s how a
parachute works. It creates a lot more resistance
from the wind and then you slow down. So the terminal velocity is the
velocity at which you no longer go faster than. So it’s the velocity at which
you stop accelerating or it’s the velocity you reach and you
don’t go any faster than that. It’s basically based on
your wind resistance. So at the terminal velocity
your acceleration is 0. So what we know is, is that
the force of the air– we could call that the air force. So we know that the force of
the air is exactly equal to the force of gravity. And how do we know that? Because the guy’s not
accelerating. It’s his terminal velocity. He’s at a very high speed. He had accelerated all the
way to this point. But the more he accelerated
and the faster he got, the more resistance the wind
provided up to a point where the wind provided so much
resistance that he stopped going any faster. And that’s the terminal
velocity. So at terminal velocity, the
force of the air is equal to the force of gravity. What’s the force of gravity? Well the force of gravity is
just the guy’s weight. So the force of gravity is equal
to the guy’s mass, 70 kilograms. And we have
our units right. 70 kilograms. Times the
acceleration of gravity. Well the acceleration of gravity
is 9.8 roughly meters per second squared. We could use a calculator
to calculate this. I feel cheap now. I could have done it
by hand anyway. 686. So it equals 686 Newtons. The second part of the
question– and this is interesting. If a sky diver pulls in their
arm and aims their body downward, so now the sky diver
looks more like this and he pulled in his arms and he
aimed his body down. So he’s diving, really. The terminal velocity can be
increased to about 180 miles per hour or 80.5 meters
per second. They give us this. We could’ve figured
it out though. So now he’s going
a lot faster. Roughly 50% faster than he
was, or maybe– well, 30% faster than he was
going before. He’s going a lot faster
and why is that? Because he’s more
aerodynamic now. We’ll do more on pressure later,
but I want you to get the intuition that when you’re
laying flat there’s just a lot of wind pressing against
your body. You have a lot of surface area
exposed to the wind. But when you’re diving in this
situation, like the sky diver is, he has a lot less
exposed to the wind. Really just his head. His head is breaking the
wind and nothing else. And that’s why it takes a lot
more speed for the force of the wind resistance to match
the force of gravity. So the question is asking, if
the sky diver pulls in their arms and aims their body
downward, the terminal velocity can be increased to
about 80.5 meters per second or 180 miles per hour. So he’s going very fast. What force does air resistance
now exert on the sky diver? And I’ll let you think about
that for a second. Maybe you want to pause it
and think of it yourself. And now that you’ve unpaused it,
I’ll tell you that this is a trick question. Because once again, the sky
diver has reached a new terminal velocity. By definition, at the terminal
velocity, the sky diver is no longer accelerating. The sky diver is not going any
faster because the wind resistance is so strong that
it completely matches the force of gravity. So once again, the wind
resistance, the force of the air or the force of the
wind, is equal to the force of gravity. And what is the force
of gravity? Well that’s his weight. And we already figured
that out. That was 686 Newtons. And I know what you’re
thinking. You’re saying, Sal, this
doesn’t make sense. He’s now going so much faster,
doesn’t the air exert more force on him? Well no. The air is exerting
the same force. If he was going that same speed,
but he was flattened out, I would agree with you. The air would be exerting
more force on him. But what’s happened now is that
he’s once again reached a state where his acceleration
is 0. It’s at a higher velocity
and I want you think a lot about this. He’s at a much higher
velocity now. On his head, for example,
there’s a lot more wind going by. But it’s pressing on a
smaller surface area. And I’m not going to go too
much into detail of pressure right now. But I want you to get
that intuition. So although the wind is going a
lot faster, it’s going a lot faster on a smaller area. And its actual force is
the exact same thing. And we know that because he’s
not accelerating anymore. Because he’s at his
terminal velocity. So think about that a bit. It’s a bit of a trick question,
but I think it gives you a good intuition on what
acceleration means, what terminal velocity means, and
it’ll start to give you a little bit of an intuition
on even wind resistance, on pressure. I’ll see you in the
next video.

26 thoughts on “Newton’s Laws Examples (part 2)

  1. ration between mph and m/s is that there is about 0.4469'4' m/s is a mph. 1mile/hour = 1609meters/hour. 1609meters/hour=0.4469'4' meters per second.

  2. There also has the bouyancy force acting on the person, therefore the gravity is not totally equals to the air resitance. the large surface area will decrease the terminal velocity.

  3. this is funny.
    the terminal velosidy on a ant isn't a speed what wil kill the ant.

    that is why ant's can't fall to there dead.

  4. Thank you SO much for posting these videos, you helped me get full marks on my physics exam!! (: Btw, can you post some videos on thermal expansion and calorimetry? Thanks a bunch (:

  5. Lol @ him saying he feels cheap for not calculating the mass*g by hand. Dude, done those type of calculation by hand since middle school.

  6. @pheonixofpeace this is in a free fall as there is nothing artificial in resisting or assisting in his fall… watch felix baumgartner's story plz

  7. why is mass 1, 8??? if you will transpose 10 to the other side of the equation it will be negative and it will give you an answer of -8 or will I just ignore the sign?

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