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# Tension in an accelerating system and pie in the face

The second part to the complicated problem. We figure out the tension in the wire connecting the two masses. Then we figure out how much we need to accelerate a pie for it to safely reach a man's face. Created by Sal Khan.

## Want to join the conversation?

- Imagine if the bald man was standing at greater distance than he is at present, then the pie must be thrown with more acceleration. But Sal did not tell the dependency of the force (required to throw the pie at the man's face) on the distance between him and the man. PLEASE HELP...(17 votes)
- He's not throwing the pie. His hand is applying a constant force on the pie until it comes in contact with said bald man's face.(51 votes)

- Was there a missing video? I did not see this problem in the previous videos (Introduction to tension part 1 & 2).(24 votes)
- me too so confusing....(4 votes)

- At6:00, why do the net forces have to be zero ? when we accelerate the pie, there is a net force, right ?(7 votes)
- The vertical component of the net force must be zero, but the horizontal component will be positive to the right.(17 votes)

- Why does he uses the 'coefficient of friction' instead of seperating it into 'coefficient of STATIC friction' and 'coefficient of KINETIC friction'?(5 votes)
- It is the coefficient of static friction because the pie is not slipping down the hand.(12 votes)

- Where is the previous video?(15 votes)
- The previous video is mild and medium tension.(0 votes)

- My intuition says that the mass of the pie is important to determine the acceleration necessary,

but Sal says that it cancels out. Wouldn't a pie with more mass need a greater acceleration?(1 vote)**Acceleration**= velocity/time.

As you can see, there is no**mass**involved here.

It is quite natural to think that more mass, more the acceleration needed, but, apparently, that's not the way it works.(11 votes)

- what is the formula for finding tension(2 votes)
- There's a no particular formula for tension. Tension is just a kind of force usually applied to a string. If you have an object of mass m hanging on a string with a downward acceleration due to gravity, g, on that object, the tension on the string, is the same as the force of gravity on the object, F = mg.

You may need to take the mass of the string into account for some problems where the string is particularly long or thick, but the mass of the string is almost always negligible for most physics problems.(10 votes)

- What is the video called that has the problem for the beginning?(6 votes)
- it was deleted, but here is the youtube link for it: https://www.youtube.com/watch?v=KdoBDjvee3U(0 votes)

- Hi just wondering where is the first part to this problem?(3 votes)
- Here's the video :

https://www.youtube.com/watch?v=KdoBDjvee3U&feature=related

I don't know why its on YouTube, though and not on Khan Academy.

Hope this helps :)(3 votes)

- At2:20, he puts 196 in for the weight of the object. Why does he put 196 in instead of 20kg?(3 votes)
- Kilograms is not a weight it is a mass. To get a weight in Newtons you need to multiply the mass of 20 kg by the acceleration of gravity which is 9.8 m/s^2 giving you 196 N.(3 votes)

## Video transcript

Welcome back. We just finished this problem
with the pulleys and the inclined plane. And I just wanted to do one
final thing on this problem just because I think
it's interesting. And then we can move onto
what seems like a pretty fun problem. So the last thing I want to
figure out is, we figured out that this 20 kilo-- actually,
the whole system will accelerate up and to
the right at 4.13 meters per second squared. And then the second part of this
question is, what is the tension in this rope
or this wire? And at first you might say,
this is complicated. You know, this thing isn't
static anymore. The thing is actually
accelerating. How do I do it? Well this is how you
think about it. Just pick one part
of the system. Let's say that all we could see
was this 20 kilogram mass. So let's say all we could see
was this 20 kilogram mass. And we know it's suspended
from a wire. And we also know that this
20 kilogram mass is not accelerating as fast
as it would if the wire wasn't there. It's accelerating only at
4.13 meters per second. If the wire wasn't there, it'd
be accelerating at 9.8 meters per second, the acceleration
of gravity. So the wire must be exerting
some upward force on the object. And that is the force
of tension. That is what's slowing-- that's
what's moderating its acceleration from being 9.8
meters per second squared to being 4.13 meters per
second squared. So essentially, what is the
net force on this object? On just this object? Well the net force is-- and
you can ignore what I said before about the net force
in all the other places. But we know that the object
is accelerating downwards. Well, we know it's 20 kilograms.
So that's its mass. And we know that it's
accelerating downwards at 4.13 meters per second squared. So the net force, 20 times--
see, times 20 is 82-- let's just say 83 Newtons. 83 Newtons down. We know that the net force
is 83 Newtons down. We also know that the tension
force plus the force of gravity-- and what's the
force of gravity? The force of gravity is just
the weight of the object. So the force of tension, which
goes up, plus the weight of -- the force of gravity is equal
to the net force. And the way I set this up,
tension's going to be a negative number. Just because I'm saying
positive numbers are downwards, so a negative number
would be upwards. So tension will be what
is 83 minus 196? Minus 196 is equal to
minus 113 Newtons. And the only reason why I got a
negative number is because I used positive numbers
for downwards. So minus 113 Newtons downwards,
which is the same thing as 113 Newtons upwards. And so that is the tension
in the rope. And you could have done the same
thing on this side of the problem, although it would
have been-- well, yeah. You could have done the
exact same thing on this side of the problem. You would've said, well what
would it have accelerated naturally if there wasn't some
force of tension on this rope going backwards? And then you're saying, oh,
well, we know it would have gone in this direction at some
acceleration, but instead it's going in the other direction. So you use that. You figure out the net force,
and then you say the tension plus all of these forces have
to equal the net force. And then you should solve
for the tension. And it would be the
same tension. Now we will do a fun and
somewhat simple, but maybe instructive problem. So I have a pie. This is the pie. This is parallel. And I have my hand. You can tell that my destiny
was really to be a great artist. This is my hand. And I'm holding a pie, and I'm
looking to smash this pie into this individual's face. I actually was a, I was the
newspaper cartoonist in high school, so I have some
minor-- but anyway. Let's make it a bald man. Well anyway, I shouldn't be
focusing on the drawing. He has a moustache. Anyway, I'm looking to throw
this pie into this guy's face. And the problem is, I need to
figure out how fast do I need to accelerate this pie for
it to not fall down? Right? Because what's happening? Well there's the force of
gravity on this pie. There's a force of gravity on
this pie and if I don't accelerate it fast enough, it's
just going to slide down. And I'll never be able
to, It'll never reach the guy's face. So I don't want this pie
to slide down at all. How fast do I have
to push on it? Well, we know that the
coefficient of friction-- you don't know this, but I know
that the coefficient of friction between my hand and
the pie, the coefficient of friction is equal to 0.8. So given that, how fast do
I have to accelerate it? Well let's see what's
happening. So we have the force of
gravity pulling down. So let's say that the mass
of the pie is m. m equals mass. So what is the force of gravity pulling down on the pie? Well the force of gravity is
just equal to m times 9.8. Right? The force of gravity is
equal to m times 9.8. In order for this pie to not
move down, what do we know about the net forces
on that pie? Well we know the net forces
on that pie have to be 0. So what would be the
offsetting force? Well, it would be the
force of friction. So we would have a force of
friction acting upwards. Right? Because the force of friction
always acts opposite to the direction that the thing
would move otherwise. So essentially, our force of
friction has to be greater than, roughly, greater
than or equal to. Because if it's greater than,
it's not like the pie is going to move up. Friction by itself will never
move something, it'll just keep something from
being moved. But let's just figure
out the minimum. I won't do the whole
inequalities. The force of friction has to
be equal similarly, to 9.8 times the mass of the pie. So if the coefficient of
friction is 0.8, what is the force that I have to apply? Well, the force I have to apply
in this case is going to be the normal force, right? That's normal to the
bottom of the pie. Right? My hand is now like the
surface of the ramp. So this is the normal force. And we know that the force of
friction is equal to the coefficient of friction times
the normal force. I'm going to switch colors
because this is getting monotonous. And the force of friction,
we know has to be 9.8 times the mass. So 9.8 meters per second
times the mass. 9.8m is the force of friction. And that has to equal to
coefficient of friction times the normal force. And remember, the normal force
is essentially the force that I'm pushing the pie with. And we know this is 0.8, so we
have 9.8 times the mass-- that's not meters, that's the
mass-- is equal to 0.8 times the normal force. So you have the normal force is
equal to 9.8 times the mass divided by 0.8. What's 9.8 divided by 0.8? 9.8 divided by 0.8 is
equal to 12.25. So the normal force that
I have to apply is 12.25 times the mass. So that's the force
I'm applying. It's time the mass. We don't know the
mass of the pie. So how fast am I accelerating
the pie? Well, force is equal to mass
times acceleration. This is the force, 12.25m--
that's the force-- is equal to the mass times the acceleration of the pie, right? And it's the same pie that we're
dealing with the whole time, so it's still m. And you can take out m from both
sides of the equation. So the acceleration, the rate at
which I have to change the velocity, or the acceleration
that I have to apply to the pie is 12.25 meters per
second squared. And so actually, I have to apply
more than 1g, right? Because g is the force
of gravity. And gravity accelerates
something at 9.8 seconds-- 9.8 meters per second squared. So I have to do something at
12-- I have to push and accelerate the pie at 12.25
meters per second squared. So it's something a little over
1g in order for that pie to not fall and in order for my
normal force to provide a force of friction so that
the pie can reach this bald man's face. I will see you in
the next video.