in this video we're looking at a vertical atwood machine it's
assumed here that the string is light and the pulley is light and frictionless
and we're told that big m is heavier than little m so we know which way the thing is accelerating
i want to compute the acceleration in terms of m and m and g and the tension in the string as well
so i'm going to go ahead and put the acceleration vectors into the problem and it's important when
we analyze each of these masses to define the positive direction as the actual direction
of acceleration that guarantees that when we write down our two equations and two unknowns
a is going to be a positive quantity in both of those so it's really important that you use these
coordinate systems then i'm going to diagram all the forces acting on each moving piece
and then we apply newton's second law to each of the masses so let's get all the forces in
the tension in the string is the same everywhere provided it's a light string
a light pulley and frictionless pulley so i'm going to show the tension pulling up
on big m and that's an unknown tension pulling up on little m
and that's an unknown that's a really common mistake to say t is equal to mg
but if that was true here there wouldn't be any acceleration because the forces would be balanced
so t is an unknown that will be solved for later in a system of equations gravity pulling down on
the big one with a rather large weight vector of mg gravity pulling down on the little one
with a rather small weight vector of little mg note something qualitative in my force diagrams
i made t bigger than little mg so the net force points upward and m accelerates upwards so that
makes sense on the other side i made big mg bigger than t so the net force points downward and that
mass accelerates downward now if you don't draw things to scale like that it's not going to break
the mathematics on you but it's just nice to have things qualitatively right in the diagram
now i analyze newton's second law on each of these masses
i'll start with little m and i'm writing down f net equals m a so f net equals little m a
and using my upwards is positive coordinate system t counts as positive little mg counts as negative
and that's equal to little m times its acceleration so there's our first equation
and because the string is not stretchy a is exactly the same for both objects
big m is also accelerating with the same magnitude but it's moving the opposite direction but we
define that as a positive for the analysis of big m so f net equals m a and i'm dealing with
big m now and the force of gravity counts as positive the way i set up my coordinate
system for this one so big mg minus t is equal to big m a we have our second equation these two
equations are already conveniently set up for elimination if i add the left hand sides the
t's cancel and i get big mg minus little mg and on the right hand side i get little m plus big m
times a and i've immediately got an answer for what the acceleration is i'm going to
go ahead and factor g out of the left-hand side so it looks like g times the quantity m minus m
and i had to divide by the coefficient of a so that's big m plus little m all
right so there's a but how do we get t we can go back to either one of the original
equations i'm going to choose the top one and replace a with this new value that we found
and then we have to do a lot of simplifying so to start things out t just adding this mg
to both sides t is m a plus mg or i could write it as m times the quantity a plus g
but a is given by this expression and i'm going to
do two things at once here i'm going to pull the g out so i have little mg
and the second thing is i'm going to get a common denominator so this term after i pull the g out
that's a 1 but i'm going to write it as m plus m divided by itself so i have m minus m over m plus
m and then to get a common denominator i turn to my left over 1 into m plus m divided by m plus m
now that i have a common denominator i can add the numerators and when i do that
i get a 2 big m and then i have a negative little m plus little m so those cancel and we're left
with mg times the quantity 2 big m over big m plus little m and just cleaning it up a little
i end up with two big m little m g divided by big m plus little m and we're done with that one
just one last note here is that any time you're dealing with an
atwood machine whether it's vertical or whether one of the masses is on a horizontal track
or a tilted track or maybe even two masses on two different tilted tracks connected by a pulley
it always comes down to force diagrams followed by newton's second law applied to each of the moving
pieces be very careful to identify the positive direction for each mass in the way that it's
actually accelerating and that means that a will be a positive number in each of the equations you
write down if you get that mixed up then a will be positive in one equation and negative in another
and that's impossible and so basically the algebra is going to fall apart on you
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