in this video we're looking at a vertical Atwood machine with two masses that are given in a very
general form one of them is 2m and one is 3M where m is just some unknown Mass quantity
we're told the pulley is light and frictionless and what this means for our Force analysis is
that the tension is going to be the same on each side of the string in part A we're asked to get
complete Force diagrams on these masses so we'll put all the force vectors into the diagram and
in our Force diagrams each Mass has the force of gravity pulling down on it that's the weight so
I wrote it as a little W and that's just mass times the acceleration of gravity well for the
mass 2m that's a 2mg for the mass 3M that's a 3mg the only other force acting on each of these
masses is the tension pulling up and that's the same on the left and right sides so that's just
a t pulling up on each of the masses notice the relative magnitudes of these forces were chosen
carefully I made 3mg longer than 2 mg because it's a larger weight and I can see which way
the Atwood machine is going to accelerate the heavier Mass goes down and the lighter Mass goes
up so I made tea so that it was bigger than 2 mg guaranteeing the net force is upward on 2m
and I made T so that it was smaller than 3 mg guaranteeing I have a net force downward on 3M
now it's not strictly mandatory that you get those relative magnitudes right in the picture but it
can help your intuition if they're accurate now in Part B we apply Newton's Second Law to each block
and then we'll set up a system of equations for this thing and one important point about setting
up Newton's second law in an Atwood machine is you have to have a consistent definition of the
direction of acceleration and because my mass 2m is going up and 3M is going down I'm just
going to call the actual direction of motion the positive direction and this guarantees that our
acceleration will have the same sign for each of these masses which is critical for the algebra
to work out so we apply Newton's Second Law to the mass 2m and that's just F net equals m a Now using
the sine convention the way we defined it t counts as positive and that 2 mg counts as negative it's
pointing in the negative Direction so there's my net force and that's equal to the mass that we're
talking about that's 2m times its acceleration we do a similar thing for 3M but for this analysis
downward was defined as positive again it's just F net equals m a this time our downward 3mg counts
as positive and our T counts as negative that's the net force and it's equal to the mass we're
talking about 3M multiplied by its acceleration so that's the system of equations that we're asking
about in Part B now in part C we want to get the acceleration of the blocks in terms of G
so like so many of these Atwood machine problems we take advantage of elimination if I add these
two equations the t's are going to cancel on the left hand side then we get a negative 2 mg plus
3mg which is just 1 mg on the right hand side we get a 5m times a our M's are going to cancel out
and we quickly solve for a here that's G over 5. so the acceleration of this system is one-fifth
the acceleration of Free Fall finally in part D we're asked to get the tension in the string in
terms of M and G now that we have the acceleration all we have to do is sub that back into either one
of the equations that we developed in Part B and we can solve for T and I'm going to go ahead and
do this in the top equation that was T minus 2 mg is equal to 2ma and I'll go ahead and solve
for T and this gives me a 2m on the right hand side factored out of a g plus a now I sub in
the acceleration that I found in part C that's G over 5 and we can Factor G out of this so I have
2 mg and what's left in those parentheses is a one plus one over five that's five over five plus one
over five in other words six fifths and as a final simplification we'll combine our constants out in
front so that's 12 fifths mg now as a final check on your work you should make sure the magnitude
of this tension makes sense I know that my mass 2m is going to accelerate upward with the force
of gravity pulling down on it with a magnitude of 2 mg my tension better be bigger than that
it's got to be bigger than 2 mg and 12 over 5 is a little bigger than 2 so that works if I do a
similar analysis on the 3M Mass I know the tension must be smaller than 3mg so this thing accelerates
downward and 12 over 5 is a little bit less than 3 so that works also if you enjoyed this video
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