in this problem we're given a general sort of atwood machine problem so there's no numbers
in this one we just know that one of the masses is m the other one is twice that so that's 2m
and the angle of incline is just given by a general angle theta
in part a we're asked to provide complete force diagrams for the blocks so my intent here is to
show all the forces acting on the blocks and the relevant components of those forces and i'll start
with m and m has gravity acting on it pulling straight down as gravity always does and we need
to decompose this into parallel and perpendicular pieces so i'm attempting to draw a perpendicular
here and if i project onto that perpendicular i can draw my perpendicular vector component as a
result we've used many times now that the angle of incline is the same as this angle between the
vertical and the perpendicular and this means my perpendicular component is given by mg cosine
of theta now the parallel component to the surface of the ramp is given by this side of the triangle
mg sine theta but we normally show that attached to the center of the object now we're not
considering friction in this problem we were told it was negligible so the normal force isn't really
relevant to the final solution of the problem but we were asked for complete force diagrams so i'm
going to go ahead and put the normal force in and you can very quickly say that's equal to mg cosine
theta because the force is perpendicular to the ramp must balance but we'll just leave it that way
and finally i have a rope attached to this thing and there's a tension in that rope and i'll just
call that t so that's it for m and then i look at 2m and gravity pulls straight down on that with
a magnitude of 2mg and the other force acting on it is the tension in the rope pulling straight up
and that tension is the same on both sides of the rope so i tried to make the vectors
about the same size and i'll go ahead and make 2mg about twice as long as mg
as long as i'm nitpicking the lengths of my vectors so there we are those are our complete
force diagrams now before i get into part b applying newton's second law to each of these
blocks we need to be clear about the coordinate system we're using to analyze the motion for
each block and this system is going to evolve in this direction 2m is going to fall m is going to
be dragged up the ramp so that's the direction of the actual acceleration and i'm going to call that
positive for the analysis of each of these masses it's important that your acceleration has the
same sign for each mass because we don't want the letter a to stand for a positive thing in
one case and a negative thing in the other now we apply newton's second law to each of these masses
and if i look at m i have t pointing in the positive direction and then i have a minus mg sine
theta that's the force that points in the negative direction for that coordinate system and that's
equal to the mass i'm looking at which is m times the acceleration a then i look at 2m that has a
force of 2mg pointing in the positive direction which is down for the analysis of that mass
minus a force of t that was my upward force which is negative for that coordinate system is equal to
m a well the is 2m and my acceleration is a that's the system of two equations and two unknowns where
the unknowns are the tangent in the string and the acceleration of these masses finally in part c
we want to solve this system so we go ahead and add these two equations together we're already
set up for elimination to work and eliminate the tension in the string and on the left hand side
when i add them i get a 2 mg minus mg sine theta and on the right hand side i get 3m times a and
the m's cancel out and i can factor g out of the left-hand side and we go ahead and solve for a
and it turns out a is g over 3 times the quantity 2 minus sine theta if you enjoyed this video or at
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