in this video we're given an atwood machine that has a bit of a complication so i have a horizontal
track with a cart rolling along with wheels that are essentially frictionless but that cart has a
block stacked on top of it and that block will accelerate together with the cart held in place
by static friction and we get into the details in parts c and d to compute the friction force
that it feels and the minimum coefficient of static friction to keep it in place but part a
is a pretty standard atwood machine question i want the acceleration of the system and the
tension in the string and the key to this is to realize that provided the block doesn't slip
those two masses on the track move as one mass that has a mass of 650 grams or
0.650 kilograms i'll go ahead and convert the mass of the hanging mass as well that's 0.355
kilograms and then to solve the outward machine we get into the standard force diagrams
the hanging mass feels a force of mg pulling down that's the force of gravity i crunch the numbers
real quick because i have plenty of room in my diagram 0.355 times 9.8 and this gives me 3.479
newtons the other force felt by that mass is the tension in the string pulling up and i'm
intentionally showing that vector shorter than the weight vector because i know the weight is
bigger than the tension because i know that block is accelerating downward and as usual
provided the pulley is light and frictionless the tension will be the same everywhere in the string
so that cart with the block on top also feels a tension of t and that's pulling to the right
it's clear that the atwood machine is going to accelerate in this direction where the hanging
mass goes down and the cart goes to the right and we're going to define that as the positive
direction for the analysis of forces on each of these masses next we apply newton's second
law to each of these masses again we're treating the block on the cart as a single
0.650 kilogram mass so for the 0.650 kilogram mass if i look at the net force in the x direction
that's just t and that's equal to the mass that object times the acceleration of that object
and if i plug in the mass of that object i arrive at my first equation t equals 0.650
times a then i look at the hanging mass 0.355 kilograms
and it has two forces acting on it and again downwards going to be the positive direction for
the hanging mass because i want the acceleration to come out to be a positive number in both of
these equations and i know the actual acceleration is downward so mg counts as a positive and the t
will count as a negative that's pointed in the negative direction and that's equal to the size
of that mass multiplied by its acceleration so i plug in the numbers and i have 3.479 newtons
pointed down minus a t pointed up that's equal to the size of that mass which is 0.355
times its acceleration again this is just applying newton's second law f equals ma to each individual
moving piece of the atwood machine so now i have two equations and two unknowns i have t
equals 0.650a and 3.479 minus t equals 0.355 a and using elimination is the quickest way to
solve this system of equations if i add the left hand sides the ts are going to cancel
and that gives me a 3.479 and then if i add the right hand sides i get 0.6508a plus 0.355a
and this gives me 1.005 times a now i can solve for the acceleration and i get 3.46
meters per second squared now one sort of crude check that you can always do on your work is to
verify that your acceleration is less than free fall because for the hanging mass the fastest it
could ever fall is free fall but there's a tension pulling up on it and it's slowing it down so we at
least pass that test next we get the tension in the string and i can go back to either one of the
original equations to get this done i'm going to go back to the first one because it's the simplest
so t is equal to 0.650 times a and now i have a and i get a tension of 2.25 newtons now in part
c we start to zoom in and focus on this 200 gram block that's being dragged along by the force of
friction if there was no friction force that block would just stay at the same x coordinate and the
cart would slip out from underneath it and the reason it's accelerating with the cart is that
there's a static friction force pointing to the right and that static friction force depends on
the normal force and that depends on the weight of that block so i have a weight of that little block
of mg so 0.2 kilograms times 9.8 meters per second squared and this gives me 1.96 newtons
there's nothing else going on in the vertical direction and i know this thing has a zero
vertical acceleration so the normal force must balance the weight so the normal force must be
1.96 okay so re-reading the question in part c i want the friction force on the block and i know
the acceleration of the block because it's accelerating with the cart and that's 3.46
meters per second squared and i know the mass of that little block that's 0.200 kilograms
and the only force acting horizontally on that block is the static friction
force so i can write down newton's second law f net equals m a
if you like you could specify we're talking about the x direction net force here and
there's only one force it's the static friction force and that's equal to the mass of the block
times its acceleration which is 3.46 meters per second squared and i find that the static
friction force is 0.0692 newtons finally in part d we're asked to get the minimum coefficient of
static friction so the block won't slip so if we pull too hard on that cart it'll slip out
from underneath that block just like pulling out the tablecloth from beneath the dishes
we're trying to figure out how big does that static friction coefficient have to be
to keep this thing locked in place accelerating with the cart
and so the key physical assumption here is to assume that the static friction force is maxed out
in other words we're right at the edge of where any faster acceleration would cause it to break
loose so we're assuming the static friction coefficient is small enough that our static
friction force is at its maximum possible value as the outward machine accelerates and there's
a nice little formula of course for the maximum static friction that's mu s times the normal force
well we have the static friction force and we have the normal force so now we can solve for
the coefficient of static friction so that's going to be at 0.692 newtons and the normal
force was already computed that's 1.96 newtons and i get a static friction coefficient of 0.353
the reason this is the minimum static friction coefficient is that if it got any smaller than
0.353 i wouldn't be able to produce this static friction force that i measured
in part c it would require more than the maximum possible and the thing would break
loose and slip off the back of the cart if you enjoyed this video or at least found it useful
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