in this video we're given four one kilogram masses that are all tied together with strings
those strings have unknown tensions that we're interested in solving for in part b
and so we're pulling on this system with a 10 newton force pointing to the right but the entire
system is moving with constant speed because all those blocks are experiencing a kinetic friction
force opposing that force to the right in part a we want to get that coefficient of kinetic
friction that all the blocks are feeling and again in part b we want to get all the unknown tensions
and the strings that are connecting the blocks now to answer part a if that was the only question we
were given there's a really quick way of getting the answer to that and i'll illustrate that first
in this solution but part b forces us into a more detailed analysis that could answer part a for us
as well so we'll look at the shortcut solution for part a and then we'll do the full blown problem
so that we get part b so for the shortcut method the key is to realize you can treat the entire
group of four one kilogram masses as a single four kilogram mass and the reason this works
is that for the analysis of net force all of those tensions are internal forces so what happens for
example with t1 is it pulls to the right on one of the blocks but it pulls to the left on another
block with exactly the same magnitude so when we're adding up the sum of a bunch of internal
forces they contribute zero to the vector sum of the forces so we treat that four kilogram
block as a single mass and it experiences an external force of 10 newtons pulling to the right
and this block also experiences a friction force pointing to the left
and that friction force is related to the normal force by the usual equation fk equals mu k times
the normal force where mu k is the coefficient of kinetic friction now to get at the normal force we
look at the force of gravity pulling down on this block and that force of gravity is given by mg
where m is 4 we're using 9.8 as an approximation for g and this gives us a weight of 39.2
newtons now the normal force pointing up that has exactly the same magnitude as the weight
pointing down because there are no other forces interfering in the vertical direction and i know
the acceleration is zero vertically so those must balance and i get a normal force of 39.2 newtons
this means my total friction force on the entire system is going to be 39.2 times the unknown
coefficient of kinetic friction mu k now we apply newton's second law f net equals m a
and the acceleration of the system is zero again because we have constant speed here
and i have a 10 newton force pointing to the right so we'll make that the positive direction
and then i have 39.2 mu k my friction force pointing to the left that takes a minus sign
because it's pointing to the left so we add 39.2 mu k to both sides and divide by its coefficient
and when we solve for mu k we get 0.255 okay so that answers part a and now to answer part b
one way to do it would be to start from the block all the way on the left it experiences an unknown
force of t1 to the right now that we have the kinetic friction coefficient we can figure out
the force of friction on that block then we apply newton's second law and we can solve for t1 and so
we'd work our way through the blocks from left to right solving for t1 then t2 then t3 but because
we have to get into the detailed analysis of looking at the forces on every single block anyway
i want to illustrate how you could attack that more general approach right from the beginning
so this is the alternative more thorough solution to part a and it naturally leads to the solutions
for part b so what we're going to do in this more systematic approach is just start from
the left work our way to the right and look at the net force on every single block and we know
the acceleration of every single block is zero so let's get the forces into the first block and that
first block is feeling a force of t1 pulling to the right and then it has a friction force
pulling to the left and again that's given by the coefficient of kinetic friction times the normal
force and again the normal force on this is equal to the weight so we'll pop those vectors in real
quick and that weight vector pointing down that's mg again but m is one so this gives us a weight of
9.8 newtons which in turn means the normal force is 9.8 newtons so this means our friction force is
just 9.8 times mu k then we apply newton's second law so f net equals m a where again everything
is moving at constant speed so we know that a is equal to zero and the net force on this block is
just t1 pointing to the right minus that friction force pointing to the left which is 9.8 mu k
now given that we've already used a trick to solve for mu k you can see that it's really quick to
solve for t1 here but again what i want to do at this stage is just illustrate the most general
detailed solution starting from the beginning so there's our first equation then we apply the same
kind of analysis to the next block for the next block the force diagram has one additional force
compared to the first block and the reason why is that friction points to the left as it strike to
the right but t1 is also pulling it to the left so i have a t1 pointing to the left and i have
an fk pointing to the left and of course i have a t2 pulling to the right the weight of this block
is exactly the same as the first one that's 9.8 newtons making the normal force again 9.8 newtons
and the friction force again 9.8 times mu k applying newton's second law to the second
block i have t2 pulling to the right and then i have two forces pointing to the left that's
t1 pulling to the left so minus t1 and then fk pointing to the left so minus mu k times
the normal force or 9.8 mu k that's all equal to zero again because the acceleration is zero
so this point the pattern is quick to recognize my third block has a force of t3 pulling to the
right it has t2 pulling to the left and a friction force which is equal to the same friction force
on the first two blocks and i can generate a third equation that's t3 pulling to the right
t2 pulling to the left and that same magnitude of friction force because i have the same mass
9.8 mu k pulling to the left and that's equal to zero and finally we get a fourth equation for the
block that's all the way on the right and that has 10 newtons pulling to the right t3 pulling to the
left 9.8 mu k pulling to the left and that's equal to zero so when we begin the problem with this
more detailed analysis we immediately see that we have a system of four equations and four unknowns
the unknowns are mu k t1 t2 and t3 in other words the answers to all the questions in the problem
so if we take this most general approach to the problem right from the beginning instead of using
this little shortcut to get mu k we can just solve everything all at once and we're conveniently set
up to use elimination on this system it's not that big of a deal to solve the system
if i add the first two equations i see that the t1s are going to cancel and it generates
another equation with a t2 in it and minus 2 times 9.8 mu k now i take that equation and sub it into
the third equation and when i do that the t2s cancel leaving me with a t3 and then i have a
negative 2 times 9.8 mu k minus another 9.8 times mu k so minus 3 times 9.8 times mu k is equal to
0 and then i sub that equation into the fourth one and the t3s cancel and i get 10 minus 4 times 9.8
times mu k is equal to zero we add the four times nine point eight times mu k to the right
hand side and divide by its coefficient this gives us exactly the same answer that we got
with the shortcut 0.255 now again if that's the only question that you were asked the shortcut is
better and faster but part b forced us into this detailed analysis anyway to find the tensions in
the strings i have to go back and sub in mu k so we're going to start by finding t1 because that's
the simplest equation t1 is just 9.8 mu k and we just plug in our mu k and this gives us a tangent
of 2.50 newtons next we've already generated an equation that has only t2 as the unknown that was
obtained by adding the first two of our original newton's second law equations so i have t2 minus 2
times 9.8 mu k equals 0. in other words t2 is just twice 9.8 mu k in other words twice the previous
answer for t1 so that's 5.00 newtons and then finally for t3 again we already created a simpler
equation for it by using elimination t3 is 3 times 9.8 mu k in other words 3 times t1 and that's 7.50
newtons so we've got mu k t1 t2 and t3 and we're done if you enjoyed this video or at least found
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