WEBVTT
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we begin this question by applying Newton's second law in
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these reference frame, these will be the Y axis
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and these will be the X axis. Then we
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will apply Newton's second law to both blocks in both
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directions. Let us begin by applying to block number
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one to on the X direction we have that the
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net force that acts on that direction is equal to
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the mass off block number one times its acceleration the
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X direction. But there is only one force that
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is acting on block number one that direction. And
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is this defection all force number one, which happens
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because someone dispassion block number two so it tends to
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move in this direction. Therefore, what happens is
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that block number one tends to move in this direction
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. We've respect to block number. True, both
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blocks were moved to the right, but these block
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tends to move in a slower fashion. Therefore,
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it tends to fall behind in orderto avoid that from
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happening the fictional for seven years and try to bring
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the way for the block number choo. Then this
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is why we have to frictional forces. One frictional
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force is acting from block number two on two block
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number one and another fictional force is acting from block
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number one on block number two, trying to make
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it not move. So for a block number one
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, we have the following. Frictional force number one
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is equals to the mass off the block number one
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times its acceleration in the X direction, then doing
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the same for the vertical direction. We got the
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net force in the Y direction is the coast the
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mask off block number one times its acceleration in the
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direction. But it's not moving, and it's not
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going to move in the vertical direction. Therefore,
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its acceleration is it goes to zero. Then there
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are two forces acting that access and one which is
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the normal forced that block number two exerts done Block
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number one on the weight forced off block number one
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. Then we have n one minus the waste number
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one meaning close to zero. So the normal force
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number one is, of course, the weight off
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block number one. No letters applying Newton's second law
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to the second block on the X direction we have
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that the net force and that the reaction is because
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the mass off block number two times its acceleration.
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That direction. There are two forces acting own blocks
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number choo on the horizontal direction, and those forces
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are these. Apply it force and the frictional force
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that block number one exerts on block number two trying
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to make it not move. Then these results in
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the following F that points to the positive direction.
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Mine is the friction number. True, that points
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to the negative direction is because of the mass off
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like number two times its acceleration. Okay, now
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let us apply in Utah State with lot block number
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two on the vertical direction. Doing that, the
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net force in the wider action is given by the
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mass off the block times its acceleration in that direction
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. Again, it's not moving, and it's not
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going to move on the wider action. So the
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acceleration is the question. Zero then noticed that the
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Net Force the weather action is composed by three forces
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to weights that points downwards on the normal force that
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the table exerts on block number two, then into
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minus w one. Miners that were true is equals
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to zero, meaning that the normal number two is
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he close to this some off the weights. OK
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, but what we want? Better mind. We
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want to get your mind. What is the maximum
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possible value for these force f such that one book
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isn't sleeping on top of the order. And what
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is the condition for this to happen? If they
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are not sleeping with respect to one another, then
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they are moving with the same acceleration. So the
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key condition is this one. It must be true
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that a one X is there close to a two
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X. This must be true. And are you
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call these accelerations eight on Lee. Then we have
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something with these equation at these equation at least it
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means that using this condition equation number one can be
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written ized. Fictional force number one is he goes
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to the mass number one times acceleration and for equation
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number two F minus frictional force number chewed is the
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cost of the mass number two times its acceleration.
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Then notice that we want to discover what is the
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maximum possible value for the force. So these means
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that we should also use the maximum possible value for
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the frictions and the friction in this context is the
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static frictional force. So the fictional forced number one
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is equals to static friction off refuge in times the
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normal force that acts on block number one. Similarly
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, the frictional force number chewed is it goes to
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the static, which no confusion times. And these
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is important to tell you their normal force that these
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block exerts on this block. So this is multiply
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it by the weight off block number one because the
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weight off block number one is the force that this
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block is exerting on these older block. Then it's
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even by these equations. But we also know that
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the normal force number one is he close to the
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weight number one. Then then we conclude that both
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frictional forces are equal. Therefore, let me write
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it like this on both frictional forces. I will
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call f f. Therefore, we can write these
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equations one and true as follows F f is equals
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to m one times eight for the second equation F
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minus. F f is equals to M Chu times
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a. Then we solve both equations wanting to for
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the acceleration so eight is equals. Two F f
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divided by m one and also a Is it close
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to F f? It's already f minus f f
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divided by the mass number Truth. Then we can
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equate both sides to get f f divided by m
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one is equals two f minus f f divided by
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m True. Now we can solve this equation for
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the force f and these gives the following send this
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term to the other side to get and true.
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Divided by m one times f f is equals two
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f minus f f and I will say this term
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to the other side to get at the force f
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Is it close to m two? Divided by m
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one times f f plus f f finally factor divisional
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force to God fictional forced times and choose divided by
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m one close one is equals two f So this
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is expression for the force f. Now let me
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clear the board to calculate what is its maximum value
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. Okay, so the maximum value off force F
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happens when the frictional force is also a maximum.
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So f is given by the static frictional coefficient times
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the normal force number one times am true, divided
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by M one plus one But remember that and one
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is because of the way to number one. Then
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f is it close to the static friction aquisitions times
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and one times geek times entry divided by M one
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was one. Remember, In that G is approximately
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9.8 meters per second squared. We get that f
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is it close to 0.6 times. Five times 9.8
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times 12 divided by five close one And these gifts
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and maximum force of approximately 100 new terms. So
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this is the answer for the first item? No
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, To solve the second item, let me do
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a little change in the drawing and keep the result
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of the first item here. So 100 new times
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. Okay. Now, instead of acting here,
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the Force F is acting dear directly on block number
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one. Then we have to calculate the maximum force
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again. For that, we have to use new
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toe Second law, let us begin by a plane
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. You don't second law on this block on the
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horizontal axis because these real involved value off the force
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f. So for block number one direction, the
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net force next direction is equals. True, the
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mass off block number one times its acceleration again.
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Both blocks need to have the same acceleration. If
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they are not sleeping on top off on an order
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, then let you write it like that, I
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would just put eight in here. The net force
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that acts on block number one is composed by true
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forces now the frictional force that tries to push this
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block ahead. Then, because of the action off
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this force block number two, we'll try to keep
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this block in place. So instead, off what
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we had before, we now have that the frictional
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force that acts on block number one is pointing to
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the left. And on the other hand, when
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Block number one tries to move to the right,
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it tries to bring block number two with it.
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So now we have a frictional force acting block number
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two to the right. Then, according to Newton's
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second Law for Block number one on the horizontal direction
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, we have plus F minus frictional force number one
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, So F minus frictional force number one is because
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the mask off the first block times its acceleration them
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remember that the maximum possible frictional force is given by
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static friction coefficient times the contact force, which in
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this case is the normal number one. It's very
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easy to see that the normal number one equals with
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the weight number one we have already don't that calculation
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actually, Then we can use the result to write
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F minus mu static times the weight number one Z
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goes to the mass number one times acceleration, then
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the acceleration according toe, the Newton's second law.
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Apply it on block number one. Is he close
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to F minus mu static times and one times G
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, divided by the mass number one now doing the
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same for the order block. We get that in
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that force. Acting eat on the X direction is
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because the mass off that block number two times its
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acceleration, which is also ate. Then I noticed
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that the Net force that acts on block number two
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is only the frictional force number two. So we
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have the frictional force number two being equal to the
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mass number two times acceleration. Then I noticed that
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the frictional force number two must be the maximum,
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so it's given by the static frictional coefficient times the
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contact force between blocks one. And truth of that
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contact force is the wait for us off block number
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one and these results in mews static times that mass
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number one past G so mu static times the mass
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number one times g Zico's true. The mass number
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True times acceleration. The reform we have the acceleration
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being given by muse Static times The mass number one
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divided by the mass number True pines acceleration off gravity
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. Then we can equate these two equations to get
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an equation for the force. F By doing that
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to get the following f divided by the mass number
206
00:13:11.759 --> 00:13:16.529 A:middle L:90%
one minus mu static times G. I have simplify
207
00:13:16.529 --> 00:13:20.509 A:middle L:90%
it. These masses on the second term is equals
208
00:13:20.509 --> 00:13:26.340 A:middle L:90%
to the static friction coefficients times G Times am one
209
00:13:26.350 --> 00:13:31.350 A:middle L:90%
divided by M. Chu Then f is the question
210
00:13:31.350 --> 00:13:35.639 A:middle L:90%
You static times Jeep times m one divided by mt
211
00:13:37.139 --> 00:13:39.799 A:middle L:90%
plus muse static times G and this is F By
212
00:13:39.799 --> 00:13:45.710 A:middle L:90%
AM one they only can factor out new static times
213
00:13:45.710 --> 00:13:48.429 A:middle L:90%
G to get f divided by M one is the
214
00:13:48.429 --> 00:13:54.480 A:middle L:90%
question You static times G times am one divided by
215
00:13:54.490 --> 00:13:58.149 A:middle L:90%
m two plus one and finally sent this term to
216
00:13:58.149 --> 00:14:01.779 A:middle L:90%
the other side. So the force is mu static
217
00:14:01.539 --> 00:14:07.039 A:middle L:90%
m one g and one divided by m. True
218
00:14:07.259 --> 00:14:11.049 A:middle L:90%
was one then we just have to plug in their
219
00:14:11.049 --> 00:14:13.690 A:middle L:90%
values that were given by the problem. And these
220
00:14:13.690 --> 00:14:16.629 A:middle L:90%
we will result in the following. F is equals
221
00:14:16.629 --> 00:14:26.149 A:middle L:90%
to 0.6 times, five times 9.8 times and 15
222
00:14:26.159 --> 00:14:30.620 A:middle L:90%
divided by 12 close one. This gives us a
223
00:14:30.620 --> 00:14:35.740 A:middle L:90%
force off approximately 41.6 neutrons so that his dance or
224
00:14:35.740 --> 00:14:41.269 A:middle L:90%
for the second item, then this is the final
225
00:14:41.269 --> 00:14:41.960 A:middle L:90%
answer for the complete problem.