WEBVTT
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the secret off. This question is treating this part
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off the rash cough in this part off the washcloth
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separately, and only then joined these two parts by
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doing that were actually treating this system as another very
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familiar system, which is much easier to deal with
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, which is this one. We have a block
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here, and this block has three forces acting only
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which would be W C a normal force. Because
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it's lying flat on some kind of table. There
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is a frictional force that tries to hold it in
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place and another force therefore forces, which will be
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the tension force. And here we have these object
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and the cable connecting this part toe. Another part
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that is here, sir, on the other parts
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that are two forces acting these force and that force
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to much easier. Now, isn't it much for
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much more familiar to us? So let the street
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this problem, which is completely equivalent to the actual
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problem off the question. So what is the maximum
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fraction off the wash cough that can lies off the
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table before it falling onto the ground? So we
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are going to treat they ceased them instead because it's
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much easier. So there are two access off interest
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here. Why access and ex access and we will
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be using both. Let me call his number one
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and these order number two by applying Newton's second law
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number one, we get the following. So for
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the vertical access we have that the net forest on
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the Y direction is it close to the mass M
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C kinds acceleration on the Y direction, which is
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the question zero because it's lying on the table.
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It's not moving and it's not going to move.
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So the net force underway. The reaction is close
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to zero and then the nets force in the Y
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Direction is composed by true forces, the normal force
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which points to the positive vertical direction minus the weight
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force that points negative vertical direction. This is equal
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to zero. So the normal force is it goes
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to the weight force as we can see from the
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beginning. Now what happens when we apply it to
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the X axis? So you're playing toe the exact
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says we have the following The net force on the
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X axis is equals to the mass. See a
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times acceleration on the exact sense which is equals to
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zero for the same reason. Acceleration on the Y
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access Waas equals 20 Then the net force on the
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X direction is actually goes to zero. But the
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net force on the accident reaction is composed by two
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forces. The frictional force which points to the negative
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X direction as the tension force which points in the
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positive direction. Then this tension force isn't close to
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the frictional force. So far, what we got
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is the following. The normal is equal to the
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weight on detention musicals to the friction. Let me
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organize the board before applying Newton's second law to the
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block number two. Okay, now, applying Newton's
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second law to block number to get the following note
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that block number chew their only vertical forces acting so
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we only have force and on the Y direction and
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then the net force in the right direction is equal
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to the mass. But now the mass off the
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table times acceleration on the Y direction for that second
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block. This is also equal to zero because it's
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not moving. Therefore, the net force on the
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Y direction is equal to zero on the second block
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to but on the second block we have only two
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forces owned the vertical direction. We have the tension
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force which points with the positive vertical direction have the
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weight force that points the negative vertical direction. Therefore
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, detention is equals to do eight force. Now
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let me organize the board and finish the question.
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We can finish this question by using these three equations
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that we have derived from the situation. So the
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second equation tells us that the tension force is the
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cause of the fictional force. But note that this
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should be the static regional force. Since nothing,
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nothing is moving and the maximum value off the static
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regional force is given by the static frictional coefficient times
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the normal force. But note that these equation was
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derived it for block number one. So the normal
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force near is this end C eight, three times
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and see eight. But now, from the first
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equation, we see that N c a. A
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is. It goes to wc eight. Therefore,
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the tension force is it goes to the static frictional
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coefficient times their weight off the washcloth that is on
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the table now from the third equation we get that
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attention, which is because of this so frictional acquisitions
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. Times wait. Force acting on the washcloth there
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is on the table is equals through the weight force
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that acts on the part off the washcloth that is
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off the table. Then remember that the weight is
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given by the mass times acceleration of gravity. So
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we have that W C A is because to m
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c a times G unravel you off easy goes to
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em off times g. We can simplify jeez here
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so that we get that the static frictional coefficient times
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m c is equals to em off. Then the
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question asked is about the following. It tells us
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to calculate what is the maximum fraction am oft divided
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by M. C A plus m oft using these
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relation, we can substitute the office here by the
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left hand side of this equation. So this is
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equals true. Um, you asked times M c
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a. Divided by m c A plus mu us
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times M c A. No. We can factor
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out and see a on the denominator to get i
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m. C. A times one plus from U
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. S. Divided by mute as times m.
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C. Eight. So these factors off EMC are
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simplify it and they get mu us divided by one
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plus mute as finally using the value for the static
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frictional coefficient. We got 0.4 divided by 1.4,
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which is approximately zero point twin tonight. So this
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is the maximum fraction off the wash cough that can
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lies off the table without bringing the world washcloth to
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the ground.