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Problem 69 Hard Difficulty

A damp washcloth is hung over the edge of a table to dry. Thus, part (mass $=m_{\mathrm{ca}} )$ of the washcloth rests on the table and part (mass $=m_{\text { oft }} )$ does not. The coefficient of static friction between the table and the washcloth is 0.40 . Determine the maximum fraction $\left[m_{\mathrm{oft}} /\left(m_{\mathrm{cn}}+m_{\mathrm{off}}\right)\right]$ that can hang over the edge without causing the whole washcloth to slide off the table.




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Video Transcript

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