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This problem uses the same concepts as Multiple-Concept Example 17. In Problem 80, an 85-kg man plans to tow a 109 000-kg airplane along a runway by pulling horizontally on a cable attached to it. Suppose that he instead attempts the feat by pulling the cable at an angle of $9.0^{\circ}$ above the horizontal. The coefficient of static friction between his shoes and the runway is 0.77. What is the greatest acceleration the man can give the airplane? Assume that the airplane is on wheels that turn without any frictional resistance.

$6.7 \times 10^{-3} \mathrm{m} / \mathrm{s}^{2}\approx 7\times10^{-3} \mathrm{m} / \mathrm{s}^{2} $

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Cornell University

Rutgers, The State University of New Jersey

Hope College

University of Sheffield

this question is very similar to Question 80. If you haven't done that question yet, I definitely recommend that you take a look at Question 80 before trying these one as diseases lightly. More complicated. Version off Question 80. That being said for this question, I will use the following reference frame. I will call this direction the Y direction on this direction. The X direction to discover what is the maximum acceleration that the men can give to the plane. We should calculate what is the acceleration off the plane and note that the plane will move in this direction and not in these direction. So it is march to begin by applying this Newton's second law to the airplane on the horizontal direction. So applying Newton's second law to the airplane results in the following on the X direction. There is only one force. Acting on this force is the X component off the tension force, which is these won t component X, then net force X directions equal to the mass off the airplane times its acceleration index direction as there is only the tension force acting to the extraction, we have that the tension force component X is the cost of the mass off the plane times its acceleration component. Acts there afford the acceleration off the plane. Component acts is a cost to the tension component X divided by the mass off the plane. Now what we have to do is that reminder is the maximum tension because the maximum tension will give the maximum acceleration. For that, we apply Newton's second law now to the man. So ah, playing You don't step in law to the men again on the X axis because the man is moving on these access, we have the following. Then that force on the X direction is equals. True, the mass off the men times It's his acceleration of the X direction. But the net force that acts on the X direction on the men is composed by two forces. The frictional force that acts to the positive X direction on detention forced component X that will act the negative extraction. This is the component off the tension force that acts on the man. Then we have that the frictional force mines attention Force component X is equal to the mass off the men times he's acceleration X direction I noticed that the man should be using a NY ideal cable to pull that airplane. It means that such cable can't be stretched more compressed then the acceleration off the plane is equal to the acceleration off the man So we can equate these a X with these. Yes, to get the following nutritional force minors attention Forced component X is a facilitation force divided by the minds of the plane times the mass off the meth. Then we can send this tension to the other side. To get fictional force is equals to the tension component X plus tension component X times the mass off the men divided by the mass off the airplane. Now we can solve it for detention component acts by factoring the tension component acts on the right hand side pension component X times one plus the mass off the men divided by the mass off the airplane. Then finally, the tension component X is he goes to the frictional force divided by one plus the mass off the men divided by the mass off the airplane. So we conclude that the maximum tension is exerted when the frictional force is out. So the maximum possible frictional force. Things were dealing with a static frictional force. We know that the maximum value for that force is given by the static frictional coefficient times the normal force that acts on the man on the men because the fictional force is acting on the men. Therefore, it is related to the normal force that also acts on the man. Now, to keep solving this question, I have to organize my board. Let me do it, Okay. By doing that, we can write the tension force component X as the static official or confusions times the normal force that acts on the men divided by one plus the mass off the men divided by the mass off the airplane. Then our next step is to deter mined. What is the normal force that acts on the men? For that we have to apply know Newton's second law on the Y direction as this is not trivial this time. So this goes as follows. Applying Newton's second law to the Y direction results in the following the net force in that direction. Is it close to the mass off the men times his acceleration on the wider action But the force that acts on the man in the wider action is composed by three forces. We have the normal force, the weight force and the Y component off the tension force, which should be something like this. So this is the contention component. Why? Then we can write it as follows. Normal force minus attention component White minus The weight force is equal to the mass off the man times his acceleration to the Y direction Notice that the men we've only moved in this direction. So he's not moving on the y axis. These means that the UAE acceleration is the cost of zero. So this is close to zero then the normal force that acts on the man is he goes to the tension force component. Why plus the weight off the man. No, let me clear the board so that they can continue serving this question. Now we conclude that the X component off the attention force is given by the static frictional coefficient times the y component off the tension force plus the weight off the man divided by one plus the mass off the man divided by the mass off the plane. Now we have to really eat the components off the tension force with the magnitude of the tension force. And we can do that by looking at the following triangle. This triangle, the hypotenuse is attention force. These angle is teeter. Here we have the Y component and here we have the X component. By using the sign off, Peter, we can calculate the relation between T y and teeth because the sign off Tita is a close to the opposite side, which is why divided by t then T white is given by tee times this sign off detail now for the X component, we can use the co sign off Peter, which is given by t X, divided by t. The reform X component of the tension force is given by the magnitude of the tension force times they co sign off, Tita. Now let me organize the board like this to know, get a relation that will tell us what is the magnitude off the tension force. And only then we can finish solving the question. So t ex is t times the co sign off Tita. On the other side of this equation, we have mu static times the Thein Sein off Tita, which is t y plus the weight off the man divided by one plus the mass off the man divided by the mass off the plane. Now we have to solve this equation for T For that, I'll have to use a lot of space on the board and unfortunately, I have to raise the drawing now. Okay, I have done nothing new address reorganizing the ward so they may have a lot of space to solve this equation for attention. No, let's do it. Let's keep the left hand side us idiots on on the right hand side letters split things. So now we have mu s divided by one plus m am divided by MP time Steep time Sign off Dita plus Mu s times Nobody u m divided by one plus m m divided by MP. Now let me send thes term to the other side so that now we have t times the co sign off detail minus team times um, US times The sign off Dita, divided by one plus m m divided by MP is equals tomb us times Noble. You am divided by one plus m m divided by MP now factor detention on the left hand side to get the times go Sign off Dita Miners mu static times The sign off Peter divided by one plus the mass off the man divided by the mass off the plane. And these these ecosystem you startac times the weight off the men. The wait. Is it close to the mass off the man times the acceleration of gravity? So he is and is divided by one plus the mass off the man divided by the mass off the plane. Now let us putting the values that were given by the problem to go sign off nine degrees miners Mute static, which is 0.77 0.77 times the sign off nine degrees divided by one plus the mass off the man 89 divided by 109 000 And these easy cost to muse static 0.77 times the mass off the man 89 times. Nine point H, which is the approximate value for G, and this is divided by one plus 89 divided by 109 000 and these results in the following tee times. 0.867 z cause true. 671 point zero for six. Then the tension is equals to 6 71.0 for six, divided by 0.867 on these results in approximately 773 0.986 New tennis. Now that you know the magnitude off the tension, we can go back and calculates the axe components to finally calculate the acceleration off the airplane. Let me clear the board once again, as we had the red easing the X component off the tension TX is given by t times the co sign off detail, then it is equals to 773 0.986 times the co sign off nine degrees, then the X component off. The tension is approximately 764 point 457 Finally, we can calculate the acceleration as t x 764.457 divided by the mass off the plane which is 109 000 and these results in an acceleration off approximately zero 0.0 seven meters per second squared, which we can write a seven time stem to the minus three readers per second squared

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