# Physics Question, Week 1

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#### Saiph

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<p>Alrighty, time to get started.&nbsp; This problem is entirely out of my own head, and I haven't worked it out yet, but it should be doable.&nbsp; As such it's pretty plain and unimaginitive, but I figured we'd start small, and build upon it.&nbsp; It's also part of a multi-step problem that I plan to use to fully evaluate the physics problem we know and love, a block on a ramp! (then on to the dreaded Atwoods machine!)</p><p>If anybody has any problem suggestions...I'll figure out a way to take those without cluttering this thread or resorting to e-mail.&nbsp; I do plan on selecting problems that demonstrate the practical side of physics as well.&nbsp; For example I've got a problem in mind later that looks at how fast a car can turn, before it flips over.&nbsp;</p><p>Feel free to discuss the problem in this thread.&nbsp; For those of you aren't familiar with the actual physical concepts behind this problem (and how to apply them) I'll be posting a "mini lecture" either later in the week, or with the "official" solution.&nbsp;</p><p>---------------------------------------------------</p><p>A 12kg Aluminum cube rests upon a wooden board.&nbsp; The coefficient of static friction (Us) is 0.2, and the coefficient of kinetic friction is (Uk) 0.1. </p><p>a) What is the required force to cause the block to begin moving?</p><p>b) What is the position of the block after 11 seconds of appling the force in part a?&nbsp;</p><p>c) If we stop applying the force, what is the final position of the block after it stops moving?&nbsp;</p><p>d) If we change the material of the block, are our calculations affected?&nbsp; What if we change the shape of the block? </p><p>You may use whatever methods you desire and find applicable.</p><p>---------------------------------</p><p>&nbsp;</p><p>Next weeks preview:&nbsp; We tip the board and introduce trigonometry!&nbsp; :::Shudder:::&nbsp;</p> <div class="Discussion_UserSignature"> <p align="center"><font color="#c0c0c0"><br /></font></p><p align="center"><font color="#999999"><em><font size="1">--------</font></em></font><font color="#999999"><em><font size="1">--------</font></em></font><font color="#999999"><em><font size="1">----</font></em></font><font color="#666699">SaiphMOD@gmail.com </font><font color="#999999"><em><font size="1">-------------------</font></em></font></p><p><font color="#999999"><em><font size="1">"This is my Timey Wimey Detector.  Goes "bing" when there's stuff.  It also fries eggs at 30 paces, wether you want it to or not actually.  I've learned to stay away from hens: It's not pretty when they blow" -- </font></em></font><font size="1" color="#999999">The Tenth Doctor, "Blink"</font></p> </div>

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#### origin

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<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>.&nbsp;---------------------------------------------------A 12kg Aluminum cube rests upon a wooden board.&nbsp; The coefficient of static friction (Us) is 0.2, and the coefficient of kinetic friction is (Uk) 0.1. a) What is the required force to cause the block to begin moving?b) What is the position of the block after 11 seconds of appling the force in part a?&nbsp;c) If we stop applying the force, what is the final position of the block after it stops moving?&nbsp;d) If we change the material of the block, are our calculations affected?&nbsp; What if we change the shape of the block? You may use whatever methods you desire and find applicable.---------------------------------&nbsp;Next weeks preview:&nbsp; We tip the board and introduce trigonometry!&nbsp; :::Shudder:::&nbsp; <br />Posted by Saiph</DIV></p><p>'Hand waves in the back of the class'.&nbsp; Are we to assume that the Kenetic Friction is constant?&nbsp; Or do we need to use a nonlinear differential equation, in which case I will never again visit the physics forum.</p><p>&nbsp;</p> <div class="Discussion_UserSignature"> </div>

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#### Saiph

##### Guest
<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>'Hand waves in the back of the class'.&nbsp; Are we to assume that the Kenetic Friction is constant?&nbsp; Or do we need to use a nonlinear differential equation, in which case I will never again visit the physics forum.&nbsp; <br /> Posted by origin</DIV></p><p>&nbsp;</p><p>:::grumbles about smart alecs:::</p><p>Assume the frictional coefficients are constant, no air resistance, standard gravity, and state explicitly any other assumptions you make, or feel you have to make to solve the problem.&nbsp;</p><p>&nbsp;</p><p>side note:&nbsp; If we have to us ea nonlinear differential equation for the coefficient of friction...I'd never visit again either &nbsp;</p> <div class="Discussion_UserSignature"> <p align="center"><font color="#c0c0c0"><br /></font></p><p align="center"><font color="#999999"><em><font size="1">--------</font></em></font><font color="#999999"><em><font size="1">--------</font></em></font><font color="#999999"><em><font size="1">----</font></em></font><font color="#666699">SaiphMOD@gmail.com </font><font color="#999999"><em><font size="1">-------------------</font></em></font></p><p><font color="#999999"><em><font size="1">"This is my Timey Wimey Detector.  Goes "bing" when there's stuff.  It also fries eggs at 30 paces, wether you want it to or not actually.  I've learned to stay away from hens: It's not pretty when they blow" -- </font></em></font><font size="1" color="#999999">The Tenth Doctor, "Blink"</font></p> </div>

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#### DrRocket

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<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>&nbsp;:::grumbles about smart alecs:::Assume the frictional coefficients are constant, no air resistance, standard gravity, and state explicitly any other assumptions you make, or feel you have to make to solve the problem.&nbsp;&nbsp;side note:&nbsp; If we have to us ea nonlinear differential equation for the coefficient of friction...I'd never visit again either &nbsp; <br />Posted by Saiph</DIV></p><p>If you would like to keep the arithmetic very simple, you might have everyone use the acceleration of gravity as 10 ms^-2.&nbsp; That is not too far off, and illustrates the physics clearly.&nbsp; Combined with your choice of coefficients of friction, it makes the problem calculator-free.<br /></p> <div class="Discussion_UserSignature"> </div>

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#### origin

##### Guest
<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>&nbsp;:::grumbles about smart alecs:::Assume the frictional coefficients are constant, no air resistance, standard gravity, and state explicitly any other assumptions you make, or feel you have to make to solve the problem.&nbsp;&nbsp;side note:&nbsp; If we have to us ea nonlinear differential equation for the coefficient of friction...I'd never visit again either &nbsp; <br />Posted by Saiph</DIV><br /><br />I was not trying to be a smart alec,&nbsp;just wanted to make sure I was answering the right question.&nbsp; My first test, in my first real Enginnering course (fluid dynamics),&nbsp;consisted of&nbsp;1 question with parts A, B, and C.&nbsp; By the time I got to part C, I realized I had misread the question!&nbsp; I got a freaking 17 on the test!&nbsp; Scared me to&nbsp;death.&nbsp; Thank goodness you got to drop your low grade or I would be flipping burgers today.</p><p>Nothing against burger filppers!</p> <div class="Discussion_UserSignature"> </div>

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#### Saiph

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<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>If you would like to keep the arithmetic very simple, you might have everyone use the acceleration of gravity as 10 ms^-2.&nbsp; That is not too far off, and illustrates the physics clearly.&nbsp; Combined with your choice of coefficients of friction, it makes the problem calculator-free. <br /> Posted by DrRocket</DIV></p><p>&nbsp;</p><p>So it has been suggested, so it shall be!&nbsp; Gravity is equal to 10 and light speed = 1...wait, scratch that last bit. &nbsp; I loved that part of my physics lectures.&nbsp; the professor claimed everything was a sphere (including, and especially cows) and whatever variable he didn't care to deal with = 1.&nbsp;</p><p>&nbsp;</p><p>Origin:&nbsp; No problem, I know the feeling.&nbsp; Of course I've also had the flipside.&nbsp; I finished the problem, figured I'd completely spaced the real question...and was the only student in class to come close (still wrong, but closer). </p> <div class="Discussion_UserSignature"> <p align="center"><font color="#c0c0c0"><br /></font></p><p align="center"><font color="#999999"><em><font size="1">--------</font></em></font><font color="#999999"><em><font size="1">--------</font></em></font><font color="#999999"><em><font size="1">----</font></em></font><font color="#666699">SaiphMOD@gmail.com </font><font color="#999999"><em><font size="1">-------------------</font></em></font></p><p><font color="#999999"><em><font size="1">"This is my Timey Wimey Detector.  Goes "bing" when there's stuff.  It also fries eggs at 30 paces, wether you want it to or not actually.  I've learned to stay away from hens: It's not pretty when they blow" -- </font></em></font><font size="1" color="#999999">The Tenth Doctor, "Blink"</font></p> </div>

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#### Mee_n_Mac

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<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>---------------------------------------------------</p><p>A 12kg Aluminum cube rests upon a wooden board.&nbsp; The coefficient of static friction (Us) is 0.2, and the coefficient of kinetic friction is (Uk) 0.1. </p><p>a) What is the required force to cause the block to begin moving?</p><p>b) What is the position of the block after 11 seconds of appling the force in part a?&nbsp;</p><p>c) If we stop applying the force, what is the final position of the block after it stops moving?&nbsp;</p><p>d) If we change the material of the block, are our calculations affected?&nbsp; What if we change the shape of the block? </p><p>You may use whatever methods you desire and find applicable.</p><p>---------------------------------&nbsp;Posted by Saiph</DIV><br /><br />So should "we" be posting the (our)&nbsp;answers&nbsp; .... answers plus derivations ... or leave the latter to the "teacher" ?&nbsp; So as not to "give it away" I'll, for the moment, just do the former.&nbsp; </p><p>a) a bit over 24 N.</p><p>b) 60.5 m&nbsp;along the way. (assuming the force above = 24 N)</p><p>c) 121 m. </p><p>d) Hmmm, if you leave the mass and co-effs of friction as stated then no, material nor shape matter.&nbsp; The material and shape can affect the co-effs though, 12 kg of teflon will certainly have different co-effs than 12 kg of aluminum on the wooden board. </p> <div class="Discussion_UserSignature"> <p>-----------------------------------------------------</p><p><font color="#ff0000">Ask not what your Forum Software can do do on you,</font></p><p><font color="#ff0000">Ask it to, please for the love of all that's Holy, <strong>STOP</strong> !</font></p> </div>

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#### Saiph

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<p>Hmm, good questions.&nbsp; I say post the answers whenever you want, and We'll wait for the halfway mark (tommorrow) to post our explainations/derivations.&nbsp; Gives people time to work on it a bit.</p><p>&nbsp;</p><p>I'm also working on a little lecture to go along with the problem, for those who aren't mathematically inclined, but would like to know more about the material and reasoning behind the problem.&nbsp; First part to be posted shortly, and it's likely a doozy, as there's a lot of ground to cover.</p><p>&nbsp;</p><p>Also, if anybody else is interested in doing the problems, but would like simpler ones to start with, let me know!&nbsp; I jumped right in with this one, but I can always scale things back a bit for a question or two.&nbsp;</p><p>&nbsp;</p> <div class="Discussion_UserSignature"> <p align="center"><font color="#c0c0c0"><br /></font></p><p align="center"><font color="#999999"><em><font size="1">--------</font></em></font><font color="#999999"><em><font size="1">--------</font></em></font><font color="#999999"><em><font size="1">----</font></em></font><font color="#666699">SaiphMOD@gmail.com </font><font color="#999999"><em><font size="1">-------------------</font></em></font></p><p><font color="#999999"><em><font size="1">"This is my Timey Wimey Detector.  Goes "bing" when there's stuff.  It also fries eggs at 30 paces, wether you want it to or not actually.  I've learned to stay away from hens: It's not pretty when they blow" -- </font></em></font><font size="1" color="#999999">The Tenth Doctor, "Blink"</font></p> </div>

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#### DrRocket

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<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>&nbsp; a) a <strong><u>bit over</u></strong> 24 N.&nbsp;&nbsp; Posted by mee_n_mac</DIV></p><p>?</p> <div class="Discussion_UserSignature"> </div>

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#### Mee_n_Mac

##### Guest
<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>? <br />Posted by <strong>DrRocket</strong></DIV></p><p>OK,&nbsp;an infinitesimal bit more that 24 newtons.&nbsp; If the net "side" forces = 0 @ 24 N applied, you need just a bit more to cause motion. </p> <div class="Discussion_UserSignature"> <p>-----------------------------------------------------</p><p><font color="#ff0000">Ask not what your Forum Software can do do on you,</font></p><p><font color="#ff0000">Ask it to, please for the love of all that's Holy, <strong>STOP</strong> !</font></p> </div>

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#### DrRocket

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<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>OK,&nbsp;an infinitesimal bit more that 24 newtons.&nbsp; If the net "side" forces = 0 @ 24 N applied, you need just a bit more to cause motion. <br />Posted by mee_n_mac</DIV><br />&nbsp;</p><p>OK, I see what you are getting at and it makes sense.&nbsp; The way that issue is usually handled is to either call the state at which the applied force just equals the resisting force the point of "impending motion" (this is what they do in engineering statics classes) or simply decide that at that point the mechanism of static friction releases the object which is then sliding and the kinetic coefficient takes over.&nbsp; The logical problem is that there is no such thing as in "infinitesimal amount". If you think about how the static coefficient is determined experimentally, it is reasonable to suppose that it is actually starting to accelerate at the "point of impending motion" --- basically you measure the force and increase it until you see motion, then note the new force level to determine the kinetic coefficient.</p><p>It is analagous to an old question.&nbsp; A man climbs on to the rail of a bridge and leaps off.&nbsp; Where was he when he jumped off of the bridge?</p><p>Ans. 1 :&nbsp; On the rail.&nbsp; No, that was before he jumped off</p><p>Ans. 2 :&nbsp; In the air above the rail.&nbsp; No, that was after he jumped off.</p><p>&nbsp;</p> <div class="Discussion_UserSignature"> </div>

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#### Saiph

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<p>Apparently my scanner is on the fritz...</p><p>So, to avoid suspense, Mee_n_Mac has the correct answers posted, my work will be uploaded ASAP.</p><p>Grr beans!&nbsp; Why doesn't technology work the first time!&nbsp;</p> <div class="Discussion_UserSignature"> <p align="center"><font color="#c0c0c0"><br /></font></p><p align="center"><font color="#999999"><em><font size="1">--------</font></em></font><font color="#999999"><em><font size="1">--------</font></em></font><font color="#999999"><em><font size="1">----</font></em></font><font color="#666699">SaiphMOD@gmail.com </font><font color="#999999"><em><font size="1">-------------------</font></em></font></p><p><font color="#999999"><em><font size="1">"This is my Timey Wimey Detector.  Goes "bing" when there's stuff.  It also fries eggs at 30 paces, wether you want it to or not actually.  I've learned to stay away from hens: It's not pretty when they blow" -- </font></em></font><font size="1" color="#999999">The Tenth Doctor, "Blink"</font></p> </div>

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#### DrRocket

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<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>Apparently my scanner is on the fritz...So, to avoid suspense, Mee_n_Mac has the correct answers posted, my work will be uploaded ASAP.Grr beans!&nbsp; Why doesn't technology work the first time!&nbsp; <br />Posted by Saiph</DIV></p><p>Let me offer this explanation without benefit of scanner.&nbsp; I can get my scanner to work but I can't upload the images.&nbsp; Maybe they are too big.&nbsp; Or maybe I don't know what I am doing.&nbsp; Any insight or &nbsp;suggestions that you have are appreciated.</p><p>Now for the problem.&nbsp;</p><p>1)&nbsp;The normal force on the block is mg, or 12Kg x 10 m/s^2 = 120 N.&nbsp; Frictional force at impending motion is 0.2 times that or 24 N.&nbsp;</p><p>2) One motion starts the kinetic coefficient of friction applies and the frictional force resisting the applied force is is 0.1 times the normal force or 12 N.&nbsp; So the net force on the block in the forward directioin is&nbsp;24 N - 12N&nbsp;= 12 N.&nbsp; NThe acceleration after motion starts is a = F/m&nbsp;= 12/12 N/Kg =&nbsp;1 m/s^2.&nbsp; That acceleration lasts for 11 s and the distance covered in that time is 1/2 a*t^2 = 0.5*1*121 m/s^2*s^2 = 60.5 m</p><p>3)&nbsp;If the applied force is removed after 11 seconds and the block allowed to coast the frictional force resisting motion remains as before 12 N in the direction opposite to the motion.&nbsp; Just by symmetry, or conservation of energy the block then slides for exactly the same distance as it moved under motive force. 60.5 m for a total distance traveled of 121m.</p><p>4) With ideallized problem as stated the coefficient of friction and the frictional force are determined by the material of the block and the material of the surface on which it slides alone.&nbsp; So only a change in block material would affect the calculation, and that effect would be seen in the coefficients of friction.&nbsp; In a real-world situation, block shape and roughness would affect the details of the interface with a deformable medium across which it slides and thereby affect the force modeled by friction.&nbsp; In addition, if air resistance were taken into account the overall shape of the body could have a major effect, particularly at high speeds.&nbsp; <br /></p> <div class="Discussion_UserSignature"> </div>

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#### Saiph

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<p>thanks for your synopsis DrRocket.</p><p>As for your #4 answer, spot on except one detail from what I recall.&nbsp; The material and texture do affect the coefficient of friction.&nbsp; However the shape of the material, specifically the footprint of the block on the board, doesn't matter.&nbsp; You could have a tall skinny block, or a short wide one.&nbsp; The frictional force is still the same, as these parameters do nothing to alter the normal force, or the friction coefficient.</p><p>Now, real life experience...is for engineers <img src="http://sitelife.space.com/ver1.0/content/scripts/tinymce/plugins/emotions/images/smiley-tongue-out.gif" border="0" alt="Tongue out" title="Tongue out" /> </p> <div class="Discussion_UserSignature"> <p align="center"><font color="#c0c0c0"><br /></font></p><p align="center"><font color="#999999"><em><font size="1">--------</font></em></font><font color="#999999"><em><font size="1">--------</font></em></font><font color="#999999"><em><font size="1">----</font></em></font><font color="#666699">SaiphMOD@gmail.com </font><font color="#999999"><em><font size="1">-------------------</font></em></font></p><p><font color="#999999"><em><font size="1">"This is my Timey Wimey Detector.  Goes "bing" when there's stuff.  It also fries eggs at 30 paces, wether you want it to or not actually.  I've learned to stay away from hens: It's not pretty when they blow" -- </font></em></font><font size="1" color="#999999">The Tenth Doctor, "Blink"</font></p> </div>

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#### DrRocket

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<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>thanks for your synopsis DrRocket.As for your #4 answer, spot on except one detail from what I recall.&nbsp; The material and texture do affect the coefficient of friction.&nbsp; However the shape of the material, specifically the footprint of the block on the board, doesn't matter.&nbsp; You could have a tall skinny block, or a short wide one.&nbsp; The frictional force is still the same, as these parameters do nothing to alter the normal force, or the friction coefficient.Now, real life experience...is for engineers <br />Posted by Saiph</DIV></p><p>For ideal coulomb friction in the textbook model, you are correct.&nbsp; But reember that I was addressing a less ideal model with deformable materials.&nbsp; Think about taking your tall skinny block to an extreme, a spike.&nbsp; That spike would impose extremely high stress on the substrate, an infinite stress in the case of an ideal point.&nbsp; Those high stresses deform the substrate, creating a dimple, as you move the object laterally that dimple move with it, creating a wave in the substrate, and that increases the necessary force.&nbsp; The macroscopic effect is an increase in the coefficient of friction, and in fact probably a more complex force relationship than that which is modeled by simple coulomb friction.<br /></p> <div class="Discussion_UserSignature"> </div>

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#### Saiph

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<p></p><p>That's why real life is for engineers...that scenario also requires the break down of the board's structural integirty (even just a bit).&nbsp; You're no longer sliding on the board...your partially in the board.</p><p>But your point is well taken.&nbsp;</p> <div class="Discussion_UserSignature"> <p align="center"><font color="#c0c0c0"><br /></font></p><p align="center"><font color="#999999"><em><font size="1">--------</font></em></font><font color="#999999"><em><font size="1">--------</font></em></font><font color="#999999"><em><font size="1">----</font></em></font><font color="#666699">SaiphMOD@gmail.com </font><font color="#999999"><em><font size="1">-------------------</font></em></font></p><p><font color="#999999"><em><font size="1">"This is my Timey Wimey Detector.  Goes "bing" when there's stuff.  It also fries eggs at 30 paces, wether you want it to or not actually.  I've learned to stay away from hens: It's not pretty when they blow" -- </font></em></font><font size="1" color="#999999">The Tenth Doctor, "Blink"</font></p> </div>

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#### DrRocket

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<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>That's why real life is for engineers...that scenario also requires the break down of the board's structural integirty (even just a bit).&nbsp; You're no longer sliding on the board...your partially in the board.But your point is well taken.&nbsp; <br />Posted by Saiph</DIV></p><p>An engineer and a mathematician are placed in a cabin in the woods.&nbsp; Outside the cabin is a stream.&nbsp; Inside the cabin is a window with curtains and an empty bucket.&nbsp; The curtain is on fire.</p><p>The engineer and the mathematician each take the empty bucket to the stream, fill it with water, return to the cabin and throw the water on the curtains, thereby extinguishing the fire.</p><p>Next, they are presented with a similar scenario, except that now the bucket in the cabin is filled with water.</p><p>The engineer picks up the bucket of water, and throws the water on the curtain, extinguishing the fire.</p><p>The mathematician picks up the bucket, pours the water on the floor, reducing the problem to the earlier one that he has already solved. </p><p>&nbsp;You did a nice job with the problem and the lectures for week 1.&nbsp; <strong><font size="3">Encore !&nbsp; Encore !</font></strong></p><h1></h1> <div class="Discussion_UserSignature"> </div>

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#### derekmcd

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<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>The mathematician picks up the bucket, pours the water on the floor, reducing the problem to the earlier one that he has already solved. <br /> Posted by DrRocket</DIV><br /></p><p>Classic!!!&nbsp;</p> <div class="Discussion_UserSignature"> <div> </div><br /><div><span style="color:#0000ff" class="Apple-style-span">"If something's hard to do, then it's not worth doing." - Homer Simpson</span></div> </div>

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