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Math Help - More optimization problems

  1. #1
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    More optimization problems

    [IMG]http://i115.photobucket.com/albums/n297/pakman134/optimizationsled.jpg[/IMG]
    Attached Thumbnails Attached Thumbnails More optimization problems-gash.jpg  
    Last edited by CaptainBlack; March 23rd 2008 at 12:13 AM. Reason: Image moved from remote server to MHF
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  2. #2
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    This really belongs in the "advance applied math section"

    For question 1. you should know that the maximum value that a friction force is \mu R where R is the normal reaction force between the object and the surface it in in contact with and \mu is the coefficient of friction.

    (if you don't follow what I say here ask me to post a diagram)

    the normal reaction force is given by R = mg - F \cos \theta. therefore the maximum friction force is \mu mg -  \mu F \cos \theta. If there is to be motion the forward must be greater than the maximum friction force (such that there is a net resultant force forwards).

    so we get F \sin \theta > \mu mg -  \mu F \cos \theta

    which rearranges to \tan \theta = \frac{1}{\mu}

    this can rearrange to give F > \frac{\mu mg }{\sin \theta +\mu \cos \theta }

    to find the max/min or F you differentiate using the chain rule. to get F' = \frac{-\mu mg ( \cos \theta - \mu \sin \theta) }{(\sin \theta +\mu \cos \theta )^2}

    we require \cos \theta - \mu \sin \theta = 0

    which rearranges to  \tan \theta = \frac{1}{\mu}

    using some simple trig you get \sin \theta = \frac{1}{\sqrt{\mu^2 + 1}} and \cos \theta = \frac{\mu}{\sqrt{\mu^2 + 1}}

    put this into the formula (with your value of /mu) I get the minimum value of F = 0.148 mg.

    you should be able to notice that F will be a maximum when \theta = \frac{\pi}{2} in which case F = \mu mg

    Edit: F is actually maximum in the case when  \theta = 0 in this case you would just be pulling the sledge vertically
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  3. #3
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    hm...

    I somewhat follow what you're saying and when I follow your process I come up with the same answers as you did, but with your answers they're still coming up as wrong. Maybe it is a simple math error but I can't seem to see anything really wrong with your process. Hm.....
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  4. #4
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    Quote Originally Posted by pakman134 View Post
    I somewhat follow what you're saying and when I follow your process I come up with the same answers as you did, but with your answers they're still coming up as wrong. Maybe it is a simple math error but I can't seem to see anything really wrong with your process. Hm.....

    Well the maximum value of F should be mg as it occurs when \theta = 0 I have check my working again, I am fairly confidient the minimum value of F is 0.148mg but maybe this software your using requires it answer to greater accuracy, have you tired 0.1483mg ?
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  5. #5
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    Quote Originally Posted by bobak View Post
    Well the maximum value of F should be mg as it occurs when \theta = 0 I have check my working again, I am fairly confidient the minimum value of F is 0.148mg but maybe this software your using requires it answer to greater accuracy, have you tired 0.1483mg ?
    ugh i hate these stupid online homework things... You were right about just having to take it out another decimal place. thank you very much!!
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