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Applying Newton's laws

  1. Mar 2, 2007 #1
    a)In terms of theta,µk and w calculate F.
    b)for w=400 N and µk=.25, calculate F and theta ranging from 0 to 90 in
    increments of 10.Graph F versus theta.
    c)From the general expression in part (a) calculate the value of theta for which
    the value of F, required to maintain constatnt speed, is a minimum.(Hint: At a
    point where a function is minimum, what are the first and second derivatives of
    the function? Here F is a function of theta.)for the special case of w=400 N and
    µk=.25,evaluate this optimal theta and compare your result to the graph you
    constructed in part b.


    2. Relevant equations

    i need help with part c

    c)From the general expression in part (a) calculate the value of theta for which
    the value of F, required to maintain constatnt speed, is a minimum.(Hint: At a
    point where a function is minimum, what are the first and second derivatives of
    the function? Here F is a function of theta.)for the special case of w=400 N and
    µk=.25,evaluate this optimal theta and compare your result to the graph you
    constructed in part b.


    3. The attempt at a solution

    this is the solution for part a:
    F = µk*w / (cos(theta) + µk*sin(theta))

    the problem is that i don't know how to get the dervitives in order to solve part c
     
  2. jcsd
  3. Mar 2, 2007 #2

    arildno

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    Just differentiate F with respect to the angle once first. What do you get?
     
  4. Mar 2, 2007 #3
    df/dtheta= w*uk*[-sin(theta)+uk*cos(theta)]/[cos(theta)+uk*sin(theta)]^2
     
  5. Mar 2, 2007 #4
    is what i did right? i mean squring the denomenator?
     
  6. Mar 2, 2007 #5

    arildno

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    Indeed!
    So, where can the extrema of F wrt. to the angle occur?

    EDIT:
    You should have a minus in front of the whole expression.
     
  7. Mar 2, 2007 #6
    i don't think there must be a minus in front of the whole expression.

    what do u mean by where can the extrema of F wrt?
     
  8. Mar 2, 2007 #7

    arildno

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    Oh yes, it should!
    [tex]\frac{d}{dx}\frac{1}{u(x)}=-\frac{u'(x)}{(u(x))^{2}}[/tex]
    by elementary application of the chain rule.

    What does "extrema" of a function refer to?
     
  9. Mar 2, 2007 #8
    is it that i have to set it equal to zero and solve for theta?
     
  10. Mar 2, 2007 #9

    arildno

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    Extreme values of a function consist of the function's minima and maxima.
    So yes, setting the expression for the derivative equal to zero, and solving for the angles lying strictly betwen 0 and 90 will give you the extrema in the interior of the reigon.
    Afterwards, you must determine whether this is a local minimum or maximum.
     
  11. Mar 2, 2007 #10
    shouldn't i first take the 2nd dervitive?
     
  12. Mar 2, 2007 #11

    arildno

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    Why?
    Can't you just determine first the simple fact that the critical point of f happens at the angle [itex]\theta_{m}[/itex], which satisfies the equation:
    [tex]\tan(\theta_{m})=\mu_{k}[/tex]
    To simplify computation of the 2.derivative, remember that AT this critical value, the expression in the 1.derivative's numerator is zero.
    Hence, any term in the 2.derivative that gets multiplied by that numerator expression will disappear when inserting [itex]\theta_{m}[/itex]
     
  13. Mar 2, 2007 #12
    i'm a bit confused. Can you simplify your explaination please?
     
  14. Mar 2, 2007 #13

    arildno

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    Well, start with determining the critical value, i.e, where the derivative of f is 0
     
  15. Mar 2, 2007 #14
    w*uk*[sin(theta)-uk*cos(theta)]/[cos(theta)+uk*sin(theta)]^2=0

    i couldn't solve for theta it's very complicated. Need your help!
     
  16. Mar 2, 2007 #15

    arildno

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    Well, multply both sides with the denominator.
     
  17. Mar 2, 2007 #16
    i see now how we get [tex]\tan(\theta_{m})=\mu_{k}[/tex]
    what's next?
     
  18. Mar 2, 2007 #17

    arildno

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    Now, remember that [itex]-\sin\theta_{m}+\mu_{k}\cos\theta_{m}=0[/itex]
    Remember when you use the rule of fractions when differentiating, one of the terms in the numerator will have as a factor the derivative of the denominator.

    However, when evaluated at [itex]\theta=\theta_{k}[/itex], that derivative is, of course, 0. Therefore, that term in the 2.derivative's numerator vanishes.

    Thus, in order to evaluate the second derivative at [itex]\theta=\theta_{k}[/itex], just keep the term including the derivative of the numerator.
     
  19. Mar 2, 2007 #18
    i don't get the follwing line:
    Thus, in order to evaluate the second derivative at [itex]\theta=\theta_{k}[/itex]
    , just keep the term including the derivative of the numerator
     
  20. Mar 2, 2007 #19

    arildno

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    Well, just find the 2.derivative, then!
    Afterwards, do not multiply out parentheses before inserting [itex]\theta_{k}[/itex], and see what simplifications occur.
     
  21. Mar 2, 2007 #20
    what's the dervitaive of u'(x)/u(x)
     
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