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Should be easy but its been a while

  1. Mar 14, 2004 #1
    Given: A force of 40 N is acting on the end of a 'L' shaped beam at angle [tex]\theta[/tex]--beam is attached to a wall at point 'A' and projects straight out and turns up at the very end--that is 8m long and 2m high, find the angle [tex]\theta[/tex] corresponding to the min and max moment at point 'A'. Given: [tex]0^\circ\leq\theta\leq180^\circ[/tex]

    This is the moment eq I came up with:

    [tex]f(\theta)=320\sin{\theta}+80\cos {\theta}[/tex]

    I took the first derivative:









    [tex]f(0)= 320\sin{\theta}+40\cos{\theta}=40[/tex]

    [tex]f(180)= 320\sin{\theta}+40\cos{\theta}=-40[/tex]

    [tex]f(76)= 320\sin{\theta}+40\cos{\theta}\approx330[/tex]

    Its been a while since I've taken calc so my question is "how do I find the angle corresponding to the minimum force?" I know its not 0 or 180 because the force is not collinear with point 'A'; moreover, I know the angle I'm looking for should be about 166 degrees (I solve the problem using trig already but my professori wants us to use calc when and where possible). How can I justify 166 degrees as an answer using calculus?

    Thanks a-bunch
    Last edited: Mar 14, 2004
  2. jcsd
  3. Mar 14, 2004 #2
    Should f'(t) = 320cos(t) - 40sin(t)?
  4. Mar 14, 2004 #3
    no, the derivative of 80 cos(t) is -80 sin(t).

    I'm confident in my solutions, I'm unsure about symbolically proving that a given angle corresponds to a minimum force at 'A'.
  5. Mar 14, 2004 #4
    Yeah, so? The function you gave was [tex]f(\theta)=320\sin{\theta}+40\cos {\theta}[/tex]. Notice the "40cos(t)"...
  6. Mar 14, 2004 #5
    My fault, typo. I'll fix that right now.
  7. Mar 14, 2004 #6

    matt grime

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    So you've found a stationary point. Why are you presuming it a maximum?

    Why when taking square roots do you ignore the other branch which might give you a second stationary point?
  8. Mar 14, 2004 #7
    I ignored the (-) sqrt because that would have produced -76 degrees which is outside the allowed range of 0 to 180 degrees. Also, 76 is the max because it is greater than both the right and left bounds. If 76 degrees had yielded a value less than or equal to the right or left bound I would have concluded that 76 degrees was the angle theta corresponding to the minimum force at 'A'. So since 76 degrees is a first derivative solution that is not the minimum, it must be the max.

    Again, I've solved this with trig, but my professor wants it done using calculus. I don't have to turn this in, but this is exactly like what's on his tests (I've had this same professor for a couple of classes already).
  9. Mar 14, 2004 #8

    matt grime

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    So if a function does not have a local minimum inside the region (ie the restrictions for theta) it must be that the minimum occurs at...?

    You had noted that the global maximum was the local one (the turning point), because of the values at the end points, so you should be able to fill in the ... in the above paragraph.

    Now, your conclusion that it must be around 166 tells us that something has gone wrong somewhere.
    Last edited: Mar 14, 2004
  10. Mar 14, 2004 #9
    matt, you edited your post...

    166 comes from 90+76, but How do I get the second solution from the original equation, or the derivative for that matter. If I use the more elegant approach of [tex]\tan\theta=4[/tex] thus [tex]\theta=\arctan 4[/tex] I still get approx 76 degrees.
    Last edited: Mar 14, 2004
  11. Mar 14, 2004 #10

    matt grime

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    Not sure when you looked at my post, but I think it was wrong on the arctan values.

    The analysis on the global min is correct though.

    Note that equations like Asinx + Bcosx are rewritable as Rcos(x+t) for some choices of R and T, so in and region of 180 degrees you aren't going to get a max and min unless both occur at the end point.
  12. Mar 14, 2004 #11
    I looked at it almost immediatly after you posted it I guess.
  13. Mar 14, 2004 #12
    Thanks for the help. Maybe I'll just change my entire approach and look for the dot and cross products of the force vector and the vector A to (8,2).
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