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Moment/Tension problem

  1. Apr 4, 2009 #1
    1. The problem statement, all variables and given/known data
    It is known that a force with a moment of 7840 lb in about D is required to straighten the fence post CD. If a=8in. ,b=52 in., and d=112 in., determine the tension that must be developed in the cable of winch puller AB to create the required moment about point D.


    2. Relevant equations
    Find angle AED = arctan (52/120) = 23.43degrees
    Use M =D x F ; M=7840lb in, D=ED=112 in, F=unknown
    Substitute
    use algebra to find T(AB)

    3. The attempt at a solution
    I have attached my attempt but i think I missed something because when i check it does not check. 176.05 x sin23.43 does NOT = 7840 (given)
    Now i am stuck:frown:
    Would someone please give me a little nudge/ where did i go wrong? What i am not seeing?
    Thanks
     

    Attached Files:

    Last edited: Apr 4, 2009
  2. jcsd
  3. Apr 5, 2009 #2

    tiny-tim

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    Homework Helper

    Hi fableblue! :smile:

    I can't see your attachment yet, but I notice that your answer is out by a factor of 11.2. which looks suspiciously like d … does that help? :redface:
     
  4. Apr 5, 2009 #3
    Not really:blushing: but thanks, tiny-tim.
    May be when you are able to see the attachment you see what/where my problem lies.

    I am going to try it again later on with the use of the law of cosins. What is throwing me off is the pole not being straight. Can I use[tex]\angle[/tex]EDC, which is at the base on the post that needs to be strightened? [tex]\angle[/tex]EDC=~59[tex]\circ[/tex], I used the law of sines for that but it does not look right; [tex]\frac{112sin(23.43)}{52}[/tex]
    Can i use that angle?
     
  5. Apr 5, 2009 #4

    Doc Al

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    Staff: Mentor

    When finding torque using T = rFsinθ, θ is the angle between r and F. Consider the angle ECD.
     
  6. Apr 5, 2009 #5
    Thank you,
    I also was looking at that angle, actually i was looking at ALL the angles:wink:. But now i can see why. And i am going to assume, from your reply, that the angles are valid no matter if the post is leaning or straight. Now that i think about it the problem is looking for the intial tension in the cable.

    So we get r=52sin81.25, to get the [tex]\angle[/tex] of 81.25[tex]\circ[/tex] i used arctan(52/8) and oppisite interior angles are =, and the used algebra and my answer is F=152.544lbs.
    Does that look about right?
     
  7. Apr 5, 2009 #6

    Doc Al

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    Staff: Mentor

    Not exactly. The way I look at it, r is the length of the post from D to C. Solve for that distance. The angle that you need is BCD. (Finding the angle of 81.25° is a useful step along the way.)
     
  8. Apr 5, 2009 #7
    OK so r=[tex]\sqrt{(8^2 + 52^2)}[/tex]=52.61 in the Mp=rFsin[tex]\Theta[/tex]; substitute- 7840 lb in = 52.61sin81.25F; do algebra and F=147.29 lbs= tension in cable ABC.
    Is this correct? How do i go about checking it?
     
  9. Apr 5, 2009 #8

    Doc Al

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    Your r is correct, but your angle is still wrong. Find angle BCD.
     
  10. Apr 5, 2009 #9
    :cry:deep breath, ok i should have looked at my drawing because then i would have seen that [tex]\angle[/tex]EDC and [tex]\angle[/tex] ECD were not oppisite interior angles:blushing:.
    What i did this time was use 180-81.25 for [tex]\angle[/tex]EDC=98.75 and then i used the law of cosins to find side EC =130.78 and from there i used the law of cosines again to find [tex]\angle[/tex]ECD=57.83, then i used Mp=rFsin[tex]\theta[/tex] [tex]\rightarrow[/tex] F=7840/(52.61sin57.83) = 176.04:uhh: Is this correct? That is my intial result way back in the begining.

    I see that the angle should be 89.78 then F=149.02 and 7840~(52.61sin89.78) * 149.02 i found this through some algebra but now i am trying to see why i can not get it trough trig

    I really do apperciate the help you are giving me.
    Thank You:smile:
     
    Last edited: Apr 5, 2009
  11. Apr 5, 2009 #10

    Doc Al

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    That looks right to me.

    I don't understand what you're doing here. Where does that angle of 89.78° come from?
     
  12. Apr 5, 2009 #11
    I did some switching around of things to try and make it fit:shy:

    So are you saying that my intial results are correct? How do i go about checking this? I substitute 176.04 in for F and they do not = each other. 7840 = ....wait a minute...:eek: SHAZAM>>>>:bugeye: I was checking my solution the WRONG way:frown: not using the whole equation:redface: i was leaving out sin57.83. WOW
    So the way that i did it intially is that a correct way? r=112sin23.43 and TAB= F and i understood what i was doing so is the way i did W A Y back there correct?

    Thanks a bunch
     
  13. Apr 5, 2009 #12

    Doc Al

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    I'd say that your initial method was unorthodox (at least to me), but correct. The "obvious" way to define rFsinθ in this problem is to have r be the distance DC, since we are finding the torque on the post. But there's nothing stopping you from measuring r from D to any point along the line of the force, since the torque will be the same. You used r = the distance DE and the associated angle. That works. :wink:
     
  14. Apr 5, 2009 #13
    Yes, that was because i was told that a force/torque is the same anywhere in the post. It seemed to easy for me that is why I did not have faith in my result and all a long is was the way i was checking it.

    Thank You
     
    Last edited: Apr 5, 2009
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