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Finding the tension force and torque

  1. Mar 14, 2014 #1
    1. Finding tension force and torque (details attached)

    http://imgur.com/cZLG9xP




    2. please see attachment

    http://imgur.com/cZLG9xP




    3.
    For part 1). I get

    ​[itex]{​T} = -\sin \alpha(i) +\cos \alpha(j)[/itex]

    is this correct?

    (Where T is the tension force)


    For part 2). I am stuck .

    What I did was get

    ​[itex]{W} = -mgj[/itex]

    Then get the position vector as

    ​[itex]\frac{1}{2}2L\sin \alpha(i) \sin \alpha(j)[/itex]

    Simplifies to

    ​[itex]L\sin \alpha(i) \sin \alpha(j)[/itex]

    So the answer is

    ​[itex]-mgL\sin \alpha(k) [/itex]


    Is this correct or am I completely off? Also will the end answer contain ​[itex] \alpha[/itex] or [itex] \theta[/itex]?

    Thanks (not sure if I can chat here but great forum and hello as I'm new here)

     
    Last edited: Mar 14, 2014
  2. jcsd
  3. Mar 14, 2014 #2

    SteamKing

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    no attachment.
     
  4. Mar 14, 2014 #3
    Sorry about that, I have added the questions now :redface:
     
  5. Mar 14, 2014 #4
    For part 1) you should multiply by W/2.

    Part 2) is a little confusing since they say "for the rope and trap door" which seems non-sensible to me in context. If that part could be dropped and they were strictly looking for "the torque caused by the weight of the door about O" then I would change your answer from α to θ.
     
  6. Mar 15, 2014 #5
    Thanks for the reply :)

    So part one becomes

    [itex]{​T} = W(-\sin \alpha(i) +\cos \alpha(j))/2[/itex]

    and part two

    ​[itex]-mgL\sin \theta(k) [/itex]

    How does that look?
     
  7. Mar 15, 2014 #6

    NascentOxygen

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    Hi Jessica8956. http://img96.imageshack.us/img96/5725/red5e5etimes5e5e45e5e25.gif [Broken]

    Unfortunately, neither answer looks right to me. You don't know the textbook answers, do you?

    Let's focus on (b). Can you show your working?
     
    Last edited by a moderator: May 6, 2017
  8. Mar 15, 2014 #7
    Thanks NascentOxygen

    We had a supply teacher who gave us questions like the one I am stuck on. They only have a list of multiple choice answers to choose from. Sadly that teacher was only there for a week and my regular teacher said to not worry about it as it is not relevant to what we're doing :uhh:



    My workings for part (b) are


    What I did was get

    ​[itex]{W} = -mgj[/itex]

    Then get the position vector as

    ​[itex]\frac{1}{2}2L\sin \alpha(i) \sin \alpha(j)[/itex]

    Simplifies to

    ​[itex]L\sin \alpha(i) \sin \alpha(j)[/itex]

    So the answer is

    ​[itex]-mgL\sin \alpha(k) [/itex]
     
    Last edited by a moderator: May 6, 2017
  9. Mar 15, 2014 #8

    NascentOxygen

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    That is the position vector to what? I can't see there should be an ##\alpha## involved, nor do I see a ##\sin()##. Please explain how you determined this.
     
  10. Mar 15, 2014 #9
    I thought that was the position vector for W

    As I thought the line OA was ​[itex]\sin \alpha(i) \sin \alpha(j)[/itex]

    and as the length was 2L it was half that as I assumed it was in the middle.

    I'm a bit lost with this so I hope that makes sense :blushing:
     
  11. Mar 15, 2014 #10
    For part 1), intuitively I would expect the tension force T to be greater than W/2. To confirm we can look at limiting cases:

    -when α is close to zero then T should be pretty close to W/2; so far your formula looks like it might work
    -when α is close to 90° then T should be much much bigger than W/2; now your formula doesn't look too healthy

    Can you rework your formula to meet the above limiting cases? You might want to explicitly do an equilibrium of joint A.
     
  12. Mar 16, 2014 #11

    NascentOxygen

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    I suggest that you start over, abandoning trying to use i and j. (If necessary, you can at the end resolve your final answer into vert and horiz components.)

    Just work it out using trigonometry, not vectors. See how you go with that. Try (b) first, it looks easier than (a).
     
    Last edited: Mar 16, 2014
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