1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Work to elongate the things

  1. Feb 20, 2017 #1
    1. The problem statement, all variables and given/known data
    both of these notes are in the same chapter . So , they are relevant to each other ... In the first picture, what's the difference between the 2 circled part ? The first one is W = 0.5 P Δ , the second one is W = P Δ


    ..
    2. Relevant equations


    3. The attempt at a solution

    For the second picture , I think we should use W = 0.5 P Δ instead of W = P Δ ..
    (For the external work, the author use 1Δ , so 1 = P so , it's clear that the author use W = P Δ and not W = 0.5 P Δ)

    I think W = 0.5 P Δ is more appropriate because W = force x displacement , which is also area under the graph .... Since the area under the graph is triangle , so 0.5 is necessary
     

    Attached Files:

    • 588.png
      588.png
      File size:
      17.2 KB
      Views:
      29
    • 589.png
      589.png
      File size:
      27.5 KB
      Views:
      25
  2. jcsd
  3. Feb 20, 2017 #2

    NascentOxygen

    User Avatar

    Staff: Mentor

    The author seems very sloppy with his maths. The first two lines should be:
    P
    P = k∆

    and he should integrate P.d∆ not P1.d∆1

    This is analogous to doing work against a spring, so the factor ½ is required and accounts for the area under the line being triangular.

    I can't figure out how the equation in your red rectangle is relevant here. (It would apply in the case where P was fixed regardless of ∆.)

    Your second image is beyond me.
     
  4. Feb 20, 2017 #3

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    In the first image, I do not understand what it means to say "P1 is located for displacement Δ1".
    If it means it is constant over that displacement then it makes sense.

    For the second image, there seems to be missing context. It refers to "the" unit load being applied, as though this is continuing from some prior description. Can you supply that background?
     
  5. Feb 20, 2017 #4

    NascentOxygen

    User Avatar

    Staff: Mentor

    If this were the result of a machine's translation into English, then I'd say located≡ fixed :smile:
    and we already know that a value fixed ≡ constant :wink:
     
  6. Feb 21, 2017 #5
    Can I say that for the trusses member, constant force is applied so that the trusses deform ( displacement occur ) ?? So, p (Delta ) is used ??
     
  7. Feb 21, 2017 #6

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    If you apply a constant force to, say, a spring that starts off relaxed then you will get a substantial initial acceleration. The KE developed would mean that you would overshoot the equilibrium position.
    However, if the situation is that there is already a significant load and we are just applying a small extra load then the force can be largely constant over that deformation.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Work to elongate the things
  1. Thermal elongation (Replies: 4)

  2. Thermal elongation-2 (Replies: 5)

Loading...