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Forces During Chin-Ups

  1. Oct 15, 2012 #1
    1. The problem statement, all variables and given/known data

    People who do chin-ups raise their chin just over a bar (the chinning bar), supporting themselves only by their arms. Typically, the body below the arms is raised by about 30cm in a time of 1.0s , starting from rest. Assume that the entire body of a 760N person who is chinning is raised this distance and that half the 1.0s is spent accelerating and the other half decelerating, uniformly in both cases.

    2. Relevant equations

    S=Vot+(1/2)at^2

    3. The attempt at a solution

    I thought I had the right answer, but mastering physics is saying it's wrong, and I don't know what's wrong.

    .3=(0*.5)+.5*a*.5^2
    .3=1/2a*.25
    2.4=a
    F=ma
    F=(77.55)(2.4)
    F=186.12N
     
  2. jcsd
  3. Oct 15, 2012 #2
    Seems like an energy-based approach may be in order.
     
  4. Oct 15, 2012 #3

    CWatters

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    Science Advisor
    Homework Helper

    Why have you used S = 0.3m ?

    As I read the question the body starts at the bottom, accelerates for 0.15m to some velocity V then decelerates for 0.15m becoming stationary at the top. It says assume uniform acceleration so I think you can use standard equations of motion.

    In addition the force required will obviously be greater than 760N. You have forgotton something else.
     
  5. Oct 15, 2012 #4
    The question's wording is confusing me, but I understand what you mean. So it would be:

    .15=(0*.5)+1/2a(.5^2)
    a=1.2
    F=(77.55)(1.2)
    F=93.06N+760N=853.06N?

    I only have 1 more attempt to answer the question before it will put a 0 as the grade, so please let me know if this is the correct answer. Thanks.
     
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