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Homework Help: Two very short questions.

  1. Jan 25, 2009 #1
    1. The problem statement, all variables and given/known data
    1) When a swimmer dives into water just before piercing himself into water, he stretches himself. Why?

    2)In a tug of war if the parties on both ends of the rope apply apply equal and opposite forces on each other then how can one party win?

    3. The attempt at a solution
    1) I think, on stretching the body the resistance force by water on swimmer would decrease. But I cannot figure out the reason for it??? Please help me.

    2)I simply cannot figure out the reason. I drew the FBD's
    the forces on the rope cancel out. the tension and the friction on each party is the same. But then how can a team win?
  2. jcsd
  3. Jan 25, 2009 #2


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    Staff Emeritus
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    Gold Member

    For question (1) the only thing that I can think of it that the reaction force from the water would tend to compress the body, stretching would counteract this.

    For question (2) whilst the forces may be equal, what about the masses of the teams?
  4. Jan 25, 2009 #3

    Doc Al

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    Staff: Mentor

    Stretching out minimizes cross-sectional area.

    Is the friction force the same? Are the masses the same?
  5. Jan 25, 2009 #4
    So is the resisting force cross section dependent?
    Last edited: Jan 25, 2009
  6. Jan 25, 2009 #5
    I am waiting for your answer sir. What kind of force is this? On what factors does it depend except for the cross section and how? Where can I read about it?
  7. Jan 25, 2009 #6
    smaller cross sectional area=less mass of water you are hitting with the same force.
  8. Jan 25, 2009 #7

    Doc Al

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    Staff: Mentor

    Look up fluid resistance or drag forces. It depends on several factors, including the nature of the fluid (viscosity), the size (cross-sectional area) and shape of the object, and the speed. There are several models used. Flip through some of these links: http://hyperphysics.phy-astr.gsu.edu/Hbase/flufri.html#c1"
    Last edited by a moderator: Apr 24, 2017
  9. Jan 25, 2009 #8
    Thanks a lot sir! I have got what I desired. Thanks again for solving my problem.
    Last edited by a moderator: Apr 24, 2017
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