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Homework Help: Forces direction question

  1. Jun 11, 2007 #1
    1. The problem statement, all variables and given/known data

    http://img263.imageshack.us/img263/1947/untitledtq5.jpg [Broken]

    3. The attempt at a solution

    Will Fus and Fsu be located on the string in equal and opposite directions?

    I can do q (i) with F=ma, but unfortunately I cannot proceed to do so because I cannot answer question h!
    Last edited by a moderator: May 2, 2017
  2. jcsd
  3. Jun 11, 2007 #2
    Let's think about it. The unicycle rider is towing the surfboard carrier with a string, this creates a tension force. Now, it is the unicycle rider accelerating, and remember that in order for there to be a force two objects have to be in contact (since it's not a non-contact force in this case). Does any of this help?
  4. Jun 11, 2007 #3


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    Yes, a massless string serves to transmit forces from its one one to the other - that is an object pulls on it and it pulls just as hard back on the object. Obviously the string cannot "generate" unequal forces at its ends - it has no internal energy source, it just transmits the force unaltered to its ends. If it does have a weight the attractive force of the earth can alter the tensions at its ends. Also if the string drags over a surface (pulley) with friction the tension at its ends will not be the same since the interaction can change them.
  5. Jun 11, 2007 #4


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    (h) It is "easier" to apply N2 to the system. The tension forces cancels each other out and you therefore do not need to know their values.
  6. Jun 11, 2007 #5
    Ok, so am I right in saying Fsu and Fus are located "on the string"?

    I'm still unsure as to whether to take the system as one.. (lol, andrevdh you got in before I even finished my post!)
    Last edited: Jun 11, 2007
  7. Jun 11, 2007 #6
    Minor double check: when applying Newton's 2nd to find acceleration I only take the horizontal forces, correct?!

    I've even forgotten the basics!

    Hence I get...

  8. Jun 11, 2007 #7


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    That is how I've got it too: 1.1 m/s^2
  9. Jun 11, 2007 #8
    cool thanks!
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