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Rope tension and normal force of scales?

  1. Sep 22, 2008 #1
    Hi, I thought I had some sort of an understanding in physics until I came across these two websites. On the first page the person in the air exerts a force on the rope but for the rope and person to stay still another force must be exerted by person number two hence the doubling of the force on the pulley and the rope compared to the person's weight hanging in the air, correct? If so, then why does the same not apply to "group number 4" ont he second page? And futhermore, if this is true, then stepping on a force meter(an arrangement that measures normal force) should give me a reading that is double my weight, right? This is because while I press down on the meter, also the earth presses against it. And as a result of this "bathroom" weighing scales must have been set to half the mass of the equivalent force applied to them. But obviously I've got something wrong here. So I would really appreciate guidance here!

    https://www.physicsforums.com/newthread.php?do=newthread&f=61

    http://dev.physicslab.org/Document.aspx?doctype=5&filename=Dynamics_StaticEquilibrium.xml

    Thanks

    br
    Mark
     
  2. jcsd
  3. Sep 22, 2008 #2

    Doc Al

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    Staff: Mentor

    Please repost the link to that "first page" so we can see what you are referring to. (Your first link has an error.)
    In order to have any tension in a (massless) rope, it must be pulled from each end.
    When you step on a scale, the earth pulls down on you with a force equal to your weight and the scale pushes up on you with an equal force. (Otherwise you would be accelerating.) The scale measures the force it exerts, which is equal to your weight.
     
  4. Sep 22, 2008 #3
    Whups! Sorry about that. Here's the first link. And about those scales... The earth makes me push down on the scales. therefore they carry my weight. But also I attract the earth making it push against hte scales. We squash the scales together with the earth from both sides hence the reading of double my weight...

    The link http://www.exo.net/~pauld/activities/physics/stringandpulley.html
     
  5. Sep 22, 2008 #4

    Doc Al

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    Think of the scale as a spring. To compress the spring, you need a force pushing at both ends (just like to tense a rope you need a force pulling at both ends). That doesn't mean the scale reads twice the force.
    I'll take a look.
     
  6. Sep 22, 2008 #5

    Doc Al

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    No, not correct. The only reason someone can exert a force on a rope is if something else is pulling on the other end. If there was just a person "pulling" on a rope in the air there would be zero tension in the rope and both would just fall to the ground. In the belayer diagram, if the belayer just lets go, the scale reading will go to zero and the climber will fall.
     
  7. Sep 22, 2008 #6

    rcgldr

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    Group 4 is about the same as the belayer and climber situation. Assume the Belayer weighs the same as the climber, with both generating 600N of tension on the line. The tension in the line is 600N everywhere in the line, even if you put a tension measuring spring anywhere in the line. However the tension between the pulley and it's mount equals 2 times the tension in the line. Like wise, in group 4, the total downforce on the platform is twice the tension in the line.
     
  8. Sep 22, 2008 #7
    A huge thanks for your effort! This is what I originally thought but then in this case I still don't get that part where the force is double on the pulley... What makes it double and in what sort of situations does this happen?
     
  9. Sep 22, 2008 #8

    Doc Al

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    Since the rope is doubled over the pulley, the rope tension pulls on it twice--thus the net force on the pulley is double the tension.

    (This might help: Picture an imaginary box surrounding the pulley. You will see two ropes "attached" to that box, each exerting a force equal to the rope tension.)
     
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