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Homework Help: Newton's Laws Problem

  1. Jun 29, 2008 #1
    1. The problem statement, all variables and given/known data

    http://img131.imageshack.us/img131/2996/physicswb9.th.png [Broken]

    Two 100N weights are attached to a spring scale as shown. What does the scale read?

    2. Relevant equations

    3. The attempt at a solution

    This concept is difficult for me to grasp. My natural instinct would suggest that the scale would read 200N, but I know this can't be right.

    What I know (or think I know) is that the system is in static equilibrium with ΣF=0. The weight force of the block on the left is 100N, same for the block on the right. Now I notice that if I were to remove one of the blocks, say the left one, and tie the loose end to a wall, the solution becomes obvious, the scale would read 100N.

    I know the answer is 100N. What I don't know is why the answer is 100N. There is a fundamental concept of Newton's laws that I am not getting here. Specifically Newton's third law. Can someone please explain this to me as if I were a 2 year old, with big colourful pictures preferably drawn in crayon?
    Last edited by a moderator: May 3, 2017
  2. jcsd
  3. Jun 29, 2008 #2
    First we have to make clear what the reading on the spring scale means. It means the current tension force in the spring scale.

    So, I would say that the "reading" of the scale in this "situation" will be the same in the usual situation where one end of the scale is fixed on a wall. The two situations are equivalent, as you already proved in your analysis; each scale in each situation is in equilibrium.
  4. Jun 29, 2008 #3
    Yes, the second block essentially 'replaces' the wall as far as forces are concerned. You can see this quite easily if you sketch a quick force diagram; The 100N the extra block exerts on the spring, is the same 100N which would be excerted by a wall, if the spring were attached to a wall.

    For the spring scale to work, it ofcourse has to be in equilibrium (otherwise it would accalerate and won't measure the correct weight). A wall simply provides a reaction force equal to the weight exerted by the object to be measured.

    (Another consequence of this is, that when you tie something to the wall and pull it, you can pull it just as hard as when you replace the wall with a second person who pulls as hard as you. If you understand this, you'll also understand the spring-scale problem)
    Last edited: Jun 29, 2008
  5. Jun 29, 2008 #4
  6. Jun 29, 2008 #5
    Thank you, the solution is clear now.
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