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Number of forces when calculating spring compression?

  1. Jan 23, 2010 #1
    Hello everyone

    I'm quite confused here and would really appreciate help.

    Consider a spring standing upright with a box on top of it. Now according to statics there are two forces acting on the spring: the weight of the box G and the support of the ground -F. But when we calcuate the compression of the spring, it is G/k not G+F/k . Why is this? I have been told that this is because F is a reaction force so it is not taken into account. But I thought the reaction force must be directed at a different body than the action force?
     
  2. jcsd
  3. Jan 23, 2010 #2
    the reaction force tells you that the system is in equilibrium. if there is no ground (no reaction force) the spring would be acceleration downwards because of the force (weight on top) acting on it.

    G and F are equal but act in opposite direction
     
  4. Jan 23, 2010 #3

    arildno

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    Let's look on this in detail!

    1. The spring is in equilibrium
    Forces on the spring:
    a) Weight of box, W, b) Force from Ground, G, c) Weight of spring (set to 0)
    Due to equilibrium, we have W+G=0, i.e, G=-W

    b) The box is in equalibrium:
    Forces on box:
    a) Weight of box, W, b) Spring force, S
    Due to equlibrium, we have S=-W

    c) Calculated compression L of spring by means of Hooke's law:
    Hooke's law states: S=-k*L

    Thus, combining b)+c), we get, L=W/k
     
  5. Jan 23, 2010 #4
    But several websites state that the reaction force must be directed at a different object than the actual force. E.g. if i push a wall the wall pushes me. That is an action-reaction pair, right? In the case of the spring both the weight of the box and the ground act on the same object, the spring... so they can't be an action-reaction pair, right? Thus the original question: why is one force ignored when calculating spring compression?
     
  6. Jan 23, 2010 #5

    arildno

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    Okay, what are the action/reaction pairs here:

    A) Box/Earth: That's concerned with the weight of the box

    B) Box/Spring: A spring force S works on the box; the box exerts -S back on the spring.
    Since the box is in equilibrium, it follows that S is equal in magnitude, of opp. dir of box weight W, and hence that the reaction force from the box onto the spring is equal to the box's own weight

    C) Ground/Spring
    The normal force from the ground must balance the reaction force from the box upon the spring, since the spring is in equilib. Thus, the spring exerts a reaction force on the ground that by the above arguments must equal the box's weight.
     
  7. Jan 23, 2010 #6
    That's what I thought, so I was misled at another forum. however i don't understand why the supporting force of the ground is not taken into account when calculating the compression of the spring? In other words, why is the force at one end of the spring(box weight) taken into account but the force at the other end of the spring is ignored(support of the ground)?
     
  8. Jan 23, 2010 #7

    arildno

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    It is not ignored at all.

    At EVERY cross-section (area A) within the spring, there is an internal stress equal to kL/A, where L is the total compression length of the spring.

    At the ground joint, the spring exerts, therefore, on the ground, a force of -kL.

    The ground responds with a reaction force on the spring of kL.

    At the joining with the box, the spring exerts a force of kL on the box, that responds with a reaction force of -kL.

    Note that the sum of external forces on the spring from box and ground equals 0, i.e, the spring is in equilibrium, with an internal stress state equal to kL/A

    (Each cross-section of the spring can readily be seen to be in equilib, too.)
     
  9. Jan 23, 2010 #8

    arildno

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    It is, however, SUFFICIENT, to calculate the MAGNITUDE of kL by looking at the box/spring interaction, and that will be seen to set kL equal to W. Thus, the ground force must ALSO have a magnitude of W
     
  10. Jan 23, 2010 #9
    I have asked this question from dozens of people in numerous phorums, but you are the first person that has been able to answer me so that I understand it! A thousand thanks.
     
  11. Jan 23, 2010 #10

    arildno

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    You're welcome! :smile:
     
  12. Jan 23, 2010 #11
    In applying Hooke's law, it is assumed that one end of the spring is anchored and virtually immovable.
     
  13. Jan 24, 2010 #12

    arildno

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    Nope.

    The spring may be anchored to two slabs, both of each is then set in motion.
     
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