1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

I Force Analysis of a compressed spring

  1. Dec 13, 2016 #1
    Hi all, I am quite confused by the forces relating to a horizontlly placed spring when someone directly exerts a force to the compress it. If I exert a force F to the right to compress the spring, which, say, is placed against a wall, the spring exerts an elastic force on me but I don't see how these two forces are related because they are exerted on different bodies. How can they be balanced out if they act on different bodies?

    Let me thank you in advance. I have been bothered by this problem for a while.
  2. jcsd
  3. Dec 13, 2016 #2
    They don't have to be balanced. As you say, they act on different bodies so their effects may be different. The spring gets compressed and you may get some acceleration from this force unless there is friction to balance the effect of the spring on you.
  4. Dec 13, 2016 #3
    Thank you for replying. I just want some clarification. If I analyze the horizontal forces on me (assume no friction), are there two forces? One from the compressed spring and the other from the reaction force by me pushing the spring with the force F?

    If I do work by exerting this force F to compress the spring, how do you calculate the work done by using force times distance formula?
  5. Dec 13, 2016 #4
    There is one force from the spring on you. You can call it the reaction to you r force on the spring but the "reaction" is not a very useful concept.
    There is the interaction between you and the spring (two bodies) and this results in a pair of equal and opposite forces, each one acting on one of the two bodies.

    You don't if the force is not constant. You need to integrate over the distance. Or just read about work of elastic force and use the formula, if you did not learn calculus yet.
  6. Dec 13, 2016 #5
    But the two forces don't have to be equal; how can they satisfy the "this results in a pair of equal and opposite forces" criteria to be action-reaction pair?
  7. Dec 13, 2016 #6
    Yes, they are. They are a pair of forces describing one interaction.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted