1. Not finding help here? Sign up for a free 30min 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!

Spring Mechanics

  1. Jul 25, 2005 #1
    hi all,

    The problem is to find out how much lower will the end of the spring be when it reaches its new equilibrium position? One end of the spring is fixed to the ceiling and the other is attached to a 1 kg mass and the spring constant is 325 N/m.

    I read this problem from text. In order to find the new equilibrium position, we must use Hooke's law, right? But, I find that the initial length is missing.

    Given quantities:

    F = m * a = 1 kg * 9.81 m/s2
    K = 325 N/m

    Unknown quantities:
    l = ? (to be calculated)
    lo (initial equilibrium) = ? (not given in the problem)

    How can i go about to find l ( new equilibrium)? Any one please help.

    regards,

    vanchin
     
  2. jcsd
  3. Jul 26, 2005 #2
    You need to find only change in length : (initial equilibrium - new equilibrium).
     
  4. Jul 26, 2005 #3
    The initial equilibrium of the spring is not specified in the problem, then how to could the new equilibrium position be computed?

    help.

    vanchin
     
  5. Jul 26, 2005 #4

    mukundpa

    User Avatar
    Homework Helper

    Think for the two eq. cases

    F1 = K DL1
    F2 = K(DL1 + DL2) D is for delta

    can you solve for DL2
     
  6. Jul 26, 2005 #5
    Hi Mukundpa,

    Yes, can solve DL2 if DL1 is given, but in the problem specification the L1 is not given. Sorry to trouble you. The problem is as follows:
     
  7. Jul 26, 2005 #6
    think about what the x in Hooke's law stands for. As I remember it is the displacement from equilibrium. If I tell you that some thing is x meters away from its equilibrium then what is the distance from the equilibrium point to x? In other words what is the equilibrium position of a spring with no mass hanging from it (no force).
     
  8. Jul 27, 2005 #7

    mukundpa

    User Avatar
    Homework Helper

    If the problem says that in new equilibrium position means(sounds) previously the spring was in equilibrium with some force and an additional load of 1 kg is added due to which the equilibrium position is changed the new tension in the spring is F2 thus F2 - F1 = 1*9.8 N

    subtracting the two equations we get F2 -F1 = KDL2

    gives DL2 =(F2 -F1)/K

    and if initially there was no load then the position of the lower point of the spring is L0
    and the tension is zero
    in second case the tension is 9.8N and the length is Lo + DL then

    9.8 = K( DL)

    I think that the confusion is only because you are trying to get the new length of the spring which is not required. the question says "how much lower will be the end of the spring" means extension.
     
    Last edited: Jul 27, 2005
  9. Jul 27, 2005 #8
    Hi all,

    Based on the guidlines, I understand that the lower end of the spring be when it reaches its new equilibrium position is 0.0301m.

    using F = k (x), where x = l - lo

    => 9.81 = 325 * x

    => x = 9.81 / 325

    => x = 0.0301m

    Am I right?
     
  10. Jul 27, 2005 #9
    looks good to me :biggrin:
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Spring Mechanics
  1. Mechanics of a spring (Replies: 0)

  2. Springs in mechanics (Replies: 0)

  3. Spring mechanics (Replies: 12)

Loading...