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!

Hooke's law

  1. Jan 18, 2013 #1
    1. Are the spring forces at both ends of a spring always equal in magnitude?

    2. If yes, then the external forces at both ends are also the same in magnitude, aren't they? (Based on Newton's third law)
    If no, how to find the two different spring forces at each of the end?

    3. If the external forces at both ends are indeed the same, then the net force acting on the spring is always zero, isn't it? Then, it will never accelerate?
     
  2. jcsd
  3. Jan 18, 2013 #2
    The forces which the spring reacts with are the same in magnitude. The external forces that act on the spring need not be same. For example, one end might be attached to a wall, and the wall's reaction force will be equal in magnitude and opposite in direction to the spring's reaction force on that end, so that end will be stationary, no matter what is happening on the other end.
     
  4. Jan 18, 2013 #3
    If the mass of the spring is negligible, the forces on both ends are the same. If the spring has mass, the situation is a little more complicated, and the forces on the two ends are not necessarily the same. You can analyze the spring response by subdividing it into incremental masses and smaller massless springs, and solving for the movement of each incremental mass. Think of the response of a slinky toy. So if the spring has mass, the tensile force, the velocity, and the acceleration very with position along the spring and time. It behaves like a sequence of masses and massless springs in series.
     
  5. Jan 19, 2013 #4
    Thank you for your insight.
     
  6. Jan 23, 2013 #5
    Let say I have a spring with spring constant of 10 N/m. I compress both ends with the same force so that each end compresses by 5 cm. Thus, the total compression is 10 cm.

    If I use F=kx where x is the total compression, then the spring force acting on each of my hand is 1 N.

    If I use F=kx where x is the displacement of each end of the spring, then I get 0.5 N as the spring force acting on each of my hand.

    Which one is true and why?
     
  7. Jan 23, 2013 #6
    You shouldn't think of the spring as being two parts in this case. The k value would be different if you would cut it up into 2 parts. Each part would then be 2k if you cut it into equal lengths.
     
  8. Jan 23, 2013 #7
    If the mass of the spring is negligible, then the term "each end compresses by 5 cm" has no meaning. The entire spring compresses uniformly. Cross sections all along the spring will get closer together to one another. If you say that the left end of the spring displaces 5 cm to the right, and the right end of the spring displaces 5 cm to the left, then, overall, the length of the spring decreases by 10 cm. This means that, if the spring constant is 10 N/m, the inward (compressive) force applied to each end of the spring will have to be 1 N. The force you apply on the right end will be 1 N, and will be directed to the left. The force you apply on the left end will be 1 N, and will be directed to the right. You could have obtained the same forces on the spring by displacing the left end of the spring 0 cm to the right, and displacing the right end of the spring 10 cm to the left. In either case, the spring gets 10 cm shorter, which is all that counts.

    But, don't forget, if the spring has mass, this analysis needs to be modified.
     
  9. Jan 25, 2013 #8
    Thanks for explaining, again.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Hooke's law
  1. Hook law (Replies: 4)

  2. Hookes law (Replies: 1)

  3. Hookes Law (Replies: 17)

  4. Hooke's Law and force (Replies: 4)

  5. Hooke's law and springs (Replies: 11)

Loading...