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 Direction of Force

  1. Apr 17, 2012 #1
    1. The problem statement, all variables and given/known data
    Hi I know for Hooke's Law F=-kx where -ve sign implies a restoring force back to equilibrium in the opposite direction of x.

    My question is:
    I have a mass being displaced attached to a spring - consider M2 and z2(t) and K2 in the link below


    So as z2 move downwards (in the image) the spring stretches and a the restoring force acts upwards. The total force is going to be the force from the spring plus the force from the displaced mass, which is

    [tex]M_{2}\ddot{z_{2}} - K_{2}(\ddot{z_{2}} - \ddot{z_{1}}) = 0[/tex]
    note the -ve sign infront of the k2.

    but why do we write a PLUS instead?

    [tex]M_{2}\ddot{z_{2}} + K_{2}(\ddot{z_{2}} - \ddot{z_{1}})) = 0[/tex]
    (see answers below)

    So, basically why do we not write

    [tex]M_{2}\ddot{z_{2}} - K_{2}(\ddot{z_{2}} - \ddot{z_{1}})) = 0[/tex]

    as doesn't that make more sense. Because then the force of the spring restoration force equals the mass force, which is when the mass is at rest.

    The answers
  2. jcsd
  3. Apr 17, 2012 #2
    come on someone surely must know, it's just a matter of signs
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Threads - Hooke's Direction Force Date
Calculating direction derivative to a line in 2D Jan 10, 2018
Hooke's Law Sep 20, 2016
Linear Algebra - Hooke's Law Problem Mar 3, 2016
Calculus 2- Hooke's Law Spring problem May 25, 2013
ODE Hooke's Law, V(x) instead of V(t) Feb 11, 2013