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!

Calculating force from a system of connected springs

  1. Dec 2, 2012 #1
    1. The problem statement, all variables and given/known data

    http://img854.imageshack.us/img854/1828/physicssprings.png [Broken]

    Two identical masses on a frictionless surface are attached to two walls by springs, each with a force constant k1, and to each other by a spring with force constant k2. (See figure, where system is at equilibrium) For simplicity, assume the springs are massless. If the left mass is moved a distance x to the left and the right mass is moved a distance x to the right, the magnitude of the force on each mass is equal to what in terms of k and x (where x signifies displacement)?

    2. Relevant equations

    The general equation for force resulting from a spring is F = -kx, where k is the force constant and x is the displacement.

    3. The attempt at a solution

    I'm taking the mass on the right. Assume it is moved a distance x to the right; this would mean that the magnitude of force due to spring 1 (the spring connected the wall) would be F1 = (k1)(x). (The direction of the force would be to the left.)

    The middle spring (spring 2) is stretched a total 2x (x from each mass); from what I learned, however, since it is stretched x to the right and x to left, this means that there would be two resultant forces--a force (k2)(x) going to left (exerted on the mass to the right), and a force (k2)(x) going to the right (exerted on the mass to the left).

    Since the question asks for the force exerted on EACH mass, I'll take the mass on the right:

    F(total) = F1 (force exerted on mass by spring 1) + F2 (force exerted on mass by spring 2) = (k1)(x) + (k2)(x) = (k1 + k2)(x)

    However, the correct answer is (k1 + 2k2)(x). How is this possible? This would mean that spring 2 exerts a force of 2(k2)(x) to the left on the right mass, but it should only be exerting a leftwards force of (k2)(x). Could somebody please explain this? Thank you so much in advance!
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Dec 2, 2012 #2

    Doc Al

    User Avatar

    Staff: Mentor

    If a spring is stretched by amount x, the force it exerts on each end is kx. The force isn't 'divided' between the two ends. (Note that you didn't divide the force from the rightmost spring in two!)

    Since the middle spring is stretched by 2x, the force it exerts will be k(2x).
     
  4. Dec 2, 2012 #3
    Thank you so much Doc Al! [Ah, finally the problem makes sense!]
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Calculating force from a system of connected springs
Loading...