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Spring and masses

  1. Oct 11, 2006 #1
    A spring of stiffness k supports a box of mass M in which is placed a block of mass m. If the system is pulled down a distancde d from the equilibrium position and released, find the force of reaction between the block and the bottom of the box as a function of time. Neglect any air resistance. For what value of d does the block just begin to leave the bottom of the box?

    What i considered for this is as follows:
    using newtons second law:
    for the overall system,
    [tex](m+M)a = kd - (m+M)g[/tex]
    then for the block of mass m:
    then i subbed in the second equation for ma in the first, and tried to rearrange for [tex]F_{normal}[/tex]
    however i am not sure how to get the force as a function of time, writing a as dv/dt doesnt help since i cannot rearrange it to integrate i think.

    thanks for the help
  2. jcsd
  3. Oct 11, 2006 #2


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    Staff: Mentor

    What would be the equation of motion without the extra block inside the first block?
  4. Oct 11, 2006 #3
    without the extra block it would just be
    [tex] Ma=kd - Mg[/tex]
    is this correct? but then how do i proceed
  5. Oct 11, 2006 #4


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    Staff: Mentor

    That is just a starting equation describing the forces. What physical motion results from those forces? What equation describes the SHM of the mass once the spring with the d distance preload is released?
  6. Oct 12, 2006 #5
    [tex] x(t)=d cos (\omega t)[/tex] is what comes to mind. so i can differentiate this twice to obtain acceleration as a function of time, and plug it into one of the equations i had earlier? is this a correct approach? if so what would do i solve for, thanks much appreciated
  7. Oct 13, 2006 #6
    no more help i can get on this one?
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