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Problem: two bodies linked via a spring

  1. Oct 11, 2008 #1

    fluidistic

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    Gold Member

    1. The problem statement, all variables and given/known data
    Two bodies [tex]A[/tex] and [tex]B[/tex] of mass [tex]m_1[/tex] and [tex]m_2[/tex] are connected via a spring of natural longitud [tex]l_0[/tex] and elastic constant [tex]k[/tex]. Both bodies are free of net force until at an instant [tex]t_i[/tex] something applies a constant force [tex]F[/tex] to the body [tex]A[/tex], in the direction of [tex]B[/tex]. (see the diagram)
    a)Calculate the initial acceleration of the center of mass of the system
    b)Calculate the initial acceleration of each of the 2 bodies
    c)Calculate the respective accelerations in the instant in which the spring is compressed by a length x.
    d)Indicate all the pars of action-reaction forces in the moment in which the spring is compressed by a length x.


    2. Relevant equations
    [tex]F_{spring}=k \varedelta x[/tex]
    [tex]\sum \vec{F}=m\vec{a}[/tex].



    3. The attempt at a solution
    a)Just for fun I calculated the center of mass to be at [tex]\frac{l_0}{m_1+m_2}[/tex] if the origin is situated at body [tex]A[/tex] in instant [tex]t_i[/tex].
    I think I've read somewhere that if an external force is applied, then it will modify the acceleration of the center of mass of the system following Newton's second law.
    So [tex]\vec{a}_{CM}=\frac{F}{m_1+m_2}i[/tex]. Am I right?
    b)Using Newton's second law, [tex]\vec{a}=\frac{F}{m_1+m_2}i[/tex] for the body [tex]A[/tex].
    And here start my problems. I'm not sure how to find the acceleration of the body [tex]B[/tex] at [tex]t_i[/tex]. I'm tempted to go against my intuition and say that it will be the same as the center of mass of the system, but I don't think it's possible. Please help me going further this.
     

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  3. Oct 11, 2008 #2

    Doc Al

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    Staff: Mentor

    Yes, you are right.
    Not exactly. What's the net force on body A? The mass of body A?
    The way to find the acceleration of any system (which could be one mass or both masses) is to apply Newton's 2nd law to that system. Start by identifying the forces acting on the system of interest at the moment of interest.
     
  4. Oct 11, 2008 #3

    fluidistic

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    Oops! I did it right but get confused when typing here.
    b) For the body [tex]A[/tex], [tex]\vec{a}=\frac{F}{m_1}i[/tex].
    So for the body [tex]B[/tex], I have to draw a F.B.D. in order to identify the forces acting on it at [tex]t_i[/tex].
    I'm not sure... but at [tex]t_i[/tex] the spring isn't compressed, right? Or it is, by a differential x? If so, then [tex]\vec{a}=\frac{k \cdot \Delta x}{m_2}[/tex] and I think I could even work out [tex]\Delta x[/tex] in function of [tex]F[/tex]. Am I in the right direction?
     
  5. Oct 11, 2008 #4

    Doc Al

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    Good.
    Right.
    Right. At that moment, the spring has not had a chance to compress.
    Let [tex]\Delta x = 0[/tex]. :wink:

    (You're doing fine.)
     
  6. Oct 11, 2008 #5

    fluidistic

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    Thank you very much Doc Al. I've completed the problem now.
    I understand better now that it's possible for a body to have an acceleration but with a velocity equal to 0. It happens in springs and pendulums for example.
    In this problem, the fact that A has no velocity in [tex]t_i[/tex] is the responsible of the no-compression of the spring and so that B cannot suffer any force thus any acceleration.
     
  7. Oct 12, 2008 #6

    Doc Al

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    Staff: Mentor

    Good.

    It's certainly possible for something to have an acceleration but (at least for the moment) zero velocity. Consider a ball tossed straight up. When it reaches the top, its velocity is zero, yet its acceleration is always g downward.

    In this problem, the trick is realizing that the force exerted by a spring is simply given by its degree of stretch or compression. No stretch or compression, no force.
     
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