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Mechanics bars of mass problem

  1. Sep 13, 2004 #1
    Here is this problem which is bugging me:
    Q. Two bars of mass m1 and m2 are connected by a non deformed light spring rest on a horizontol plane . coefficent of friction between bars and plane is k . what minimum constant horizontol force has to be applied to bar of mass m1 inorder to move other bar.?

    A this is what is was trying to do
    at the time bar m2 moves maximum static friction acts and is equal to LX where X is extension of spring and L Hookes constant of spring
    for m1 similarly we write force equation in horizontol direction and put LX value from equation for m2.
    the we get relation in acceleration of block 2 and f the we apply chain rule and write a as -(dv/dx).(dx/dt) as velocity is decreasing as hookes force increases with extension.integrating both sides we get we get relation for kinetic energy of m1. then taking m1 and m2 + spring as system we by using work enery thworem write change in total mechanical energy bis equal to net work done by external force friction and F.
    well by doing this i get answer
    F = kg(m1+3/4m2) while correct answer is F= kg(m1+m2/2)
    i dont know ehere i am doing wrong
    please help?
     
  2. jcsd
  3. Sep 15, 2004 #2
    Hello Ambuj

    Let's assume that m2 is the mass on the left and m1 is the mass to the right of the spring. Hence the spring is sandwiched between the masses. You need to compute the minimum force on m1 in order to cause motion of m2. Hence, you need to find that critical value of F which causes m2 to start moving from rest (this is what I figured from the wording of your problem..please clarify).

    Let the spring constant be [tex]\mu[/tex]

    Suppose I apply F on m1, its force equation is

    [tex]F - \mu x - km_{1}g = m_{1}\frac{dv_{1}}{dt}[/tex]

    and for m2, it is

    [tex]kx - km_{2}g = m_{2}\frac{dv_{2}}{dt}[/tex]

    Oops...the tex isn't working....I wonder why. SO I'll continue this in a while after you reply back. I am trying to figure it out using force equations first and not work energy. However, if I do figure it out using work energy, my equation could be something like

    [tex]\frac{1}{2}m_{1}v_{1}^2 + \frac{1}{2}m_{2}v_{2}^2 + \frac{1}{2}\mu x^{2} = [/tex]Work done by friction+externally applied force.

    For the minimum force, I think we can set v2 = 0 if I understand your reasoning correctly...
     
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