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One-dimensional locomotion

  1. Oct 3, 2009 #1
    hi, im trying to figure out how to approach this question. i know its a relatively simple problem, but i just need to do the setup appropriately.
    two blocks of m1=2 kg and m2=1 kg are in contact on a horizontal frictionless surface in the respective order, left to right. A force of F=3N is applied to m1, towards the right. Calculate the magnitude 'f' between the two blocks. Also, find the magnitude if the force was applied in the opposite direction.
    Please help me get started with the setup for this question. Thanks!!
     
    Last edited: Oct 4, 2009
  2. jcsd
  3. Oct 3, 2009 #2
    hi everyone,
    the problem is a simple one as danishtanwar said. Though I have an explanation, I am not quite sure if it is right because I created it myself. And the action-reaction law seems to invalidate my result. Here is my calculation.
    First, Newton's second law for the whole system gives:
    F=(m1+m2)a
    Let a1 be the acceleration of m1 if it were alone, i.e., no m2 and the F is the same. Then:
    F=m1*a1
    In the system with 2 masses, m1 and m2, the acceleration of both masses is a. As the acceleration of m1 is also a, there must be a force acting in opposite direction of the given F, i.e., to the left. This force is the reaction force of m2 on m1. That force is the 'f' danishtanwar gave. Looking at only the mass m1, summation of all forces acting on m1 is :
    F - f, and this equals m1*a.
    F - f = m1*a
    Solving for f, I get
    f = F - m1*a
    Plugging in F, I get: f = m2*a, as desired.
    Can somebody please confirm or disprove my derivation?
    Thanks
     
  4. Oct 3, 2009 #3
    that actually helped, i was able to draw the right diagram for it
    except, the formula at the end can be simplified a bit more
    It would be: a=f/m2
    Then combining it with your formula we have F-f = (m1/m2)*f
    And finally f = (F*m2)/(m1+m2)
    I checked and the results make sense, Thanks a lot!!
     
    Last edited: Oct 4, 2009
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