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Out of curiosity how is the contact force forumla derived?

  1. Sep 26, 2011 #1
    Hello!

    I know that the contact force formula is
    Fc = F(m2/(m1+m2))

    but I was wondering, how did this equation come into being? How is it derived? No this is not a homework question, I was just curious to know.

    Thanks!
     
  2. jcsd
  3. Sep 26, 2011 #2
    Consider two blocks with masses m1 and m2 placed side by side on a frictionless surface.

    _-F->__|m1||m2|____
    If there's a horizontal force F applied to m1, and m1 and m2 remain in contact, then the blocks will be accelerating together, as if they were one object with a combined mass. (Newton's 2nd law)

    (equation 1) F = (m1+m2)*a

    Meanwhile, block 2 is being pushed on by block 1 (Newton's 3rd law), let's call this force F_21. Since block 2 is accelerating, and F_21 is the only force acting on it, we have (Newton's 2nd law, again)

    (equation 2) F_21 = m2*a

    Take equation 1, solve for a, then substitute into equation 2, and you get

    F_21 = [itex]\frac{m2}{m1+m2}[/itex]F

    Also, block 2 is pushing back on block 1 with a force F_12=-F_21 (Newton's 3rd law)
    However, this is not the total force acting on block 1 because we still have F from the left.

    So F1(total) = F - F_21 = F - [itex]\frac{m2}{m1+m2}[/itex]F = m1*a, as expected.
     
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