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Newton's Laws with Friction: Two Blocks

  1. Sep 19, 2009 #1
    1. The problem statement, all variables and given/known data

    Part 1

    Two blocks are in contact on a frictionless table. A horizontal force F is applied to M2, as shown. If M1 = 1.03 kg, M2 = 3.40 kg, and F = 5.10 N, find the size of the contact force between the two blocks.

    http://loncapa1.fsu.edu/enc/85/c70c507c1fdfce450f24d473053587465ad535c2c20eedbf8845ed0a22afb4e439270ed6dfaf6472da2a2cac5a4ec3a06f1770c443f1ffcb6e5f0ece685f8ea7803066440f7d55b844760e01ac475fea48c6e88e4c25cdda0806fb9fcbe8bb4cfae3d0436dc5c98e9e32a9ec3bcf2bb4.gif

    Part 2

    If instead an equal but oppositely directed force is applied to M1 rather than M2, find the size of the contact force between the two blocks.


    2. Relevant equations

    I'm pretty sure F=MA is the equation to use.


    3. The attempt at a solution

    For this, I looked at the system as a whole, summed the masses and found the acceleration of the entire system.

    I found this to be a=1.151m/s^2

    I then isolated mass 2 and mass 1 to find the forces for part 1 and 2 respectively. I obtained the answers : 3.914N for part 1 and 1.85N for part 2.

    Where am I going wrong?
     
    Last edited by a moderator: Apr 24, 2017
  2. jcsd
  3. Sep 19, 2009 #2

    rl.bhat

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    Homework Helper

    Net force acting on M1 is (the applied force F1 - reaction F2 from the mass M1)
    Your acceleration is correct, and it is common for both.
    So the acceleration of M1 = (F1 - F2)/M1 = a.
    F1 is given. Find F2. Repeat the same thing in the opposite direction.
     
  4. Sep 19, 2009 #3
    Ok, I got the first part, thanks!

    The second part is still tripping me up though. What do you mean when you say do the opposite?
     
  5. Sep 19, 2009 #4

    rl.bhat

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    Homework Helper

    Force is acted on M1 first.
     
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