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Forces Help

  1. Mar 29, 2007 #1
    1. The problem statement, all variables and given/known data

    http://img255.imageshack.us/img255/9655/untitled1vq1.jpg [Broken]

    3. The attempt at a solution

    Okay, so I have no idea how to do this problem. Well first of all, the net force of the whole system is F - f = (m1 + m2)a. I have no idea what a is. Also, since friction is involved we need to find the normal force. How do I find the normal force?
    Last edited by a moderator: May 2, 2017
  2. jcsd
  3. Mar 29, 2007 #2
    wow thats a wild and wooly problem. first, word normal here is in the perpendicular sense. so when two bodies are in contact, its the reactive force (newtons third law) in response to our weight on the floor which keeps us from falling thru it. The floor pushes back. Curiously common sense fails us here as frictional forces have nothing to do with surface area, only the magnitude of the force keeping them in contact times that coefficient, mu.

    So here we have two normal forces at right angles.

    Lets start with the easy half, the normal force between the big block and its supporting surface. Thats easy, Fn=mg*mu=9.8*4.2*0.24= 9.87 (weight times friction)

    Now the other half of the problem is less intuitive. But in principle no different, the opposite of the force of the big block on small block.

    (Here I am a bit fuzzy, do we have to worry about both blocks or just the big one.) No worries, If wrong someone will come to our rescue.

    Lets take the case where we only concern ourselves with the big block;

    now we know that in order for the smaller block not to fall, the force impressed on it must be equal to .7kg*9.8*.38=2.61N

    the Sum of forces=ma must then be 2.61

    So Fx-Fn=2.61 where Fn=9.87

    Adding the two gives 12.49. May be dead wrong, but hope its of some help. Seems like a very complex problem for someone with no idea how to compute Normal force.:rolleyes:
  4. Mar 29, 2007 #3
    Oh I get it now. Thanks alot. I have to slove for normal forces on both blocks. Thanks alot.
  5. Mar 29, 2007 #4
    NP. let me know if the answer was close. First time I seen one like that.
  6. Mar 30, 2007 #5


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    If you look at a FBD of the small block, the friction force between the 2 blocks must be balanced by the small block's weight, and therefore equal to 0.7g newtons. You should now be able to calculate the Normal force (N) between the blocks in the positive x direction, and then use Newton 2 to solve for the acceleration.
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