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Brick not Falling?

  1. Jul 19, 2004 #1
    Lets say I have a brick of mass m, length L and height h which is resting on a table in such a way that L - r of it is in the air (not on the table) and the rest is on the table. What is the minimum value of r if the brick is not to fall?

    Just by looking at this, it seems the minimum value is L/2 right? Suppose this is the situation. Let N be the normal force on the brick. Clearly N = mg. If I calculate the torque about the edge of the table, I get NL/4. Since the brick isn't falling, the torque is zero, but that would imply that NL/4 = 0! How is this possible? Does this mean the brick falls? Suppose r > L/2 (so more than half of the brick is on the table). Calculating the torque about the edge of the table again gives me mg(L/2 - r). If the torque is zero (which I hope it is), then r = L/2. This clearly doesn't make any sense. What is going on here?
     
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  3. Jul 19, 2004 #2

    plover

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    The net torque is zero (for r>L/2). Have you accounted for all the forces on the brick?
     
  4. Jul 19, 2004 #3

    Gokul43201

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    The mistake you're making is in figuring the line of action of the normal force. When r = L/2, the brick is on the verge of falling and so the normal force actually acts upwards from the only true point of contact - the edge. So the torque from the normal force about the center of the brick (which is directly above the edge of the table) = 0.

    Or you could look at it this way : The unbalanced torque makes the brick JUST BARELY start to tip, at which instant the line of action of the normal force moves to the point of contact - the edge of the table - making the net torque go to zero.
     
    Last edited: Jul 19, 2004
  5. Jul 19, 2004 #4
    What other forces are there besides gravity and the normal force.
     
  6. Jul 19, 2004 #5
    OK, that makes sense. But what about the case when r > L/2?
     
  7. Jul 19, 2004 #6

    Gokul43201

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    Okay, there too, a similar thing happens. The mistake you are making is in assuming that the normal force acts up from the mid-point of the "r section". This would be true if the force with which the brick presses on the table were uniform through this section. That is not true. The force distribution gets weighted towards the table edge so that the normal reaction ends up acting up through the center of mass of the brick.

    Again, the other way of looking at this (just like we did before) is as follows : Assume the normal reaction results in a net unbalanced torque. This torque makes the outer end of the brick (the portion of the brick above the table edge) press down on the table harder than the inner end. In other words, the normal reaction gets shifted outwards till it lines up with the weight vector.
     
  8. Jul 19, 2004 #7

    plover

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    None.

    I'm sorry, I wasn't reading your post carefully enough.

    When you say "Clearly N = mg.", that is true when considered in relation to the vertical force through the center of mass, but this force is not a torque.

    The torque can obviously be derived from that gravitational force applied at the center of mass though. How does the torque change when the position of the center of mass changes with respect to the table edge?
     
  9. Jul 20, 2004 #8
    I knew it had something to do with this. I see know. I wish my physics book would have an explanation on how the normal force is distributed through the area of contact. Thanks.
     
  10. Jul 20, 2004 #9

    Gokul43201

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    You can't rely on books to tell you everything.

    When you first found you had an unbalanced net torque, you probably suspected there was something wrong with your normal force. Sit down and think about it. The resolution I provided had no "new" physics in it. You could have figured that out yourself.

    I only say this because I feel you have a pretty good understanding of the concepts that your book teaches you.
     
  11. Jul 20, 2004 #10
    Let's make things a bit more interesting. Suppose there are now two bricks, with one on top of the other such that half of the top brick is on the bottom brick and the other half is in the air. How far can I move the bottom brick from the edge of the table before the bricks fall?

    So the bottom brick now has an extra force acting on it, i.e. the force due to the top brick. This contact force is acting at the edge of the bottom brick (since the bottom brick exerts a normal force at the bottom and through the center of the top brick and by Netwon's 3. law, there is an equal force acting in the opposite direction on the bottom brick). I derived an expression for the torque about the edge of the table but it came out rather ugly (meaning that I can't simplify for r). Is there a better way of approaching this problem?
     
  12. Jul 20, 2004 #11

    Doc Al

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    living on the edge

    The center of mass of the top brick must be located over the bottom brick (not in the air). Similarly, the center of mass of the two bricks must be located over the table and not in the air. To find the maximum extension, put the c.m.'s right at the edges.
     
  13. Jul 20, 2004 #12
    What an excellent simple idea. Thanks.
     
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