1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Center of mass of two blocks

  1. Oct 21, 2005 #1
    Two blocks each of length L are piled near the end of the table. What is the maximum distance between the edge of the table and the edge of the outer block before the system topples? I understand that if the center of mass of the system must be to the left of the edge of the table for the system to be balanced. Is this correct? However, I don't understand how to do the calculations.

    Let x1 be the center of the lower block, and x2 be the center of the upper block. The average location of the center of these two blocks is then (x1+x2)/2. This is a distance (L/2-(x1+x2)/2) away from the center of the lower block. I can calculate the distances between the centers of the blocks and the edge of the table, but I don't understand what is the condition for the system to be balanced. Is it that the center of mass of the entire system be to the left of the edge of the table?


    The answer is: 3/4*L.
     
    Last edited: Oct 21, 2005
  2. jcsd
  3. Oct 21, 2005 #2

    Diane_

    User Avatar
    Homework Helper

    Yes, that's right. The COM could be right on the edge and it still wouldn't fall over, but it'd be shaky. Literally.

    Now - given the answer you quote, I think I'm not picturing this correctly. I assumed that "piled" meant you laid one on top of the other with the same orientation. In that case, the COM will move upwards but not longitudinally, so the second block makes relatively little difference. (There is a difference in the torques involved as the block is near the edge, meaning that the two will be a bit less stable than one by itself, but I don't think that's important here.) Given the "3L/4" answer, though, I suspect that the upper is placed crosswise on the end of the lower. Is that correct, or am I missing something here?
     
  4. Oct 21, 2005 #3
    Step 1:
    Lets analyize block number 1. This block is on TOP of block number 2. There are two forces acting on it. The uppward normal force and the weight. The sum of forces in the Y direction demands that W=N. Now, lets sum the moments about an edge of the top block, left edge, right edge, it makes no difference. This yields: [tex] \sum F_y = -w(L/2) + Ny=0 [/tex]. I am calling y the distance from the edge to the normal force. It should be fairly clear that the maxium distance will be y=L/2, other wise the normal force required will be greater than W, and the forces in the Y direction will not balance.
    Step 2.
    Lets analyze block (1,2) and the table. We have now established that block 1 is L/2 away from the top of block two. Sum the forces in the Y direction again yields the normal forces is equal to two times the weight. Now pick the edge of the 2nd bottom block, where its furtherst from the edge of the table. We will call the distance the normal force acts y again, to be consistent. Sum the moments about the edge of the bottom block furthest from the edge of the table and you get:
    -Ny + w(L/2) + w(L) = 0
    Now, N = 2w.
    Simplify this and you get y=3L/4
    Dont jump to conclusions just yet. This is the distance from the edge of the bottom block to the edge of the table. So this means that the distance from the edge of the table to the outter edge of the top block is,
    (L- 3L/4)+(L/2) = (3/4)L
    Hence it is 3/4 L away.
     
    Last edited: Dec 26, 2005
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Center of mass of two blocks
Loading...