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Contact friction

  1. Sep 29, 2008 #1
    two blocks are in contact on a frictionless table. a horizontal force F is applied to M2. if M1=1.75kg, M2-3.38kg, and F=6.10N, find the size of the contact force between the two blocks.

    So i am just not sure how to find contact force cause my teacher never taught it to us. thanks. if you can just give the equation that is good enough, i dont expect you to do the problem for me, just help would be cool
  2. jcsd
  3. Sep 29, 2008 #2
    The blocks are connected, so they must accelerate at the same rate. How fast does a block with mass M1+M2 accelerate from a force F? Now, imagine that the blocks are seperated, and you want to accelerate them to that same rate. How will you distribute the 6.10 N between the blocks?
  4. Sep 30, 2008 #3
    wait i still dont get it
  5. Sep 30, 2008 #4
    Sorry, I should have defined contact force. Basically, when you push the two blocks when they are touching, the resulting acceleration will be the same as if you were pushing a block of the combined mass of the two blocks. So, the acceleration for a block of 5.13 kg will be the same as the acceleration for the system of 1.75 kg and 3.38 kg blocks.

    Here is what will happen when you push on the first block with a force of 6.10N. It will try to accelerate but run into the second block. In order to accelerate, the first block will need to push the second block to the same rate of acceleration. Therefore, there will be some force between the first block and the second block. This is the contact force. Keep in mind that any force the first block "uses" to push the second block will no longer be affected the first block, because the second block will be pushing back on the first block (Newton's third law), canceling out that force on the first block. You can think of the first block as "transferring" force you are using to push it to the second block.

    So, an equivalent problem to this would be this: How can you distribute 6.10 N of force between two blocks of masses 1.75 kg and 3.38 kg such that they will accelerate at the same rate? The force that you apply to the 3.38 kg block would be the contact force, and your answer for this problem. I'll leave it to you to figure out why this would be an equivalent problem.
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