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Easy physics question friction related

  1. Jun 28, 2011 #1
    1. The problem statement, all variables and given/known data
    This is not a homework question, but rather a design problem that has been reduced to a simple problem by me.
    We have a chamber with a diameter of 46mm with a sliding lid. The chamber has a vacuum (assumed negligible pressure) which sucks the lid down. The lid is required to slide off using the force from a piston and cylinder using the vacuum of the chamber itself (in actual fact it needs to be slightly less than the force required to slide, so that the final opening can be provided by a further application of force). So we need to find the force exherted by the vacuum on the sliding lid, then the static friction from the sliding of the lid, and finally use that to spec the cylinder used to help opening the lid. The lid is only in contact with a ring of teflon (0.1 coef. static friction against the polised steel lid) of ID 46, OD 50 (although, does that matter?) Also, there will be 6 of these chambers with the lids all joined, so the suction force is greater, but will the friction be greater? If my calculations are right, probably not...


    2. Relevant equations
    deltaP=F/A, Fmax = coef. static friction x normal force


    3. The attempt at a solution
    We have a suction force of 168N per chamber, from deltaP * A = F and deltaP is assumed to be 101000Pa, therefore 101000 * (pi*0.0232) =168N
    Overall force is then, 6*168N = 1007N
    Sliding force required = coef. static friction * Normal force,
    Sliding F reqd. = 0.1 * 1007 = 101N

    So therefore, rearranging deltaP = F/A and assuming deltaP is again 101000Pa we arrived at a cylinder diameter of 1.26mm - this seems very small, and hard to make!

    Hope you can help

    Alan
     
    Last edited: Jun 28, 2011
  2. jcsd
  3. Jun 28, 2011 #2

    uart

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    Science Advisor

    Alan, it seems you got everything right until the last step in finding the radius given the area!

    r = sqrt(A/pi)

    I got r = 17.9mm using your figures and A = F/deltaP
     
  4. Jun 30, 2011 #3
    ahh yes I can see my problem now! Ran through the calculation at the end again and I must have continually forgotten to square root! I now get the same
    Thanks for the help
    Alan
     
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