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Insights Frequently Made Errors in Mechanics - Friction - Comments

  1. May 8, 2015 #1

    haruspex

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  3. May 8, 2015 #2

    robphy

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    With regards to the work done by forces,
    I often hear blanket statements like
    "friction always does negative work" and "normal force always does zero work".
     
  4. May 9, 2015 #3

    haruspex

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    Yes.. the one about the normal force belongs in the "Forces" post. Unfortunately, my permission to edit it seems to have been taken away again. I'm forming the impression that a blog system (where an item is published and becomes cast in stone, only modifiable by a comment chain) is not suitable for what I thought this was about. It doesn't make a for usable reference text.
     
  5. May 9, 2015 #4
    If a disk slides on a rough surface, then friction will make the disk roll right? Because its the only force that can create a torque. Am I correct?
     
  6. May 9, 2015 #5

    jbriggs444

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    A wrench can produce torque without friction. So can an electric motor. Or tidal gravity. Or a windmill [where, to be clear, it can be the "lift" that is producing torque and the "drag" is not essential].
     
  7. May 10, 2015 #6
    I was talking about one particular case. Also, If the wrench was frictionless, It will simply slide instead of rotating the object right?
     
  8. May 10, 2015 #7

    jbriggs444

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    Most wrenches (box end, open end, adjustable, socket, etc) push on the faces of hexagonal and square nuts and bolts at a position offset from the center of the face. Friction is unimportant to their use.

    Edit: Hard to hold them without friction, of course. Though not completely impossible.
     
  9. May 10, 2015 #8
    Understood. If the bolt was circular, then friction will be important right?
    Also, for a ring, if a force is applied tangentially at its topmost point, then no friction is required to make it roll. And this true only for a ring.
    Force equation:
    $$F=ma$$
    Torque eqation:
    $$FR-fR=MR^2\alpha$$
    $$F-f=MR\alpha$$
    And for pure rolling, ##R\alpha=a##
    So $$F-f=Ma$$
    Since F=Ma, f=0.
     
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