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How to Keep a Sign from Collapsing

  1. Jan 19, 2015 #1
    1. The problem statement, all variables and given/known data
    . The sign shown below consists of two uniform legs attached by a frictionless hinge. The coefficient of friction
    between the ground and the legs is µ. Which of the following gives the maximum value of θ such that the sign will
    not collapse?

    (A) sin θ = 2µ
    (B) sin θ/2 = µ/2
    (C) tan θ/2 = µ
    (D) tan θ = 2µ
    (E) tan θ/2 = 2µ

    Diagram here : http://www.aapt.org/physicsteam/2014/upload/exam1-2013-1-6-unlocked.pdf
    Page 2, Diagram 6

    2. Relevant equations


    3. The attempt at a solution
    I drew a FBD and got a force of gravity for each leg with force mg down (relative to the surface on the ground) , normal force of mgcos((180-theta)/2) up (relative to the surface of the leg), and friction of mu*N towards the interior of the sign. The question is where do I go from here?
     
  2. jcsd
  3. Jan 19, 2015 #2

    phinds

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    Since the two forces you have computed don't oppose each other, you aren't going to get far that way. How about you find the force that the friction force IS opposing?
     
  4. Jan 19, 2015 #3
    Friction opposes the horizontal component of the normal force which is 1/2*mg*sin(180-theta), right?
     
  5. Jan 19, 2015 #4

    phinds

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    Well, what do you think?
     
  6. Jan 19, 2015 #5
    Well I am pretty sure that that's it, but I want to make sure.
     
  7. Jan 19, 2015 #6

    phinds

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    I didn't look at your figures but you have the right idea. The friction is a horizontal force so obviously the only force that can directly oppose it is a horizontal force.
     
  8. Jan 19, 2015 #7
    So is the next same just to equate the horizontal component of the normal force with the force of friction?
     
  9. Jan 19, 2015 #8

    phinds

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    You seem very hesitant to embrace the obvious. I suggest that you spend some time studying force vectors in various kinds of problems, not just ones where the forces that matter happen to be horizontal.
     
  10. Jan 19, 2015 #9
    I'm hesitant because I already equated it with each other and got mu=cos(90-theta/2) which is not one of the answer choices.
     
  11. Jan 19, 2015 #10

    phinds

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    That could be due to a calculation error, but can't be due to the forces not cancelling each other when the end is right at the point of moving.

    My point is that once you've done enough of these problems you'll have an instinctive feel for what forces have to cancel each other and an answer that seems wrong will just lead you to recheck your calculations, not your logic.
     
  12. Jan 19, 2015 #11

    BvU

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    I always learned that normal forces are just that: normal forces. Normal (perpendicular) to the horizontal plane in this case. So I don't think there is such a thing as a horizontal component of the normal force in this exercise.
     
  13. Jan 19, 2015 #12
    Ok so in that case m*g*sin(theta) = mu*m*g*cos(90-theta/2) ?
    If so anyway to simplify cos(90-theta/2)?
     
  14. Jan 19, 2015 #13

    phinds

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    That's correct. But there IS a horizontal component to the ladder's force on the floor. There are two resultant forces from the ladder's force on the floor, the normal force and the horizontal force.
     
  15. Jan 19, 2015 #14
    OK I'm confused.
     
  16. Jan 19, 2015 #15

    phinds

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    To such an utterly vague statement, I can only say, well that's unfortunate. I'm not. Do you have a question? Do you understand force vectors at all? If you don't then that's where to start.
     
  17. Jan 19, 2015 #16
    You two are telling me conflicting things.
     
  18. Jan 19, 2015 #17

    phinds

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    OK, I certainly don't mean to be. What two things are you talking about? Do you want me to guess or would you like to tell me?
     
  19. Jan 19, 2015 #18
    You are telling me to use the horizontal component of the normal force but BvU is saying to just use the normal force without decomposing it.
     
  20. Jan 19, 2015 #19

    phinds

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    No, I most emphatically am NOT telling you to use a component of the normal force. Please re-read what I said in posts #6 and #13
     
  21. Jan 19, 2015 #20
    I was using the horizontal component of the normal force.
    And you said ...
     
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