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Homework Help: Please help a few probs

  1. Apr 19, 2005 #1
    I have a few problems where I just need a push in the right direction:

    (1) A tightly stretched "high wire" is 46.5 m long. It sags 3.49 m when a 59.7 kg tightrope walker stands at its center. What is the tension in the wire? I have no idea what to do on this problem...

    (2)A 36.6 kg round table is supported by three legs placed equal distances apart on the edge. What minimum mass, placed on the table's edge, will cause the table to overturn? Neglect the mass of the legs. Wouldn't I need to know some other information like radius or something?

    (3)A cube of side l = 120 cm rests on a rough floor. It is subjected to a steady horizontal pull, F, exerted a distance h = 83.0 cm above the floor. As F is increased, the block will either begin to slide, or begin to tip over. What is the maximum coefficient of static friction for which the block begins to slide rather than tip? Wouldn't I ned to know F to figure this out? Keep in mind my answer has to be a numerical value, not a formula.

    Any tips on any problem would be greatly appreciated. Thank you so much for helping...
  2. jcsd
  3. Apr 19, 2005 #2

    James R

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    (1) The shape of the wire is an isosceles triangle with "height" 3.49 m. You can work out the angles at the two sides (on either side of the tightrope walker), and then resolve forces on the walker (two tension forces + gravity).
  4. Apr 19, 2005 #3
    Thanks, any ideas on the other questions?
    Last edited: Apr 20, 2005
  5. Apr 20, 2005 #4
    I'm stuck on those last two too...
  6. Apr 20, 2005 #5
    did you get numbers 5 or 6? My brain just wont work today. The last two questions regarding the damn ladder?
  7. Apr 20, 2005 #6
    Yeah... I set x and y axis as they are assigned in the prob, then just found torques. Torques are not perpendicular to the things they are acting on, so you must convert. Most the conversions were cos(theta) i think... although I think one was sin(90-theta). That help?
  8. Apr 20, 2005 #7
    Haha, I am seriously having a rediculous amount of difficulty with this. Did you use the center of the ladder as your pivot point? Then for torque you're using rFsin(theta) right? Is that what you meant by converting? Is F just the weight of the ladder?... I think I'm just having a difficult time identifying the forces acting on the ladder that contribute to torque.
  9. Apr 20, 2005 #8
    If you want some help, a problem, or some pictures wouldnt hurt.

    Torque = rFsin(theta) is the torque equation

    cos(x) = sin(90-x) is a trig identity
  10. Apr 20, 2005 #9
    The problems I'm most concerned with are in the first post, I just got sidetracked.

    See https://www.physicsforums.com/showthread.php?t=72065 for the ladder problem

    The problems are pretty self explanitory, but if you want some pics i'd be happy to upload some.
    Last edited: Apr 20, 2005
  11. Apr 20, 2005 #10
    I dont see any ladders in the first three problems.
  12. Apr 20, 2005 #11
    I just changed my last post, sorry
  13. Apr 20, 2005 #12
    You've got me on #2 and #3.
  14. Apr 20, 2005 #13
    yeah they're crazy... I don't know what my professor is thinking.
  15. Apr 20, 2005 #14
    Ok, the ladder will pivot about its ground contact. The center of mass would just be at the center, or 3.87m from the pivot point. You can find the torque in the counterclockwise direction from there, but keep in mind that the force is not just mg. Now since the torques sum to zero, the wall is providing the same torque but in the clockwise direction.

    The forces in the x and y directions both equal 0. The force towards the ground (-y) is F = mgsin(71.4), what can you say about the normal force (in x and y) applied on the ladder?

    A similar argument for the wall works.
  16. Apr 20, 2005 #15
    I picked the point against the wall as the rotation point
  17. Apr 20, 2005 #16
    squib: Either one works. Now find the torque in the clockwise direction from that point, can you find the force, and radius?
  18. Apr 20, 2005 #17
    Thanks a lot. I really appreciate all the help, now if only we can find someone that can do the second two probs. I think I'm going to repost them.
  19. Apr 20, 2005 #18
    On the cube problem I seem to come up with u = (a*h)/(g*(l/2))

    My problem is that I can't figure out what a is...

    this could also be written u = (F*h)/(mg*(l/2)), F being the constant force...
  20. Apr 20, 2005 #19
    yeah, I can't find a way to express u without using F or A. Bah.
  21. Apr 20, 2005 #20
    do you think maximizing F or A would give us an answer? Eg: putting all the constants in, taking the derivative of the whole thing and setting it equal to zero? Then we could use this force to put back into your equations. Does that sound plausible? I think I'll try it.

    bah ... darned equation has no max (duh). That doesnt work
    Last edited: Apr 20, 2005
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