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Physics question

Discussion in 'Off-Topic Discussion' started by svdude, Apr 12, 2019.

  1. Apr 13, 2019 at 4:23 PM
    #81
    AxisCab

    AxisCab Well-Known Member

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    DSCN0661.jpg
    DSCN0662.jpg

    If you want to solve this problem, you hafta understand:
    algebra
    sine and cosine
    force as a vector


    Peeps who light up math books are why we can't have nice things.
     
  2. Apr 13, 2019 at 5:03 PM
    #82
    takoashihaisen

    takoashihaisen Well-Known Member

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    No need to use sin or cos on this problem. If you understand the triangle ABC and triangle DBE are similar (and understand the length CE = length EB), you do not have to use sin or cos. You do not need to calculate the length CE or EB
    diagram.jpg
     
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  3. Apr 13, 2019 at 5:05 PM
    #83
    Skydvrr

    Skydvrr Well-Known Member

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    Man I suck at geology.
     
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  4. Apr 13, 2019 at 5:07 PM
    #84
    sunchip89

    sunchip89 Member

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    Sum of the moments about a point bud :)

    IMG_20190413_190615913.jpg
     
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  5. Apr 13, 2019 at 5:14 PM
    #85
    mynewtoy

    mynewtoy Didn't know this was a thing

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    I don't understand this thread. You don't need math. We are talking about 50lbs.

    Use this it won't break

    https://www.e-rigging.com/3-16-galvanized-cable

    [​IMG][​IMG]
    Property Value
    Breaking Strength: 4,200 lbs
    Working Load Limit
    (5:1) Design Factor: 840 lbs
    Weight per foot: 0.065 lbs
     
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  6. Apr 13, 2019 at 5:25 PM
    #86
    Jaysonn

    Jaysonn Well-Known Member

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    I was reading through the thread and i agree that the tension will be equal depending on whether your rigging points on the 50 pound beam are at the center of gravity on the vertical plane of the beam and the horizontal plane of the beam. The tension difference will be extremely small if the rigging points are on the top of the beam.
     
  7. Apr 13, 2019 at 6:32 PM
    #87
    svdude

    svdude [OP] Well-Known Member

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    My point in this isn't the strength, I'm using 1/8 cable to support about ~200 lbs, the breaking point on the cable I'm using is almost 4 times that. The point is that I'm using pulleys and would like to use a single cable through all pulleys on one side. I do agree with the post above that each cable will both support 25 lbs since the cg is still centered on the beam and is equal distance from both vertical cables.

    My hope was to use pulleys on both sides of a system and use a single cable through the pulleys on both sides. Since both pulleys will always be supporting the same weight, if one side raises faster than the other and the object getting lifted gets tilted, it won't "self level" itself since both pulley systems are still supporting the same weight. Knowing the above, I'll use a seperate cable for each corner if the object I'm hoisting up.
     
  8. Apr 13, 2019 at 6:35 PM
    #88
    svdude

    svdude [OP] Well-Known Member

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    Thanks! This makes sense, I just was drawing blanks on whether there would be any horizontal forces acting on the cables since the beam is at an angle. However, since both cables are completely parallel with each other the only force is gravity and the cg is still centered on the beam, each cable still holds 25 lbs.
     
  9. Apr 13, 2019 at 6:45 PM
    #89
    slnvail

    slnvail Well-Known Member

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    This thread made my head hurt.
     
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  10. Apr 13, 2019 at 7:29 PM
    #90
    svdude

    svdude [OP] Well-Known Member

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    Haha .. this is the too bad. It's actually a pretty simple problem to solve if you know the fundamentals. This is proof however that I clearly don't know the fundamentals.
     
  11. Apr 15, 2019 at 10:36 AM
    #91
    salladh

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    I didn’t read this whole thing, and maybe someone beat me too it, but This isn’t a static’s problem, as the starting point given by the OP is not a static state.

    The bar is going to equalize once you hang it. Because the bar is being hung at an angle, the attachment points in the ceiling will be closer than 5ft.Once the bar “equalizes” itself it will impart horizontal forces on its support structure (and a compressive force on the bar) because the cables are angled. The tension in the cables is function of those horizontal forces.

    As you hang weight on the bar, it will “equalize” more, and the tensile forces seen by the cables will be amplified by the additional horizontal force.

    Edited cuz I Forgot to add that the horizontal forces are gonna be tiny So nbd

    image.jpg
     
    Last edited: Apr 15, 2019 at 2:54 PM
  12. Apr 15, 2019 at 12:56 PM
    #92
    phdog

    phdog Well-Known Member

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    Has no one ever moved a sofa? On flat ground both people are holding equal weight if the sofa is consistent. Once you start going up or down stairs the weight shifts. Same thing here. Level beam and equal weight distribution. Angle the beam and the weight shifts some depending on the degree to which one end is higher than the other. Math will tell you how weight is distributed.
     
  13. Apr 15, 2019 at 3:28 PM
    #93
    mynewtoy

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    that's a little different because your moving the sofa up an incline. So the center of mass moves away from the direction of travel. This is static

    going up stairs it's alot harder for the person on top to grip and pull up than the person on the bottom
     
  14. Apr 15, 2019 at 3:29 PM
    #94
    takoashihaisen

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    I thought if AE=squareroot (24), angles BAE, AED, ABD, and BDE are 90 degree. In other words, AE = squareroot (24) is the requirement to have AB and ED normal to the earth (or you can say perpendicular to the ceiling). When AE is not squareroot (24), these 4 angles are no longer 90 degree. What do you think?

    OP did not specify the length AE in the initial setup - only 5 ft of 50 lb beam and 2 ft support and 3 ft support. OP can make these AB and ED normal to the earth if he picks root 24 as the distance AE.

    diagram 2.jpg
     
    Last edited: Apr 15, 2019 at 3:43 PM
  15. Apr 15, 2019 at 3:30 PM
    #95
    mynewtoy

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    Please someone with scales go do this. Anyone have two fish scales and a board?
     
  16. Apr 15, 2019 at 4:53 PM
    #96
    phdog

    phdog Well-Known Member

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    Okay, not a perfect analogy. So now stop and hold it. Now it's static and the person on the bottom is holding more weight.
     
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  17. Apr 15, 2019 at 5:34 PM
    #97
    takoashihaisen

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    Okay, probably best to wait for the better diagram.

    @salladh



    Somehow you estimated as Ds=squareroot (24), if that is the case, the beam length should the cross point between Fy1 and the beam, and Fy2 and the beam. In this case, theta 1 = theta 2 = 0. Therefore Fx1=Fx2=0. Are you saying that the beam will move from this position (theta 1 = theta 2 = 0) and theta 1 and theta 2 will be no longer zero?. If the beam somehow moves to the right, vector Fx1 + vector Fx2 will have the right direction. If the beam somehow moves to the left, vector Fx1 + vector Fx2 will have the left direction. Either case, vector Fx1+vector Fx2 will no longer be zero, that is not equilibrium.
    image11.jpg
     
  18. Apr 15, 2019 at 11:40 PM
    #98
    Fuergrissa

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  19. Apr 16, 2019 at 1:35 AM
    #99
    Skydvrr

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    God damn geography. I bow out of this thread.
     

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