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Homework Help: Force angle question (i think)

  1. Dec 11, 2008 #1
    I'm not sure i know enough to know how to ask my question, but here goes.

    I want to figure out how the force pushing up on a pivot gets transferred to a cylinder also pushing on that pivot, for lack of better description, here are pictures of kinda what I am asking, the numbers are more for discussion, i am really looking for "how" to find this rather than a specific answer

    http://02ab560.netsolhost.com/joe/phys/phys1.jpg [Broken]
    http://02ab560.netsolhost.com/joe/phys/phys2.jpg [Broken]
    I haven't taken physics yet (next semester).
    Last edited by a moderator: May 3, 2017
  2. jcsd
  3. Dec 24, 2008 #2
    ok bear with me, and stop me if ive messed something up here (very possible)

    I am assuming a few things here, because i never got an answer.
    1.The angle of force that the vertical load is compared to is the angle of the rigid arm
    http://www.woodrell.com/joe/phys/physb-1.jpg [Broken]
    2.The force applied at an angle is the cosine of the angle delta between force and load
    http://www.woodrell.com/joe/phys/physb-2.jpg [Broken]

    ok with that let me label the things i am looking at
    a = 9 length of rigid arm
    b = 6 distance between fixed pivots
    c = height of movable pivot at rest above bottom fixed pivot
    d = height of top fixed pivot above movable pivot at rest
    e = horizontal distance between fixed and movable pivots
    f = pi/12 angle of rigid arm at rest
    g = angle delta of cylinder to angle of force on rigid arm of the force
    http://www.woodrell.com/joe/phys/physb-3.jpg [Broken]

    so with this
    c = Sin(f) * a
    d = b - c
    e = Cos(f) * a
    g = pi/2 - arctan( d / e ) - f

    so bringing those things together i get
    g = pi/2 - (arctan((6 - sin(pi/12) * 9) / (cos(pi/12) * 9))) - pi/12

    with that equation for the angle of the cylinder, i want to find how the force changes as the mounting point of the cylinder moves down the rigid arm closer to its fixed pivot, aligning with the angle of force, but moving in on the lever.

    this melted my brain for a while, but i collapsed it down to an equation with the length of the rigid arm as x, and scaling with where the pivot is on the am

    pi/2 - (arctan((6 - sin(pi/12) * x) / (cos(pi/12) * x))) - pi/12)
    http://www.woodrell.com/joe/phys/physb-4.jpg [Broken]

    and the force it would generate at this delta angle
    cos(pi/2 - (arctan((6 - sin(pi/12) * x) / (cos(pi/12) * x))) - pi/12))
    http://www.woodrell.com/joe/phys/physb-5.jpg [Broken]

    and multiplying that by the proportion of the mounting position on the rigid arm
    cos(pi/2 - (arctan((6 - sin(pi/12) * x) / (cos(pi/12) * x))) - pi/12)) * x/9
    http://www.woodrell.com/joe/phys/physb-6.jpg [Broken]

    basically what graphing this told me was that the greatest force is achieved when the cylinder attached to the end of the arm even though the cylinder was at more of an oblique angle

    the reason for all this thought is just a personal project, that will probably result in nothing more than a thought process, but i at least want to make sure i am getting it right so far

    anyway... thanks for your help guys, just trying to understand this stuff better
    Last edited by a moderator: May 3, 2017
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