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  1. #1
    New Member JeffD's Avatar
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    Suspension Force Calculation - Need Mathletes

    I know how to calculate the suspension forces if you're hanging from two vertical points alone, but what if you connect to two eye bolts in the ceiling and run the suspension down through two eye bolts in the wall to achieve 6' attachment points. How do you calculate the force on each of the four bolts?

    The reason I'm asking is because I'm considering an indoor hang and don't want to depend on two eye bolts in the wall studs alone. If the vertical hang relieves some of that pressure then maybe....

    Thanks.
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  2. #2
    Senior Member JohnSawyer's Avatar
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    That particular design won't spread the load, much. The load point on the wall is acting like a pulley, transferring some of the inward load into vertical.

    Now if you tied to BOTH at the same time, I would think that you'd have a more even load distribution.

    I'm waiting for Grizz to take this on...
    "Do or do not, there is no try." -- Yoda


  3. #3
    New Member JeffD's Avatar
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    By both at the same time do you mean one continuous loop of suspension running from one end of the hammock to the other through all 4 bolts?

  4. #4
    Senior Member GrizzlyAdams's Avatar
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    what is really need here is someone who remembers their mechanical statics...

    But I'll take a whack at it.

    Yep, this is a pulley system. In the pulley diagram you usually think of, there is
    a downward force on the axle of the pulley, and that force in magnitude is split equally between the sides of the line through the pulley.

    In this picture, I think, the downward force is the force on the line from the
    hammock to the ring on the wall. We know this to be T = (W/2)(1/sin theta)
    where we like theta to be 30 degrees, that angle.

    If my guess is right, that means that the horizontal stress on the vertical ring is
    T/2, and the magnitude of the force on the line going to the ceiling ring is also T/2. You'll get the vertical and horizontal components of that on the ceiling ring by doing the appropriate trig projections (which depends on the placement on the ceiling), one of them referring to the downward pull that's trying to get the bolt out of the ceiling, the other referring to the force trying to break the bolt sideways.
    Grizz
    (alias ProfessorHammock on youtube)

  5. #5
    New Member JeffD's Avatar
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    Thanks Grizz.

    So if I weighed an even 200 pounds, the force from hammock to wall bolt is 100 pounds(per side) and the force from wall bolt to ceiling bolt is 100 pounds(per side).

    Note: The only attachment point for the suspension line is the ceiling bolts, then the line is just threaded through the wall bolt to create a turn that simulates attaching to two trees at 6 feet high.

  6. #6
    Senior Member GrizzlyAdams's Avatar
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    Quote Originally Posted by JeffD View Post
    Thanks Grizz.

    So if I weighed an even 200 pounds, the force from hammock to wall bolt is 100 pounds(per side) and the force from wall bolt to ceiling bolt is 100 pounds(per side).

    Note: The only attachment point for the suspension line is the ceiling bolts, then the line is just threaded through the wall bolt to create a turn that simulates attaching to two trees at 6 feet high.
    That's my guess (assuming a 30 degree angle from hammock to wall ring), but there are much better mechanical engineers around here than me. Make sure your life insurance is paid up before you test.


    Caveat---what is definitely true is that the sum of the force vectors at the bolt on the wall is zero. I read that with a pulley system even when the lines from the pulley are not in line with the opposing force, the tensions on the lines are half the weight of the pulley. I have not verified that the this gives the sum of force vectors being zero.

    But the idea you float should decrease the force on the wall bolt.
    Last edited by GrizzlyAdams; 08-14-2011 at 09:21.
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  7. #7
    Senior Member GrizzlyAdams's Avatar
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    Reset.

    This is not a pulley system. Pulleys are free to move along the cord, and that's why the tension on both cords is the same---when hung, the pulley will move to a place where the angle of the cords with respect to the vertical is the same.

    For the problem at hand, the sum of forces at the wall bolt being zero is the right way to think about this, and the case of the ceiling bolt being directly above the wall bolt makes the answer plain.

    For the sum of the force vector from the hammock and the vertical vector has to be equal to the horizontal vector at the bolt, but since the vertical vector has no horizontal component, that means the horizontal vector is the horizontal component of the hammock force vector, which we know already as the shear vector at the tree.

    The same reasoning shows that the vertical vector from bolt to ceiling has to
    be the same as the vertical component of the hammock vector, to wit, half the weight.

    This means that the ceiling bolt takes the vertical component of the force from the hammock, W/2 where W is the weight. The shear force on the wall remains the same, (W/2) cot(theta).

    So the wall bolt has less overall force on it, but not the part that is trying to tear it out of the wall.
    Last edited by GrizzlyAdams; 08-14-2011 at 22:56. Reason: more explanation
    Grizz
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  8. #8
    New Member JeffD's Avatar
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    Thanks for the time and effort. This place is great.

  9. #9
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    Quote Originally Posted by GrizzlyAdams View Post
    Reset.

    This is not a pulley system. Pulleys are free to move along the cord, and that's why the tension on both cords is the same---when hung, the pulley will move to a place where the angle of the cords with respect to the vertical is the same.
    I disagree with the assumption this is not a pulley system. The pulley does not need to be free to move along the line, the line needs to be free to move through the pulley. The only difference between this eyebolt and an ideal pulley is the friction force between the suspension line and the eye. If the suspension were tied at the wall eyebolt it would no longer be a pulley system, but would instead become statically indeterminate. The indeterminate value would be the tension in the line from the ceiling to the wall eyebolt. With the right amount of pretension it could very much help with the wall eyebolt's forces, but trying to get and keep the right amount of pretension would likely be impossible.

    If this does behave as an ideal pulley system there is no difference in the net force on the eye bolt in the wall for a 30 degree hang with or without the vertical eyebolt. The horizontal component is unchanged, and the vertical component has gone from w/2 up to w/2 down. The line tension will remain W. Shallower angles (closer to horizontal) will actually result in a greater force on the wall eye bolt, steeper angles should result in a reduced force on the eye bolt.

    If you want to reduce the force on the wall bolts, place the ceiling eye bolts directly over where you want each end of the hammock to be. Tie one line from the ceiling bolt to the hammock, and a separate line from the wall eye bolt to the hammock. In this case the wall to hammock line can be horizontal. Each line will take the directional component of the force, and that component will be a direct withdrawal load from the holding member. If you know the type of wood, the eyebolt thread, and the embedment length there are tables to calculate allowable withdrawal strengths.

    -Mike

  10. #10
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    Quote Originally Posted by JeffD View Post
    Thanks Grizz.

    So if I weighed an even 200 pounds, the force from hammock to wall bolt is 100 pounds(per side) and the force from wall bolt to ceiling bolt is 100 pounds(per side).

    Note: The only attachment point for the suspension line is the ceiling bolts, then the line is just threaded through the wall bolt to create a turn that simulates attaching to two trees at 6 feet high.
    Each ceiling bolt will be subjected to T, the tension in the cord. Assuming 30 degree hang angle that T is W, the weight of the occupant. In your example, 200 lbs. The wall bolts happen to be subjected to a force of T, but it is based on the free body diagram of the wall bolt. Sum the forces of the lines coming into the bolt, and the bolt must handle any unbalanced forces to remain at rest. There are two forces acting on the bolt, one from the line to the ceiling, and one from the line to the hammock. The force in each is T or W (200 lbs). One is directed straight up, so Fy=+200lbs. One is directed towards the hammock, so Fy=-100lbs and Fx=+173lbs. Summing the forces we have Fy= +200-100 = +100 lbs. Fx = +173 lbs. The bolt must oppose these net forces, so 100 lbs down and 173 lbs towards the wall. Vector sum of these forces is sqrt(100^2+173^2) = 200 lbs.

    The net result being W is a coincidence of the angles chosen...

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