Physics help - cantelievers

[raftingtigger]

I'm trying to calculate some forces and am stuck. I have checked the hang calculators and gotten some of what I need. Below is a diagram simplified the best I can. Large rectangle in a wall, with the smaller rectangle a pole. For calculations figure the wall and pole do not move or bend.

Capital letter are forces, lower case are angles. If S, a, b, and c are known what are the formulas for C and T?

Canteliver forces.jpg

I would like to make a forces calculator to share for hanging a hammock off a guyed pole.

[aero-hiker]

Hi Raftingtigger,
I'll bite! (I am an aerospace engineer and love Newtonian Physics!)

First, however, I'd like to ask/clarify something about your picture:

1. Will the vertical member (large rectangle, i.e. wall) always be perfectly vertical and perpendicular to the ground?
2. Will the guy pole always be perpendicular to the large member/wall?
3. Is angle "c" to account for the hang angle? (hangle)
4. Will the hammock be set up so that the axial direction is in/out of the page in your drawing? (i.e. will it be setup along/parallel the wall in the picture or bridging across to another wall, out of frame but on the right of your picture? If the former, then you need an additional guy line to keep the two guyed poles from folding in on each other.)
4. You know how to calculate the tension S given a hangers weight and the hang angle. (Hint: it's not 1/2 the hangers weight!)

I'll post a solution now assuming that:

1. The answer to 1 is "yes" the wall/large vertical member is always perfectly vericle, and
2. c is a hang angle for a hammock hung in the plane of the drawing (i.e. the other end is attached somewhere on the right out of frame; questions 3 and 4), and
3. The guy pole is attached with a hinge or pin, not welded or rigidly fixed to the wall, and able to carry a bending moment (torque).

Also, consistent with these assumptions I can write the equations in terms of the hangers weight in addition to the tension S.

1. Calculate the vertical and horizontal components of the tension being carried in S:
a. If we know the hang weight, then the vertical component, S_v, is simply 1/2 the hang weight, and the horizontal component, S_h, is 1/2 the hang weight divided by tan(pi - c) (in radians) or tan(180 - c) in degrees. (Also, tan(pi - c) = -tan(c) and the problem is simple enough to not have to set a sign convention and pay close attention...)
To summarize: S_v = 0.5*W and S_h = 0.5*W/|tan(c)| = 0.5*W*|cot(c)|
b. If we know the tension, S, then S_v = S*sin(c) and S_h = S*|cos(c)| (because sin(pi - c) = sin(c) and cos(pi - c) = -cos(c) and I'd rather mentally keep track of the direction of each force rather than worry about a formal sign carrying system...)
2. Since this system is at rest, we know that the forces must balance each other. If angle "b" is pi/2 (90 deg) then this means that T_v = S_v, and is the easiest case. If angle "b" is not pi/2 then a system of equations (two equations and two unknowns) is the result. Let's start with the simpler case where b is pi/2:
a. For b = pi/2 or 90 degrees: C_v = 0, therefore: T_v = S_v = 0.5*W = S*sin(c). T = T_v/sin(pi/2 - a) = S_v/cos(a). Then, T_h = T*cos(pi/2 - a) = T*sin(a) = ( S_v/cos(a) )*sin(a) = S_v*tan(a). C = C_h = T_h - S_h. Therefore, C = S_v*tan(a) - S_h = S_v*tan(a) - S*|cos(c)| = S*( sin(c)*tan(a) - cos(c) ) or in terms of hang weight, C = 0.5*W*( tan(a) - cot(c) ). The resulting equations in terms of the knowns are: T = 0.5*W/cos(a) = S*sin(c)/cos(a) and C = 0.5*W*( tan(a) - cot(c) ) = S*( sin(c)*tan(a) - cos(c) )

Hopefully I've done an OK job explaining... If I've made any mistakes let me know, I'll work through on paper and add diagrams...

Also, for the case where angle "b" is not 90 degrees, answer my questions about problem setup and assumptions, just so I don't work through more math only to discover I've solved the wrong problem.

One last note: Buckling problems are pretty complicated from what I remember, so make sure that the guyed pole has a sufficiently large diameter and strength to prevent buckling. Tensile forces are usually easier to deal with.

I hope this helps! (And I hope I haven't made any foolish mistakes!)

[gunner76]

Why does my head hurt....

[Bike-N-Hike]

Thank goodness I don't have to get out my Statics book!!!!!!!!

[Canoefor2]

I love it!!!!! I told my kids that this is where the cool guys hang out.

[dbass]

And I thought helping my 6th grader with her math was an achievement. <note to self: never gloat about superior mathematic/physics abilities until after consulting hammockforums>

[Sunny Bear]

AeroHiker- This is why I love this site!! The amount of gathered knowledge and experience contained here is second to none!

[aero-hiker]

Thanks Sunny Bear... All math is high-school level though... I think with drawings my text could be simplified and wouldn't seem so complicated... It's just trigonometry and geometry. A free body diagram of where the guy pole, guy line, and hammock line meet would go a long way in simplifying the explanation, but I was too lazy to scan something...

[Zilla]

[QUOTE=aero-hiker;1532909]Hi Raftingtigger,
I'll bite! (I am an aerospace engineer and love Newtonian Physics!)

First, however, I'd like to ask/clarify something about your picture:

1. Will the vertical member (large rectangle, i.e. wall) always be perfectly vertical and perpendicular to the ground?
2. Will the guy pole always be perpendicular to the large member/wall?
3. Is angle "c" to account for the hang angle? (hangle)
4. Will the hammock be set up so that the axial direction is in/out of the page in your drawing? (i.e. will it be setup along/parallel the wall in the picture or bridging across to another wall, out of frame but on the right of your picture? If the former, then you need an additional guy line to keep the two guyed poles from folding in on each other.)
4. You know how to calculate the tension S given a hangers weight and the hang angle. (Hint: it's not 1/2 the hangers weight!)

I'll post a solution now assuming that:

1. The answer to 1 is "yes" the wall/large vertical member is always perfectly vericle, and
2. c is a hang angle for a hammock hung in the plane of the drawing (i.e. the other end is attached somewhere on the right out of frame; questions 3 and 4), and
3. The guy pole is attached with a hinge or pin, not welded or rigidly fixed to the wall, and able to carry a bending moment (torque).

Also, consistent with these assumptions I can write the equations in terms of the hangers weight in addition to the tension S.

1. Calculate the vertical and horizontal components of the tension being carried in S:
a. If we know the hang weight, then the vertical component, S_v, is simply 1/2 the hang weight, and the horizontal component, S_h, is 1/2 the hang weight divided by tan(pi - c) (in radians) or tan(180 - c) in degrees. (Also, tan(pi - c) = -tan(c) and the problem is simple enough to not have to set a sign convention and pay close attention...)
To summarize: S_v = 0.5*W and S_h = 0.5*W/|tan(c)| = 0.5*W*|cot(c)|
b. If we know the tension, S, then S_v = S*sin(c) and S_h = S*|cos(c)| (because sin(pi - c) = sin(c) and cos(pi - c) = -cos(c) and I'd rather mentally keep track of the direction of each force rather than worry about a formal sign carrying system...)
2. Since this system is at rest, we know that the forces must balance each other. If angle "b" is pi/2 (90 deg) then this means that T_v = S_v, and is the easiest case. If angle "b" is not pi/2 then a system of equations (two equations and two unknowns) is the result. Let's start with the simpler case where b is pi/2:
a. For b = pi/2 or 90 degrees: C_v = 0, therefore: T_v = S_v = 0.5*W = S*sin(c). T = T_v/sin(pi/2 - a) = S_v/cos(a). Then, T_h = T*cos(pi/2 - a) = T*sin(a) = ( S_v/cos(a) )*sin(a) = S_v*tan(a). C = C_h = T_h - S_h. Therefore, C = S_v*tan(a) - S_h = S_v*tan(a) - S*|cos(c)| = S*( sin(c)*tan(a) - cos(c) ) or in terms of hang weight, C = 0.5*W*( tan(a) - cot(c) ). The resulting equations in terms of the knowns are: T = 0.5*W/cos(a) = S*sin(c)/cos(a) and C = 0.5*W*( tan(a) - cot(c) ) = S*( sin(c)*tan(a) - cos(c) )

Hopefully I've done an OK job explaining... If I've made any mistakes let me know, I'll work through on paper and add diagrams...

Also, for the case where angle "b" is not 90 degrees, answer my questions about problem setup and assumptions, just so I don't work through more math only to discover I've solved the wrong problem.

One last note: Buckling problems are pretty complicated from what I remember, so make sure that the guyed pole has a sufficiently large diameter and strength to prevent buckling. Tensile forces are usually easier to deal with.

I hope this helps! (And I hope I haven't made any foolish mistakes!)[/Q


That looks correct.

[raftingtigger]

Thanks aero-hiker.

I'll re-start with the full diagram.
hammock forces.jpg

This is 1/2 the system. The other half is a mirror image. W and all angles can be measured and each angle may vary from 0-90 degrees. Consider the only significant mass to be W. Pole taking force C is rigid. Consider it a static system.

The point of the exercise is to see how each force varies with changes in the different angles (hammock hang, pole, and anchor) so that a pole supported hammock system can be optimized.

If I have forgotten any forces, such as rotational on the pole, please add them.

I am over my head with the physics, and am just doing empirical experiments for now, but the information should be interesting to many hammockers.