Engineering 101
July 21, 2015 9:45 AM Subscribe
What is the load on each of these suspension points?
I just finished fabricating and installing this light fixture. It is suspended by the three cords of the same length and spaced equally. Say the fixture weighs 30 pounds. What is the weight suspended by each of the three cords? A neighbor told me that each of the three cords holds the entire 30 pounds.....I'm having trouble understanding why.
I just finished fabricating and installing this light fixture. It is suspended by the three cords of the same length and spaced equally. Say the fixture weighs 30 pounds. What is the weight suspended by each of the three cords? A neighbor told me that each of the three cords holds the entire 30 pounds.....I'm having trouble understanding why.
Each cord would need to be designed to take the full (dynamic) load, as if one failed, the others would have to be able to support everything.
posted by scruss at 10:04 AM on July 21, 2015 [1 favorite]
posted by scruss at 10:04 AM on July 21, 2015 [1 favorite]
But that doesn't mean that when all three are functional each one is supporting everything (which doesn't even make sense).
posted by kenko at 10:06 AM on July 21, 2015 [1 favorite]
posted by kenko at 10:06 AM on July 21, 2015 [1 favorite]
Best answer: Your neighbor is wrong. The sum of the loads at the three ceiling points is equal to the total weight of the fixture, but because the two of the suspension points are slightly inbound of the ends it's also not true that they are each supporting 10 pounds. (disclaimer: EE)
Also, practically, because there are three cords, any length differences could easily give rise to large tension differences. e.g. the middle one could carry almost zero load, even though it looks like it's the same length.
posted by achrise at 10:07 AM on July 21, 2015 [2 favorites]
Also, practically, because there are three cords, any length differences could easily give rise to large tension differences. e.g. the middle one could carry almost zero load, even though it looks like it's the same length.
posted by achrise at 10:07 AM on July 21, 2015 [2 favorites]
Best answer: seconding archrise. you're not going to get three cords equal. so it depends on how much flex there is and how close you were to making it balanced.
the total for all three is 30 pounds, but exactly how they are distributed isn't clear.
for example, if the middle support is longer than the other two to a fair degree (ie slack) the weight will be divided almost 50/50 between the ends, so about 15 lb each end.
or, if the middle is tighter than the others, and it's fairly well balanced, the middle could be taking almost all the weight (near to 30lb) while the ends don't do much at all.
in practice you'll be somewhere between those extremes.
one way in which your neighbour could be interpreted as being right is if they actually meant "you need to build it as though any of the supports could be taking the full weight" - that's a good, conservative rule of thumb.
posted by andrewcooke at 10:12 AM on July 21, 2015 [4 favorites]
the total for all three is 30 pounds, but exactly how they are distributed isn't clear.
for example, if the middle support is longer than the other two to a fair degree (ie slack) the weight will be divided almost 50/50 between the ends, so about 15 lb each end.
or, if the middle is tighter than the others, and it's fairly well balanced, the middle could be taking almost all the weight (near to 30lb) while the ends don't do much at all.
in practice you'll be somewhere between those extremes.
one way in which your neighbour could be interpreted as being right is if they actually meant "you need to build it as though any of the supports could be taking the full weight" - that's a good, conservative rule of thumb.
posted by andrewcooke at 10:12 AM on July 21, 2015 [4 favorites]
Response by poster: " a good, conservative rule of thumb"
That describes my neighbor to a Tee!
Thanks for all the great answers.
posted by Mr.Me at 10:16 AM on July 21, 2015
That describes my neighbor to a Tee!
Thanks for all the great answers.
posted by Mr.Me at 10:16 AM on July 21, 2015
This reminds me of first year engineering, and why I took Chem E (where it was all thermodynamics and transport phenomena and vectors are a dirty word).
Situations like this can lead to all sorts of non-intuitive results due to, as muddgirl pointed out, balancing moments. If you assume all the cords are of equal length (in a simplified "spherical cow" universe) then by varying the lever arms (where along the length of the fixture the cords are attached) you can end up with all sorts of different loads on the cords, anywhere from zero to 30lbsf(*)
Consider, for example, if the center cord is located exactly at the center of mass of the fixture (again in the simplified one-dimensional case). In that case the load on the center cord is 30lbs and the two outer cords take zero load. Not only is the tension in the center cord balancing the load of the fixture, but the moments also balance as the forces are equal but opposite and acting at exactly the same point.
(*) As you say in America-units. Right-thinking individuals naturally use the kilogram-force and metric-slug.
posted by selenized at 10:48 AM on July 21, 2015
Situations like this can lead to all sorts of non-intuitive results due to, as muddgirl pointed out, balancing moments. If you assume all the cords are of equal length (in a simplified "spherical cow" universe) then by varying the lever arms (where along the length of the fixture the cords are attached) you can end up with all sorts of different loads on the cords, anywhere from zero to 30lbsf(*)
Consider, for example, if the center cord is located exactly at the center of mass of the fixture (again in the simplified one-dimensional case). In that case the load on the center cord is 30lbs and the two outer cords take zero load. Not only is the tension in the center cord balancing the load of the fixture, but the moments also balance as the forces are equal but opposite and acting at exactly the same point.
(*) As you say in America-units. Right-thinking individuals naturally use the kilogram-force and metric-slug.
posted by selenized at 10:48 AM on July 21, 2015
I think the third cable makes it a statically indeterminate problem.
Also, your neighbor isn't necessarily being conservative enough. The suspension system should be built to hold 6 times the weight, since it's suspended over peoples' heads.
posted by Huffy Puffy at 1:38 PM on July 21, 2015 [3 favorites]
Also, your neighbor isn't necessarily being conservative enough. The suspension system should be built to hold 6 times the weight, since it's suspended over peoples' heads.
posted by Huffy Puffy at 1:38 PM on July 21, 2015 [3 favorites]
Huffy Puffy is right. It is a statically indeterminate problem. You don't have enough equations to solve for the force in each member using force and moment balance alone. You must introduce information about the way the cords deform to acquire enough equations to solve the system. To get this additional equation, you need to use the constitutive equation (to relate deformation to stress). This is a classic statically indeterminate problem from graduate level mechanical engineering. Here is how it is solved. The cords in the example are all dissimilar materials, but the method is the same for any case.
posted by incolorinred at 6:42 PM on July 21, 2015
posted by incolorinred at 6:42 PM on July 21, 2015
Wait!
It's only statically indeterminate if it's a 2-D problem.
The wires do all appear to be in line together. If they weren't, you could also consider the side view, and sum moments in that direction.
The difference is similar to one of a table with 3 legs in a line, vs. a table with 3 legs in any other configuration.
You have 3 unknowns you want: let's call them tensions A, B, C. In 2 dimensions, you can get 2 independent equations by summing the forces in vertical direction (they all add up to the weight) and by summing moments at a point of your choice (so, for example, the outside tensions are the same).
If you have a third dimension, you can get another independent equation, and you're good to go.
Look from the side, at an angle where 2 of the wires line up. Sum moments around the attachment point of those 2 wires. The relative lever arms of the weight (distributed force) and the third attachment point gives you the tension on that third wire. Your other equations (all 3 add up to weight, and outside tensions are equal) give you the other two.
posted by Huffy Puffy at 3:45 AM on July 22, 2015 [1 favorite]
It's only statically indeterminate if it's a 2-D problem.
The wires do all appear to be in line together. If they weren't, you could also consider the side view, and sum moments in that direction.
The difference is similar to one of a table with 3 legs in a line, vs. a table with 3 legs in any other configuration.
You have 3 unknowns you want: let's call them tensions A, B, C. In 2 dimensions, you can get 2 independent equations by summing the forces in vertical direction (they all add up to the weight) and by summing moments at a point of your choice (so, for example, the outside tensions are the same).
If you have a third dimension, you can get another independent equation, and you're good to go.
Look from the side, at an angle where 2 of the wires line up. Sum moments around the attachment point of those 2 wires. The relative lever arms of the weight (distributed force) and the third attachment point gives you the tension on that third wire. Your other equations (all 3 add up to weight, and outside tensions are equal) give you the other two.
posted by Huffy Puffy at 3:45 AM on July 22, 2015 [1 favorite]
Mr.Me: "A neighbor told me that each of the three cords holds the entire 30 pounds.....I'm having trouble understanding why."
Speaking as not-a-structural-engineer, this is a crazy idea. Take it to its logical extreme: A 10lb light fixture supported by a thousand cords would be putting 5 tons of force on the joists above.
posted by Plutor at 1:10 PM on July 22, 2015
Speaking as not-a-structural-engineer, this is a crazy idea. Take it to its logical extreme: A 10lb light fixture supported by a thousand cords would be putting 5 tons of force on the joists above.
posted by Plutor at 1:10 PM on July 22, 2015
Huffy Puffy, that is not a correction analysis. The problem remains invariant whether considering it in 2 dimensions or 3. The 3rd dimension in this case does not provide useful equations because it provides zero identity statements. If we consider the sum of the forces in the z-direction (we will say the z axis is the orientation perpendicular to the line of supports) we get [Sigma]Fz = 0. Since there is no component of force acting along the z direction, this expression becomes 0 = 0. Furthermore, in the 2-dimensional treatment, only the moment about the z-axis is considered. If we consider the other two moment balances about the x and y axes, My and Mz, these also produce 0 = 0 identities. 0 = 0 does not help you solve for the variables of interest.
It is an axiom of physics that system behavior not change depending on choice of coordinate system. If solving the problem in one system produces different answers than solving in another, we must conclude that governing equations, solution scheme, or both is invalid. In this case, the governing equations (Newton's law) is valid so the solution scheme must be invalid.
posted by incolorinred at 11:13 AM on July 24, 2015
It is an axiom of physics that system behavior not change depending on choice of coordinate system. If solving the problem in one system produces different answers than solving in another, we must conclude that governing equations, solution scheme, or both is invalid. In this case, the governing equations (Newton's law) is valid so the solution scheme must be invalid.
posted by incolorinred at 11:13 AM on July 24, 2015
You are right that the forces are the forces no matter how you look at them.
If the attachment points are all collinear (as they appear to be), it's a statically indeterminate problem.
If they are not collinear, then that third dimension matters, and you can solve it with statics.
Or you can just assume that some 2 of the 3 attachment points take 15 pounds, and design them all to do so.
posted by Huffy Puffy at 6:37 AM on July 26, 2015
If the attachment points are all collinear (as they appear to be), it's a statically indeterminate problem.
If they are not collinear, then that third dimension matters, and you can solve it with statics.
Or you can just assume that some 2 of the 3 attachment points take 15 pounds, and design them all to do so.
posted by Huffy Puffy at 6:37 AM on July 26, 2015
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Force_from_weight = Tension_Cord_1 + Tension_Cord_2 + Tension_Cord_3.
Then to calculate the exact tension, I'd balance the moments, but I'd need to know the length of each moment arm. Eyeballing the picture and assuming the whole thing was designed properly, it's safe to assume that each cord is sharing the weight equally.
posted by muddgirl at 10:03 AM on July 21, 2015 [2 favorites]