Physics question

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

 

Man I suck at geology.

 

Sum of the moments about a point bud

 

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


Property Value
Breaking Strength: 4,200 lbs
Working Load Limit
(5:1) Design Factor: 840 lbs
Weight per foot: 0.065 lbs

 

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.

 

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.

 

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.

 

This thread made my head hurt.

 

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.

 

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

 

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.

 

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

 

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.

 

Please someone with scales go do this. Anyone have two fish scales and a board?

 

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.

 

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.

 

God damn geography. I bow out of this thread.

 

Just came across this thread, and it was a fun read, haha. Some things were correct, some were not. Most answers used some unstated assumptions, others didn't exactly follow the given parameters, etc.,

In the unequally loaded crane example that was posted, the centre of gravity was not in the centre of the beam (when horizontal or in the angled position), and so they are measuring the horizontal distance to the COG. That is the real question in all of this, because one of the stipulations in the original post was that the cables make a right angle with the ceiling in the final loaded position. The easiest (physical) way to do that is with rollers, assumed frictionless in the analysis to ensure that they are perfectly perpendicular.

Assuming it is a 2-dimensional problem, constant cross-section, and constant material properties in the beam, the COG will be at exactly the geometric centre of the beam (half way between each end, halfway between the top and bottom), it is a purely static problem. If the height of the beam is very thin compared to the length (or the rigging is attached at the centre of the beam), you can assume the COG is in the centre of the two cables and that they will have the same tension in each cable, regardless of the angle of the beam provided the difference in cable length is less than the length of the beam. If you don't make that assumption, you need to consider the thickness any offset thickness of the beam and how much it will move the centre of gravity horizontally (x-direction).

I apologize for any illegibility of these drawings or non-simple things I did with the geometry (e.g. sinϕ doesn't need to show up, it was just a bridge I used to show the similarity), or steps I didn't show, haha. I'm pretty tired.




EDIT: I also assumed the difference in cable length to be at equilibrium, i.e. they don't change length under load.

Jeff

 

Fantastic thread.