# Calculate tilt angle of a bar lifted via two ropes on fixed points

• Pp122
In summary, the equilibrium angle of a bar that is lifted on its two ends with ropes attached to fixed lifting points can be calculated by solving for the angles and then for the tensions.
Pp122
Summary: How to calculate the equilibrium angle of a bar that is lifted on its two ends with ropes attached to fixed lifting points?

Hello and good day all,

First of all I would like to apologize in advance for my english, I am not a native speaker so some grammar errors may be left.

My question today is about the tandem lifting of an object (i.e. lifting simultaneously with two lifting devices), I would like to eventually calculate the load share in the rigging legs. I expected the question to be fairly easy but it seems way harder to solve than I thought initially, maybe that's just me though. It's been a while that I am thinking about this problem without having a clear definitive way to formulate a general answer.

Starting with the basic example of a plane problem with two lifting points of assumed known fixed coordinates connected to a simple bar via slings (able to rotate, I guess we can consider them rigid in this model) I realize that my main challenge is to find what will be the equilibrium position of the bar, after that it would be simple in-plane statics with 2 unknown forces amplitudes but known direction hence a determinate problem, nothing too difficult when this stage is reached.

For the model, we have:

• lA, lB lengths of the slings A and B
• L length of the bar
• d distance of the COG of the bar (0<d/L<1)
• mA, mB masses of the slings (I choose to ignore them for a start for simplicity)
• m mass of the bar
• θA, θB angles of the slings with vertical axis
My first take at it is to make a kinematic closure and make projection on horizontal and vertical axes.

This gives:

lAsin(θA)+Lcos(α)+lBsin(θB)−(xB−xA)=0
−lAcos(θA)+Lsin(α)+lBcos(θB)−(yB−yA)=0

I have 3 unknown angles that are linked by 2 equations, my system has one degree of freedom.

The next step would be to find a physical equation or constraint to block this last DOF and determine the values of the 3 angles.

I tried to explore two ways to do so with limited success.

1. Use statics, rotational equilibrium gives a third equation and allows to express α in terms of θA and θB. α=arctan((1−d/L)tan(θB)−d/Ltan(θA)). Which I can simply substitue in my system. I don't really know how to solve this system precisely by hand. I had a little bit of success with this method, using a solver in excel, changing the variables θA and θB aiming to minimize the least squares for the two equations of my system. However Excel solver cannot solve the system if my lifting points are too close to each other (approaching the length of the bar). I am not really satisfied by this, if anybody could guide me to solve this system that would be much appreciated!
2. There would be another option which is to express that the potential energy of the system will be minimized at equilibrium. So I would need to find the minimum for mAgyGA+mgyG+mBgyGB, so as I consider that I can neglect the masses of the slings, I would only need to find the angle α that makes the potential energy of the bar the lowest, i.e. lowest vertical position. Easier said than done! I believe that the best method for this kind of question is Newton-Raphson, that I would like to apply on yG(α)=yA−lAcos(θA)+dsin(α). I still have a term in cos(θA) therefore I still have to plug my system in the Newton-Raphson method, I don't really know how.
I'm desperate, confused, distressed, please somebody help!

#### Attachments

• Bar Angle - Fig 1.pdf
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Delta2
Paging @haruspex, @berkeman

I probably won't be much of a help(that's why I paged more knowledgeable members than me on this kind of problems) but I think this system doesn't have a unique equilibrium position.

That is because we have 1 equation from the sum of torques=0
another 2 equations from the sum of forces in x and y axes=0

But there are 4 unknowns, The two tensions and the two angles they make with the vertical (or horizontal).

EDIT: Ah ok now I noticed your two equations that are some sort of constraints of the system. Hmmmm, I think I better not comment further on this I got confused too... It turns out 5 equations and 5 unknowns (The magnitudes of the two tensions plus the three angles) it might have a unique solution after all.

EDIT 2: I think this problem solution consists of a system of 5 equations (1 from torque equilibrium, 2 from force equilibrium and 2 the constraints which you write in the OP) with 5 unknowns (two tensions + 3 angles). And no I do not believe you can decouple the equations so you first solve for the angles and then for the tensions. It is obvious at least to my intuition, that the angles depend on the magnitude of tensions and vice versa.

Last edited:
Delta2 said:
But there are 4 unknowns, The two tensions and the two angles they make with the vertical (or horizontal).
We know the vertical components of the tensions sum to the mass of the bar.
We know that the horizontal components of the tensions are equal and opposite.
This is a 3 bar linkage. Knowing one angle will give you the other.

Baluncore said:
We know the vertical components of the tensions sum to the mass of the bar.
We know that the horizontal components of the tensions are equal and opposite.
This is a 3 bar linkage. Knowing one angle will give you the other.
That's fine but after an overall consideration, I think the mistake the OP does is that he wants to decouple the 5 equations and solve first for the angles, and then for the tensions, I do not think this is correct approach.

Hi @Pp122. This might help.

There are three forces acting on the bar. Since the bar is in equilibrium, the lines of action (LoAs) of these forces must be concurrent, i.e. pass through a single point. (Otherwise there would be a net torque on the bar.)

The bar’s weight acts vertically downwards through it centre of gravity. Therefore the point of intersection of the LoAs of the two tensions must be vertically below the bar’s centre of gravity.

Edit - minor only.

Lnewqban and Delta2
I would go with option 2. Energy methods are often significantly simpler to apply if relevant to the question you want to answer. If the result has a complicated functional form with energy methods this will not change by using equilibrium equations.

Delta2
@Orodruin he seems to have come up with three equations regarding three unknowns, the three angles, I don't think the angles can be decoupled from the magnitudes of tensions can they?

Delta2 said:
@Orodruin he seems to have come up with three equations regarding three unknowns, the three angles, I don't think the angles can be decoupled from the magnitudes of tensions can they?
My point is that you only need to solve one equation for one unknown if you use energy methods. Doing the force and torque equilibrium equations is a lot of extra work to find the potential minimum.

Lnewqban and Delta2

## 1. How do I calculate the tilt angle of a bar lifted via two ropes on fixed points?

The tilt angle of a bar lifted via two ropes on fixed points can be calculated using the formula: tan θ = (2Fsinα) / (mg + 2Fcosα), where θ is the tilt angle, F is the force applied by each rope, α is the angle between the ropes and the horizontal, and m is the mass of the bar.

## 2. What is the significance of calculating the tilt angle of a bar lifted via two ropes on fixed points?

Calculating the tilt angle of a bar lifted via two ropes on fixed points is important in understanding the stability and equilibrium of the system. It can also help in determining the appropriate forces and angles needed to keep the bar in a balanced position.

## 3. Can I use this formula for any type of bar or only for a specific shape?

This formula can be used for any type of bar as long as the forces and angles are measured accurately. However, the mass of the bar must be evenly distributed and the ropes must be attached at fixed points.

## 4. What are the units of measurement for the variables in the formula?

The units of measurement for the variables in the formula are as follows: θ (degrees), F (newtons), α (degrees), m (kilograms), and g (meters per second squared).

## 5. Are there any limitations to using this formula?

One limitation of using this formula is that it assumes a perfectly rigid bar and ropes with no stretching or bending. It also does not take into account any external forces acting on the bar, such as wind or friction.

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