Tension in 4 strings suspending object at 4 different points

In summary: If the cables are installed without careful adjustment then one of the four is always likely to be slack. You can use this fact to advantage by assuming that each of the cables in turn is slack and solving for tensions in the other three. Find worst case.In summary, The grid suspended by its four corners with strings is held in equilibrium by applying tension to the strings. The grid is represented by four uniform poles in the x-y-z direction. The mass of the grid, gravity, and length of the strings used to suspend the grid are unknown. The attempt at a solution assumes that the grid is held fairly taut, but the grid shown in the original installation was not as tight. If the cables are not installed
  • #1
Chandan Pednekar
12
0

Homework Statement



A grid made of ropes is suspended by its four corners with strings tied to four different points on a building. Three of which are fixed and fourth is attached to a pulley which is in turn fixed to the building.

Given data :
Grid mass is 25 kilogram suspended at a height of 6m
Gravity is 9.8 m/s^2
Length of strings used to suspend grid are 5m, 5m, 6m and 9m(tied to pulley).
X and Y components of strings are given.
https://imgur.com/a/7MHwl (Diagram)
upload_2017-12-14_11-58-58.png

Calculate tension in each string.

Homework Equations



I know how to find tension in two strings which suspend an object at a common point. How can I calculate
tension which object is suspended at different points?

The Attempt at a Solution


I am not from physics background and am a software developer. I understood basic tension calculations but couldn't work out for four strings suspending object at different points.

P.S. -> I am new to physicsforums. I read the rules and checked existing posts but couldn't find a post I could understand and borrow from for my given problem.

This is for an art installation. I need to calculate tension in strings as I have to order motors with appropriate stall torque for pulley winding/ unwinding. If I have created this post in the wrong section please let me know and I will move it to the appropriate section.
 

Attachments

  • upload_2017-12-14_11-58-58.png
    upload_2017-12-14_11-58-58.png
    101.6 KB · Views: 1,082
Last edited by a moderator:
Physics news on Phys.org
  • #2
Your question is not clear,
You can establish equilibrium condition in x-y-z directions,
 
  • #3
There is not enough information. You need the x,y coordinates of the suspension points relative to the grid.
You will also need to make some assumption about the grid behaviour. I assume it is held fairly taut, so can be treated as a uniform flat plate.
 
  • #4
haruspex said:
There is not enough information. You need the x,y coordinates of the suspension points relative to the grid.
You will also need to make some assumption about the grid behaviour. I assume it is held fairly taut, so can be treated as a uniform flat plate.

I think its not going to be as taut to be treated as a uniform flat plate unfortunately. This is how the original installation looked like. Though it won't be as loose as shown in image. https://static.wixstatic.com/media/.../a18de6_aa9c4fbd418b47d0a71a81dbf69e4bc1.webp

x, y co-ordinates of suspension points relative to grid? What do I consider the origin in this?
 
  • #5
I love physics said:
Your question is not clear,

What data is missing that would allow me to solve this problem?

I love physics said:
You can establish equilibrium condition in x-y-z directions

I would still require forces acting on the strings in x-y-z direction for that, right? That is what I am trying to figure.
 
  • #6
Chandan Pednekar said:
I think its not going to be as taut to be treated as a uniform flat plate unfortunately.
Ok, but representing it as four uniform poles might not be too far off. That way it can flex. And I think you can ignore the horizontal displacements that would occur in the flexing.
Chandan Pednekar said:
x, y co-ordinates of suspension points relative to grid?
Pick one corner of the grid as origin and set the other corners at (L,0,z1), (L,L,z2), (0,L,z3).
 
  • #7
haruspex said:
Pick one corner of the grid as origin and set the other corners at (L,0,z1), (L,L,z2), (0,L,z3).

Hi. I have upgraded the diagram according to your guidance.
Link to diagram : https://imgur.com/a/LflI1

All points are measured with respect to origin at point A.(bottom left corner of grid)
Mass of grid is 25kg.
Gravity is 9.8m/s^2.

I tried to calculate Tension using matrix and solving equations simultaneously as following.
Link to calculations : https://imgur.com/a/yz4cQ
But couldn't solve it for any string as there are always at least two non zero entries in each row.

Are my co-ordinate positions correct relative to the grid? Am I solving it the right way?
I solved similar problems of particle equilibrium and was able to solve them. Though, in all those examples there was a common point at which the load was suspended. In my case load is suspended at different 4 corners. Correct me wherever wrong.
 
  • #8
eVYQWtx.png

There are many different solutions to this problem . How the loads distribute between the cables depends on how they are adjusted for length during installation .

If the cables are installed without careful adjustment then one of the four is always likely to be slack . You can use this fact to advantage by assuming that each of the cables in turn is slack and solving for tensions in the other three . Find worst case .
 

Attachments

  • eVYQWtx.png
    eVYQWtx.png
    9.2 KB · Views: 1,202
Last edited:
  • #9
Nidum said:
View attachment 217488
There are many different solutions to this problem.

Could you please point me to similar or same problems?

Nidum said:
View attachment 217488
If the cables are installed without careful adjustment then one of the four is always likely to be slack . You can use this fact to advantage by assuming that each of the cables in turn is slack and solving for tensions in the other three .

Three cables are going to be fixed to structure. Remaining one will be wound to motorized pulley where pulley is fixed to fourth point on structure.
Wouldn't that mean those 3 fixed point cables are always going to be under tension? Am I right?
If yes, Only the fourth cable attached to motorized pulley would be having slack. Problem is, most of the times it would be taut enough to contain tension.
Also, I have been told that the pulley's location might change to any point during installation. That is why I wanted to be sure about how do I calculate tensions since I might have to calculate it over installation site.
 
  • #10
Nidum said:
If the cables are installed without careful adjustment then one of the four is always likely to be slack
If treating the square as a rigid plate (and ignoring the weights of the cables), yes. But in post #6 I recommended treating the square as four poles freely jointed, though maybe supposing the plan view remains squarish.
 
  • #11
haruspex said:
If treating the square as a rigid plate (and ignoring the weights of the cables), yes. But in post #6 I recommended treating the square as four poles freely jointed, though maybe supposing the plan view remains squarish.

Hi. For now let's assume that square is rigid. I want to know, what I did in calculations, is it right? Or are there any other ways to do this?
 
  • #12
haruspex said:
If treating the square as a rigid plate (and ignoring the weights of the cables), yes. But in post #6 I recommended treating the square as four poles freely jointed, though maybe supposing the plan view remains squarish.

Update : We built a scale model yesterday of 1:10 ratio to check grids behavior. It was observed that the grid will be taut enough to be considered as a uniform plate. Also, the the cables suspending the grid from the structure will be taut enough to have very little slack in them.

Any other ways or correction in the methods I am doing wrong would be appreciated.

Thanks.
 
  • #13
Chandan Pednekar said:
the grid will be taut enough to be considered as a uniform plate. Also, the the cables suspending the grid from the structure will be taut enough to have very little slack in them.
As @Nidum notes, that is a bit contradictory. In general, three cables would be enough to hold a rigid plate steady, so one of the four should be relatively slack. That might be obscured by the weight of the cables themselves providing some tautness.
This is not just a technical matter. It means that equations based on four light taut strings attached to a rigid square plate will be overspecified and lead to a contradiction. Something has to give.
 
  • #14
Chandan Pednekar said:
We built a scale model yesterday of 1:10 ratio to check grids behavior
That is close to impossible: mass and stiffness will scale with different factors
 
  • #15
haruspex said:
As @Nidum notes, that is a bit contradictory. In general, three cables would be enough to hold a rigid plate steady, so one of the four should be relatively slack. That might be obscured by the weight of the cables themselves providing some tautness.
This is not just a technical matter. It means that equations based on four light taut strings attached to a rigid square plate will be overspecified and lead to a contradiction. Something has to give.

Yes you are right that three taut strings should hold a rigid plate. So i suppose it cannot be considered a rigid plate, since the grid is made of ropes and won't act like a rigid plate unless pulled taut from all four corners. I'll have to make some more considerations.
 
  • #16
BvU said:
That is close to impossible: mass and stiffness will scale with different factors
Thanks. I'll take this into account for further calculations.
 
  • #17
The original statement of the problem described an all rope construction . That effectively means a very flexible construction . Obtaining answers for tensions and deflections in all the rope segments is theoretically not difficult - especially if a few simplifying assumptions are made - but in practice the large number of simultaneous equations that even simple constructions like this generate means that computer methods would have to be used .

For a simplistic analysis that is possible by hand calculation you could assume that the cradle is flexible enough to compensate for any problems with imperfectly adjusted support cables but not so flexible that it changes shape by any large amount . This is similar in principle to @haruspex's four rods idea .

With this assumption the problem becomes one of simple statics . You could solve the problem in full form with each suspension rope's different length being taken into account but since you are making approximations anyway I think it would be reasonable to make the further approximation that all four suspension ropes are identical(1) . Doing this means that calculation of tension in the suspension ropes becomes a trivial exercise . Whatever answer you get apply a reasonable safety factor and you have the information needed to choose a suitable hoist system and motor .

Note (1) : You could probably get slightly better results by assuming that only the three fixed suspension ropes are identical and that the hoisting rope is different . Personally I don't think that it is worth adding this extra complexity .
 
Last edited:
  • #18
Nidum said:
For a simplistic analysis that is possible by hand calculation you could assume that the cradle is flexible enough to compensate for any problems with imperfectly adjusted support cables but not so flexible that it changes shape by any large amount . This is similar in principle to @haruspex's four rods idea .

With this assumption the problem becomes one of simple statics . You could solve the problem in full form with each suspension rope's different length being taken into account but since you are making approximations anyway I think it would be reasonable to make the further approximation that all four suspension ropes are identical(1) . Doing this means that calculation of tension in the suspension ropes becomes a trivial exercise . Whatever answer you get apply a reasonable safety factor and you have the information needed to choose a suitable hoist system and motor .

Note (1) : You could probably get slightly better results by assuming that only the three fixed suspension ropes are identical and that the hoisting rope is different . Personally I don't think that it is worth adding this extra complexity .

Hi Nidium. Thanks for simplifying what others have stated. I have made approximations in suspension cable lengths and calculated for both cases.
1. Actual (where the dimensions of cable are roughly similar to the actual dimensions)
2. Approx (where I have considered all the cables to be in symmetry and of equal length.)

Here is the link to diagrams.
https://imgur.com/a/QF16i

Here is the link to calculations.
https://imgur.com/a/LLQbN

Are they correct? For actual part worst case torque is around 1.8Nm and for approx part its around 1.2Nm.
The motor I have been looking at provides 4.9Nm. So Factor of Safety for actual is around 2.5 and for approx is 4.0.
Please correct me wherever I am wrong.

Thanks.
 
  • #19
Chandan Pednekar said:
The motor I have been looking at provides 4.9Nm. So Factor of Safety for actual is around 2.5 and for approx is 4.0.

Guys. I was wrong about the motor safety factor. Seems stall torque is not a good measurement for safety factor as actual torque while motor is rotating is much lower.

How can I calculate torque for a 60rpm motor, 1:270 gear ratio, given 75pwm value through arduino while rotating?

Also, I am unsure about my cable tension calculations in my previous post. Could someone please validate?
 
Last edited:
  • #20
Please ignore calculation link of post #18 as those are incorrect. I tried to solve the simultaneous equations without first reducing vectors to unit vectors.

Here is the new link. https://imgur.com/a/OF35b
I have tried to solve equations by considering all suspension cables to be identical. Yet, I am unable to reduce any row in the matrix to only a single non-zero value. Any other way to do this?

I have also calculated angles cable AB makes with x,y and z axes. But what do I consider weight to be lifted if I am calculating using angles?
Is it correct to calculate tension as given in this page. https://www.wikihow.com/Calculate-Tension-in-Physics (Method 2, Type 3, Use multiple strands to support a hanging object)?
each cable makes an angle 45.67, 45.67 and 81.08 degrees with the x, y, and z-axis respectively. (since cables are identical)
For cable AB
tension along x-axis = cos(45.67) * m * g = 171.203 N
tension along y-axis = cos(45.67) * m * g = 171.203 N
tension along z-axis = cos(81.08) * m * g = 37.988 N
Is this correct? I am lost.
 

Related to Tension in 4 strings suspending object at 4 different points

1. What is tension in 4 strings suspending an object?

Tension is a force that is created when an object is pulled or stretched. In the case of 4 strings suspending an object at 4 different points, the tension is the force that is pulling the object upwards.

2. How is tension calculated in this scenario?

Tension can be calculated using the equation T = F/L, where T is the tension, F is the force applied, and L is the length of the string. In this scenario, the tension in each string can be calculated separately and then added together to find the total tension.

3. What factors can affect the tension in the strings?

The tension in the strings can be affected by several factors, including the weight of the object, the length and thickness of the strings, and the distance between the points of suspension. The angle at which the strings are attached to the object can also affect the tension.

4. How does the tension in each string affect the stability of the suspended object?

The tension in each string plays a crucial role in the stability of the suspended object. If the tension in one string is significantly different from the others, it can cause the object to tilt or even fall. Therefore, it is essential to ensure that the tension is evenly distributed among all the strings for the object to remain stable.

5. Can the tension in the strings be adjusted?

Yes, the tension in the strings can be adjusted by changing the length of the strings or by adding or removing weights from the object. However, it is crucial to maintain equal tension in all the strings to keep the object stable.

Similar threads

  • Introductory Physics Homework Help
Replies
5
Views
923
  • Introductory Physics Homework Help
Replies
18
Views
3K
  • Introductory Physics Homework Help
Replies
3
Views
2K
  • Introductory Physics Homework Help
Replies
17
Views
3K
  • Introductory Physics Homework Help
Replies
13
Views
1K
  • Introductory Physics Homework Help
Replies
18
Views
178
  • Introductory Physics Homework Help
Replies
10
Views
2K
  • Introductory Physics Homework Help
Replies
26
Views
6K
  • Introductory Physics Homework Help
Replies
2
Views
3K
  • Introductory Physics Homework Help
Replies
19
Views
895
Back
Top