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Calculating tension in a rope

  • Thread starter madsam
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  • #1
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Hey, im having difficulty with this question, thanks guys.

Homework Statement


Two chain slings, one at each end, are used to support a rectangular box girder of mass 8 tonne. The length of each sling chain is 5m. The rectangular cross sectional shape of the box girder has dimensions of 1.5m by 0.5m

In part a of the equation it is lying width-side down (as to be longer horizontally) and each sling goes completely around the girder and reconnects with itself, to form a shape similar to this
/\
|_|

In part b however, it has been rolled over so the girder is taller (and less stable) than previously. Similar to:

/\
| |
|_|


Homework Equations





The Attempt at a Solution


Im unsure how to calculate the tension, but i did find this (However it probably has no relevance to the answer at all);

the top section of the rope = 2.5m
hence each part is 1.25m long

yeah and um, thats it. quite pathetic really =\
is there a formula for calculating tension?

And if there is a formula for finding compression as well (in say, a brick lodged between two walls), could you say that too?

Thanks, Sam
 
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Answers and Replies

  • #2
PhanthomJay
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Sam, welcome to PF. Your initial approach is correct, the top section of the rope will be 2.5 meters long (1.25m long for each leg of the triangle, but only 1.5m long (0.75m each leg) when the girder is rotated. You have to then calculate, for each case, the rope tensions under the 4 tonne load (the 8 tonnes divides equally to each sling arrangemnt), by determining how the load gets distributed to each leg of the sling using Newton's first law and trig to determine the components of the load in the x and y directions. Note that in the latter case, the angle between the ropes and girder will be much flatter, hence, the tensions will be greater. Is this a homework question, and are you familiar with the method to be used? As for your 2nd question, I don't quite understand what you are asking.
 
  • #3
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Hey, thanks for the answer, i do engineering studies in year 11 and we were given a sheet with this question on it, earlier in the year, except our teacher asked us to find something else/used it as an example. So im not too sure whether or not it will be in our exams (he sets the exam) and i figured its best to be safe and find out.

I'm semi-familiar with the method, however i dont understand how you detect the actual tension within the rope. Is it the same method which you use to calculate normal vectors?

using Newton's first law and trig to determine the components of the load in the x and y directions.
What i dont understand is well, all of that. Our teacher is sort of vague about everything, and well, we must discover everything for ourselves pretty much. I know if you put the top section into an x y graph, then you can find the lengths to be 1.25, .75, and 1, with angles of 53, 90, and 37 degrees respectively. But well, then what? (we'll just work with the first part of the question because that seems easiest) How do you determine how much tension and/or load is contained within each rope?

I read somewhere on the internet tension = mass (gravity - accelleration) but i'm not entirely sure how reliable that is, and whether that only works on vertical 'ropes' as such.

Just ignore the second question, its not really relevant or necessary to anything i need to know, now at least.
 
  • #4
tiny-tim
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Welcome to PF!

I'm semi-familiar with the method, however i dont understand how you detect the actual tension within the rope. Is it the same method which you use to calculate normal vectors?
Hi madsam! Welcome to PF! :smile:

Are you familiar with force diagrams?

They work the same way as vector diagrams.

You draw all the forces, which are the weight and the two tensions.

Then, as PhanthomJay says, you use trig to calculate the tension. :smile:
I read somewhere on the internet tension = mass (gravity - accelleration) but i'm not entirely sure how reliable that is, and whether that only works on vertical 'ropes' as such.
That (it's a vector equation of course) is only for when the object is moving, in which case you have to apply Newton's second law to deal with the acceleration. :smile:
 

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