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Object Lifted By Cord (Stress/Tension Problem)

  • Thread starter kritzy
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  • #1
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Homework Statement


A 25kg object is being lifted by pulling on the ends of a 1.10mm diameter nylon cord that goes over two 3.50m high poles that are 4.8 m apart. How high above the floor will the object be when the cord breaks?

Homework Equations


F=EA
F=ma

The Attempt at a Solution


I thought that by using this equation, I could find the tension in the cord.
F=(500 x 10^6)(pi)(.00055)^2=475

X-dir force
Tcos θ + Tcos θ = 2Tcos θ=0

Y-dir force
Tsin θ + Tsin θ -mg=0
2Tsin θ=mg

I think my equations are incorrect. I didn't know how to account for the height and weight of the cord. Help would be much appreciated.
 

Answers and Replies

  • #2
tiny-tim
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A 25kg object is being lifted by pulling on the ends of a 1.10mm diameter nylon cord that goes over two 3.50m high poles that are 4.8 m apart. How high above the floor will the object be when the cord breaks?

I didn't know how to account for the height and weight of the cord.
Hi kritzy! :smile:

(what does the weight have to do with it? :confused:)

Use the triangle whose length is 3.5 - h and whose width is 2.4 to find cosθ and sinθ :wink:
 
  • #3
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Hi kritzy! :smile:

(what does the weight have to do with it? :confused:)

Use the triangle whose length is 3.5 - h and whose width is 2.4 to find cosθ and sinθ :wink:
I tried to solve for sinθ and cosθ but I end up with to many unknowns.
T=tension
cosθ=2.4/T
sinθ=(3.5-h)/T
I tried to substitute for T. I got
tanθ=(3.5-h)/2.4
I still have two unknowns and only one equation.
Can you give me another hint?
I attached a file which shows the diagram. Maybe it will help
 

Attachments

  • #4
tiny-tim
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I tried to substitute for T.
I'm confused … isn't T the breaking strength, which I assume is given in the question? :confused:

and I'm off to bed now :zzz: …
 

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