# Object Lifted By Cord (Stress/Tension Problem)

• kritzy
In summary, 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. The question asks for the height of the object above the floor when the cord breaks. The equations used are F=EA and F=ma, and the attempt at a solution involves using the triangle whose length is 3.5 - h and whose width is 2.4 to find cosθ and sinθ. However, the weight of the cord is not taken into account and there are still two unknowns in one equation.
kritzy

## 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?

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.

kritzy said:
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!

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

Use the triangle whose length is 3.5 - h and whose width is 2.4 to find cosθ and sinθ

tiny-tim said:
Hi kritzy!

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

Use the triangle whose length is 3.5 - h and whose width is 2.4 to find cosθ and sinθ

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

• Physics Problem.doc
53 KB · Views: 349
kritzy said:
I tried to substitute for T.

I'm confused … isn't T the breaking strength, which I assume is given in the question?

and I'm off to bed now :zzz: …

## 1. How does the weight of the object affect the stress on the cord?

The weight of the object directly affects the amount of stress on the cord. The heavier the object, the more force is pulling on the cord, causing more stress.

## 2. What is the relationship between the length of the cord and the stress on the cord?

The longer the cord, the more it will stretch and the more stress it will experience. This is because the weight of the object is distributed over a greater distance, causing more tension on the cord.

## 3. Does the material of the cord affect the amount of stress it can withstand?

Yes, the material of the cord plays a significant role in determining how much stress it can handle. Some materials, like steel, are stronger and more durable than others, like nylon, and can withstand higher levels of stress.

## 4. How can I calculate the stress on the cord?

The stress on the cord can be calculated using the formula stress = force/area. This means that the amount of stress is directly proportional to the force being applied and inversely proportional to the cross-sectional area of the cord.

## 5. What safety precautions should be taken when lifting an object with a cord?

When lifting an object with a cord, it is important to make sure that the cord is strong enough to handle the weight of the object. It is also important to inspect the cord regularly for any signs of wear or damage. Additionally, proper lifting techniques should be used to distribute the weight evenly and minimize stress on the cord.

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