Struggling to make relation between elastic force and height

[renobueno4153]

Homework Statement

A 25 kg object is being lifted by two people pulling on the ends of a 1.15 mm diameter
nylon rope (UTS = 500 x 106 N/m2) that goes over two 3 m high poles placed at a distance
of 4 m from each other, as shown in Figure 1.
How high above the floor will the object be when the rope breaks?

Relevant Equations

hooks law?

Hello guys this is what I tried so far. I used the UTS to calculate the force it needs when the rope tears.
My idea was to make a relationship/ function that would give me the force depending on height. Yeah i couldnt find a way to solve it.

I also thought about how I could use hooks law (how it was given to me in my script) with the thought of instead of having two part of a rope id have one singular rope from the middle to the top where I could find the difference in height.

But the problem here is that I wasnt given the youngs modulus to make idea work.

Since we did some similar exercise (pulling a rope and looking at the forces acting on it) with trigonometry I thought Id try that. But i dont know how and where to apply the forces I have found on the picture. Also in lecture we never looked at tension forces. I was just desperate and found a video explaining how tension forces would act on a rope. So maybe the pic is flawed.

Im super confused about the question too since F doesnt depend on height in comparision to for example potential energy. So im struggeling to imagine how, when and where the rope tears. Is there context which I miss in my understanding?

 

[Steve4Physics]

1.15mm is a bit thin to call a rope! A ‘cord’ might be a better term.

Nylon is stretchy, but this won’t affect the answer to the question. You don’t need the Young modulus and you don’t need to use Hooke’s (note the spelling!) law. If you know how to answer the question correctly (see below), this should become clear.

As the rope gets stretched it diameter will reduce. So we need to assume that the rope’s diameter at the moment it would break, is 1.15mm. Assuming this is true, I agree with your value for tension.

Do you know how to resolve a force into components? This is needed to answer the question.

The object’s weight (downwards) is balanced by vertical components (acting upwards) of the 2 tensions. This allows you to write an equation and then find $\theta$.

When you’ve found $\theta$ you should be able to find the required height using simple trig’.

 

[Herman Trivilino]

I used the UTS to calculate the force it needs when the rope tears.

Can you show us that calculation? Or at least the value of the force?

 

[renobueno4153]

its in the first picture. all the way to the right. :)

 

[Herman Trivilino]

Start with a free-body diagram of the point where the 25 kg object hangs from the nylon rope.

 

[kuruman]

As @Herman Trivilino said

Start with a free-body diagram of the point where the 25 kg object hangs from the nylon rope.

And use it to find an expression for the tension in the string as a function of height $y$ from the ground. Note that the higher the weight rises, the higher the tension becomes. (Why?)

 

[PeroK]

As @Herman Trivilino said

And use it to find an expression for the tension in the string as a function of height $y$ from the ground. Note that the higher the weight rises, the higher the tension becomes. (Why?)

It seems more logical to me to calculate the tension as a function of the distance from the top. Then the height is 3m minus this distance.

 

[kuruman]

It seems more logical to me to calculate the tension as a function of the distance from the top. Then the height is 3m minus this distance.

Perhaps. The question asks for the height above ground. The distance from the top can enter the calculation in the form $d=(3~\text{m}-y)$ as a reminder of what needs to be calculated.

 

[Lnewqban]

Im super confused about the question too since F doesnt depend on height in comparision to for example potential energy. So im struggeling to imagine how, when and where the rope tears. Is there context which I miss in my understanding?

The pulling force of each person and the tension force along the whole nylon string must be the same for each instant.

In this specific type of situation, the magnitude of those forces increases as the object is lifted (and not in a linear manner).

Please, see:
https://demonstrations.wolfram.com/TensionOfARopeWithAHangingMass/

The reason is that for the object to remain in equilibrium (not moving vertically for each analyzed instant), a vertical force of 25g must counteract its weight of 25g.

Because of the above, the vertical component of each string tension must be 12.5g, value which must remain the same for any angle adopted by each side of the string.

As the object is being lifted by two people pulling on the ends, each side of the string located between both poles tends to be horizontal.

In order to keep the vertical component of the tension in each side of the string constant for any angle, the horizontal component of the tension must change.

Therefore, progressively increasing the pulling force of each person and the tension force along the whole nylon string as the object is lifted, is the only way the explained above can happen.

 

[Herman Trivilino]

It seems more logical to me to calculate the tension as a function of the distance from the top. Then the height is 3m minus this distance.

That's the only way to do it! After all, the 3 m height of the poles has no effect on the distance from the top.