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.