Solving Nylon Rope Fracture Problem Up to 25 kg Lift Height

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SUMMARY

The discussion centers on calculating the height at which a 25 kg object will be lifted before a 1.00 mm diameter nylon string breaks, given its tensile strength of 1570N. Participants clarify that the forces acting on the string (Fa and Fb) must equal the tensile strength for equilibrium, rather than exceeding it. The correct approach involves using the Pythagorean theorem and understanding the relationship between the angle of the string and the forces acting on it. The final consensus emphasizes the need to calculate the forces accurately without assuming they exceed the tensile strength.

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Homework Statement


A 25 kg object is being lifted by pulling on the ends of a 1.00 mm diameter nylon string that goes over two 3.10 m high poles that are 4.5 m apart, as shown in Figure 9-87. How high above the floor will the object be when the string breaks?
*see attached picture*

The foreces on the rope (called Fa and Fb) are equal.

The Attempt at a Solution


I know that the forces to fracture the rope are in the directions of the rope. In this case, Fa and Fb, and they have to be greater then the tensile strength of the nylon, which is 1570N.

So, the Forces in the y direction are the Fg of the box (25kg x 9.8m/s^2)=245N, and the combined forces of the the string (Fay and Fby) = 245N. If the forces Fa and Fb are equal, then the forces Fay and Fby are equal, each at 122.5N, correct?

I don't really know how to set-up the problem.
If Fa+Fb>1570N, then I see Fa and Fb=786N (added up to be greater than 1570N).
Now, using the Pythagorean Thm, I know the hypotenuse and a leg of the right triangles made by the forces on the rope. I find the angle to be 8.96 degrees.

I then look at the right triangle made by the lengths of the poles. If the bottom leg is 1/2 the total length, 4.5m, then a leg on on triangle is 2.25m. To find the heght at 8.96 degrees, I take the tan(8.96) x 2.25m = 0.35m. But I know that is not the right answer.

What am I doing wrong? (Sorry for having a lot to read, but this is how I did the problem)

Thanks.
 

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Where did you get the tensile strength from? It should be a stress, not a force.

Good question though: I wonder if they expect you to take into account the strain - that will make the cord stretch before it breaks...
 
ssjdbz101 said:

Homework Statement


A 25 kg object is being lifted by pulling on the ends of a 1.00 mm diameter nylon string that goes over two 3.10 m high poles that are 4.5 m apart, as shown in Figure 9-87. How high above the floor will the object be when the string breaks?
*see attached picture*

The foreces on the rope (called Fa and Fb) are equal.

The Attempt at a Solution


I know that the forces to fracture the rope are in the directions of the rope. In this case, Fa and Fb, and they have to be greater then the tensile strength of the nylon, which is 1570N.

So, the Forces in the y direction are the Fg of the box (25kg x 9.8m/s^2)=245N, and the combined forces of the the string (Fay and Fby) = 245N. If the forces Fa and Fb are equal, then the forces Fay and Fby are equal, each at 122.5N, correct?

I don't really know how to set-up the problem.
If Fa+Fb>1570N, then I see Fa and Fb=786N (added up to be greater than 1570N).
Now, using the Pythagorean Thm, I know the hypotenuse and a leg of the right triangles made by the forces on the rope. I find the angle to be 8.96 degrees.

I then look at the right triangle made by the lengths of the poles. If the bottom leg is 1/2 the total length, 4.5m, then a leg on on triangle is 2.25m. To find the heght at 8.96 degrees, I take the tan(8.96) x 2.25m = 0.35m. But I know that is not the right answer.

What am I doing wrong? (Sorry for having a lot to read, but this is how I did the problem)

Thanks.
Your error is in the your statement that Fa + Fb >1570N. It should be Fa = Fb = 1570N. Supposing I pull horizontally on one end of a horizontal nylon string with a force of 1570N, and a person on the other end also pulls on it with a force of 1570N. The string is in equilibrium. The tension in the string is 1570N, not 3140N, and, having a fracture strength of 1571N, it will not break. Do you see why?
 
Work with just one of the sides: Fa

You have shown that Fay is always equal to 122.5 N

How can you calculate Fa (for any angle) from this information? (find a formula that relates Fa and theta)

How does the angle change with height of the package? (find a formula that relates angle with height of package)

Assume that they are not taking into account strain and set Fa to the ultimate (breaking) force. (calculated from the ultimate stress and the area of the cord)
 
Thanks, I see what I was doing wrong. You guys were a lot of help!
Thanks again.
 

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