Tension in Each Rope: Calculating Force and Work

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Homework Help Overview

The problem involves a muscle exerciser with two steel ropes and a spring. The spring's force and work done during compression are calculated based on its length change. The scenario includes determining the tension in the ropes and the forces acting at specific points in the setup.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the calculation of force and work done on the spring, noting that the force is not constant and questioning the potential energy stored in a compressed spring.
  • There is exploration of the relationship between the pulling force and the tension in the ropes, with some participants suggesting that the tension should be halved due to the two ropes.
  • Questions arise regarding the correct application of angles and components of forces in the context of the problem.

Discussion Status

Participants are actively engaging with the problem, providing insights and corrections to each other's calculations. There is recognition of the need to recalculate certain values, particularly the spring constant and potential energy. Multiple interpretations of the forces involved are being explored, with no explicit consensus reached yet.

Contextual Notes

Some participants note that the compression of the spring is only 0.2 m, which affects the calculations. There is also mention of the horizontal forces acting at specific points, indicating a focus on the geometry of the problem.

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Homework Statement


A muscle exerciser consists of two steel ropes attached to the end of a strong spring contained in a telescopic tube. When the ropes are pulled sideways in opposite directions, the spring is compressed.

The spring has an uncompressed length of 0.80 m. The force F (in N) required to compress the spring to a length x (in m) is calculated from the equation F = 500 (0.80 - x).

The ropes are pulled with equal and opposite forces, P, so that the spring is compressed to a length of 0.60 m and the ropes make an angle of 30º with the length of the spring.

0017ab74c2a6.jpg


(a) Calculate: (i) the force, F, (ii) the work done in compressing the spring.
(b) By considering the forces at A or B, calculate the tension in each rope.
(c) By considering the forces at C or D, calculate the force, P.

Answers: (a) (i) 100 N, (ii) 10 J, (b) 57.7 N, (c) 57.7 N

2. The attempt at a solution
(a) (i) F = 500 * (0.8 - 0.6) = 100 N
(a) (ii) W = Fs = 100 * (0.8 - 0.6) = 20 J, which is wrong.

I made a graph but I am not sure whether I noted the force correctly.
9f9d5acc2f0d.jpg


(b) I am looking for AC and I have AO (where O is the middle point) = 100 N and the angle is equal to 30 degrees. So AC = 100 / cos 30 = 115.5 N
(c) Since PCA and PCA are same steel ropes, therefore they have the same forces which are equal to 115.5 N.

What do I miss here?
 
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moenste said:
(a) (i) F = 500 * (0.8 - 0.6) = 100 N
Good!

moenste said:
(a) (ii) W = Fs = 100 * (0.8 - 0.6) = 20 J, which is wrong.
Careful. The force is not constant. Instead: What's the spring potential energy stored in a compressed spring?

moenste said:
(b) I am looking for AC and I have AO (where O is the middle point) = 100 N and the angle is equal to 30 degrees. So AC = 100 / cos 30 = 115.5 N
Almost. Realize that two ropes pull down, so each one accounts for half of that total.

moenste said:
(c) Since PCA and PCA are same steel ropes, therefore they have the same forces which are equal to 115.5 N.
The pulling force does not equal the tension in the ropes. What component of the tension force equals the force P? (There are two ropes there as well.)
 
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Doc Al said:
Careful. The force is not constant. Instead: What's the spring potential energy stored in a compressed spring?
PE = 1/2 kx2? k = 100 N / 0.8 m = 125 N/m so PE = 0.5 * 125 * 0.22 = 2.5 J?

Doc Al said:
Almost. Realize that two ropes pull down, so each one accounts for half of that total.
So AC = 100 / cos 30 = 115.5 N. But it is the combined force of the two ropes, and the tension in one rope is 115.5 N / 2 = 57.75 N.

Doc Al said:
The pulling force does not equal the tension in the ropes. What component of the tension force equals the force P? (There are two ropes there as well.)
Maybe CO and OD? CO = AC * cos 60 = 28.9 N. And since we also have CB therefore P = 28.9*2 (due to AC and CB) = 57.75 N. Is this the right logic?
 
moenste said:
PE = 1/2 kx2? k = 100 N / 0.8 m = 125 N/m so PE = 0.5 * 125 * 0.22 = 2.5 J?
Right idea, but recalculate that value of k. The compression is only 0.2 m.

moenste said:
So AC = 100 / cos 30 = 115.5 N. But it is the combined force of the two ropes, and the tension in one rope is 115.5 N / 2 = 57.75 N.
Good!

moenste said:
Maybe CO and OD?
The force P is horizontal. Consider the horizontal forces acting at point C.
 
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Doc Al said:
Right idea, but recalculate that value of k. The compression is only 0.2 m.
k is equal to the tension required to produce unit of extension. So m in N / m is equal to the compressed part of the spring and not to it's normal length (as I wrongly calculated). As a result: PE = 1/2 * (100 / 0.2) * 0.22 = 10 J.

Doc Al said:
The force P is horizontal. Consider the horizontal forces acting at point C.
Maybe CO and OD? CO = AC * cos 60 = 28.9 N. And since we also have CB therefore P = 28.9*2 (due to AC and CB) = 57.75 N. Is this the right logic?
 
moenste said:
k is equal to the tension required to produce unit of extension. So m in N / m is equal to the compressed part of the spring and not to it's normal length (as I wrongly calculated). As a result: PE = 1/2 * (100 / 0.2) * 0.22 = 10 J.
Good!

moenste said:
Maybe CO and OD? CO = AC * cos 60 = 28.9 N. And since we also have CB therefore P = 28.9*2 (due to AC and CB) = 57.75 N. Is this the right logic?
Good!
 
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Likes   Reactions: moenste

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