# Tension in Each Rope: Calculating Force and Work

• moenste
In summary: Just remember that the force P is equal to both CO and OD, so it's actually P = 28.9 * 2 = 57.8 N. In summary, a muscle exerciser contains two steel ropes attached to a strong spring in a telescopic tube. The spring has an uncompressed length of 0.80 m and can be compressed by pulling the ropes in opposite directions. The force required to compress the spring can be calculated using the equation F = 500 (0.80 - x). In a specific scenario where the ropes are pulled with equal and opposite forces, the spring is compressed to a length of 0.60 m and the ropes make a 30º angle with the spring. Using this information,
moenste

## 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.

(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.

(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?

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.)

moenste
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.

moenste
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!

moenste

## 1. What is tension?

Tension is a pulling force that is applied to an object by a rope, cable, or other similar object. It is a force that acts in a direction opposite to the direction of the pull.

## 2. How is tension calculated in each rope?

Tension can be calculated by dividing the force applied to the rope by the cross-sectional area of the rope. This can be represented by the formula T = F/A, where T is tension, F is the force, and A is the cross-sectional area of the rope.

## 3. What factors affect the tension in each rope?

The tension in each rope is affected by the magnitude of the force applied, the angle of the rope relative to the object it is attached to, and the properties of the rope itself, such as its length, thickness, and material.

## 4. Can tension in each rope be negative?

No, tension cannot be negative. It is always a positive value, as it represents the force applied to the rope. However, the direction of the tension can be negative if it is acting in the opposite direction of the pull.

## 5. How is work related to tension in each rope?

Work is the product of force and displacement, and tension is a type of force. Therefore, the tension in each rope can be used to calculate the work done by the rope. This can be represented by the formula W = Fd, where W is work, F is the force, and d is the displacement.

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