Tension in Each Rope: Calculating Force and Work

  • Thread starter Thread starter moenste
  • Start date Start date
  • Tags Tags
    Rope Tension
Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
5 replies · 3K views
moenste
Messages
711
Reaction score
12

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?
 
Physics news on Phys.org
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.)
 
  • Like
Likes   Reactions: 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.
 
  • Like
Likes   Reactions: 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!
 
  • Like
Likes   Reactions: moenste