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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.

(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?