Help Needed: Solving Work Done on String Question

In summary, The question asks for the work done when the midpoint of a string, with natural length 2a and modulus of elasticity b, is pulled from the line AB to create an equilateral triangle ABC. The midpoint represents half the natural length, and when stretched to the equilateral triangle, the extension is 2a. The work done is found by subtracting the work required to pull the string to AC1 from the work required to pull it to AC2, using the formula Work = (b(x)^2)/2a. However, this does not account for the tension from BC, so the correct answer is found by using the formula 2(b(2a)^2)/2a, which equals 4ab
  • #1
GregA
210
0
Hi folks...can someone tell me where I'm going wrong with this question, or back me up if I'm right?

A string of natural length 2a and modulus of elasticity b has its ends attached to fixed points A and B where AB = 3a. Find the work done when the midpoint C of the string is pulled away from the line AB to a position where triangle ABC is equilateral. My attempt to solve is as follows:

I ignore B and focus just on A and C
if the natural length of the string = 2a then the midpoint represents half this length and therefore = a
AB = 3a, and so AC = 3a/2... the extention x = a/2
AC when streched to the equilateral triangle = 3a and so...x = 2a
The question asks me how much work is done from pulling the midpoint from its initial position on the line AB to its final position

Surely I want to subtract the work taken to pull the string to from its natural length to AC1 from the work required to pull the string to AC2..using Work = (b(x)^2)/2a...
(b((2a)^2-(a/2)^2)/2a = 15ab/8.
The books answer is given as 4ab however...I've tried to find a problem with my working and cannot. :frown:
 
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  • #2
I think I have it...There are two tensions opposing the movement of that string not just AC but BC :redface:
If the question means how much work is done to extend from its natural length to its final position it is simply 2(b(2a)^2)/2a...=4ab
 
  • #3


It appears that your calculations are correct. It is possible that the book's answer is incorrect or that there is a mistake in the question. It would be helpful if you could provide more context or information about the source of the question and the book's answer. Also, double-check your calculations and make sure you are using the correct units for the modulus of elasticity (b). If you are still unsure, it may be beneficial to consult with a colleague or professor for a second opinion. Ultimately, it is important to trust your own calculations and reasoning, but it is always good to have a second set of eyes to catch any potential errors.
 

1. What is work done on a string?

Work done on a string refers to the amount of energy transferred to the string by an external force. It is equal to the force applied to the string multiplied by the distance over which the force is applied.

2. How is work done on a string calculated?

The equation for calculating work done on a string is W = Fd, where W is work, F is the force applied, and d is the distance over which the force is applied. This assumes that the force and distance are in the same direction.

3. What factors affect the work done on a string?

The work done on a string is affected by the magnitude of the force applied and the distance over which the force is applied. Additionally, the angle between the force and the direction of motion of the string can also affect the work done.

4. Can negative work be done on a string?

Yes, negative work can be done on a string if the force applied is in the opposite direction of the movement of the string. This means that the string is losing energy, rather than gaining it.

5. How does work done on a string relate to the tension in the string?

The tension in a string is directly proportional to the work done on the string. This means that as the tension increases, the work done on the string also increases. Alternatively, if the tension decreases, less work is done on the string.

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