- #1

gnits

- 137

- 46

- Homework Statement
- To find the equilibrium position of a mass attached to an elastic string

- Relevant Equations
- F=ma

Hi, can anyone see if I have made an error in answering this simple question, my answer is not the one given in the textbook.

Here's my diagram of the system in equilibrium. So the mass has dropped a distance x below the midpoint of PQ:

So the forces balance as:

T1 = g + T2

So, using Hooke's law T = Yx/a (where 'Y' is the modulus of elasticity of the string and 'a' the natural length of the string) we have:

( 4g( (1/2) + x) ) / (1/2) = g + ( 4g( (1/2) - x) ) / (1/2)

Because both parts of the string above and below the mass have a natural length of 1/2 and the part above is of length 1 + x and so has an extention of (1/2) + x and that below the mass is of length 1 - x and so has anextention of (1/2) - x

This solves to give x = 1/16 and so the height above Q would be 15/16.

The book answer if 3/4 = 12 / 16.

Thanks for any help,

Mitch.

Here's my diagram of the system in equilibrium. So the mass has dropped a distance x below the midpoint of PQ:

So the forces balance as:

T1 = g + T2

So, using Hooke's law T = Yx/a (where 'Y' is the modulus of elasticity of the string and 'a' the natural length of the string) we have:

( 4g( (1/2) + x) ) / (1/2) = g + ( 4g( (1/2) - x) ) / (1/2)

Because both parts of the string above and below the mass have a natural length of 1/2 and the part above is of length 1 + x and so has an extention of (1/2) + x and that below the mass is of length 1 - x and so has anextention of (1/2) - x

This solves to give x = 1/16 and so the height above Q would be 15/16.

The book answer if 3/4 = 12 / 16.

Thanks for any help,

Mitch.