# Elastic Spring

1. Jan 26, 2005

### kingyof2thejring

Hi There,
Iam having trouble with this question:

A block B of mass 5 kg is fastened to one end of each of two springs. The other ends of the springs are attached to fixed points A and C, 4 metres apart on a smooth horizontal surface, as shown in the diagram <the link contains the file>.

http://s3.yousendit.com/d.aspx?id=2ZAI1AKPYV05L1MUR0PJY6RRMP

Spring AB has natural length 2 metres and modulus of elasticity 30 N, while BC has natural length 1 metre and modulus 40 N (you may assume that the springs meet at the centre of B).

(i)Find the length of AB when B is in its natural equilibrium position.

now if extension = length of the spring - natural length
Iam i required to find out the length of the spring?
If i am then using hooke's law T=kx, 5g=30x/2 i get x= 3.26 as the extension.

I would appreciate some help

2. Jan 26, 2005

### quasar987

Not an easy one. There are a few things to think about. Here's how I would proceed.

1° The total system, while stretched, measures 4 m, but the relaxed lenght of AB is 2m, and the relaxed lenght of BC is 1m.. SOoo if we call $\Delta l_{AB}$ and $\Delta l_{BC}$ the lenght by which each spring is streched, we get the equation

$$4 = 2 + 1 + \Delta l_{AB} - \Delta l_{BC}$$

(Why $- \Delta l_{BC}$ ? Because the $\Delta l$s represent stretching in two oposite directions, so $\Delta l_{BC}$ itself will be negative. Thefor, if we want the sum to add up to 4, we must substract $\Delta l_{BC}$, not add it.)

Or, more simply,

$$\Delta l_{BC} = \Delta l_{AB} - 1 \ \ (*)$$

2° In the case we're interested in, the mass B is in equilibrium position. This is fancy words to say that $\Sigma F = 0$, so we also have the equation

$$0 = -k_{AB}x_{AB} - k_{BC}x_{BC}$$

Or, since at equilibrium position, these x represent what we have called $\Delta l_{AB}$ and $\Delta l_{BC}$, we get, making use of relation (*):

$$0= -30\Delta l_{AB} - 40(\Delta l_{AB} - 1)$$

Allowing you to find what you were looking for.

Side note: We could also have supposed $\Delta l_{AB}$ and $\Delta l_{BC}$ to be both positive, but it would have made a mess because $\Delta l_{AB}$ and $\Delta l_{BC}$ still represent stretching in two oposite directions: towards the right for AB, and towards the left for BC. So, this means that we would be considering a stretching of the spring AB to the right as positive, left as negative. But also considering a stretching of spring BC to the left as positive, right as negative! The consequence of this subtelty would be apparent in the equation of motion $$0 = -k_{AB}x_{AB} + k_{BC}x_{BC}$$ where we would have taken opposite signs for the force of the different springs. Again, that is because our setting $\Delta l_{AB}$ and $\Delta l_{BC}$ as positive implied that displacement to the right of the mass means a positive $x_{AB}$ but a negative $x_{BC}$. But the forces by both springs are not opposite; they are in the same direction. Hence we must adjust the sign of k.

3. Jan 26, 2005

### kingyof2thejring

cheers thx for sharing your ideas
as a sidenote should the -30 be -15 if k= lambda/l

4. Jan 26, 2005

### quasar987

what are lambda and l?

5. Jan 26, 2005

### kingyof2thejring

l is the natural length
lambda is modulus of elasticity

6. Jan 26, 2005

### quasar987

I don't really know what the modulus of elasticty is. I thought it was the same as the spring constant k.

If you know what the relation between the modulus of elasticity and the spring constant is, then there shoudln't be a problem.

Out of curiosity, what is the relation between the modulus of elasticity and the spring constant ?

Thanks

7. Jan 26, 2005

### vincentchan

spring constant depends on the length of the spring, if the 2 springs made by the same material, the longer one has a smaller spring constant

the elasticity modulus is the property of a material, it doesn't not depend on its physical size...

8. Jan 26, 2005

### quasar987

How are they related mathematically in temrs of k, lambda and relaxed lenght l ?

9. Jan 26, 2005

k=lambda/l