# Homework Help: Hookes Law

1. Oct 24, 2012

### Moninder

1. The problem statement, all variables and given/known data

Three identical 6.00kg masses are hung by three identical springs, as shown in the figure. Each spring has a force constant of 8.50kN/m and was 11.0cm long before any masses were attached to it. How long is each spring when hanging as shown? (Hint: First isolate only the bottom mass. Then treat the bottom two masses as a system. Finally, treat all three masses as a system.)

2. Relevant equations
F=-ma
F=-kx

3. The attempt at a solution
I started with the bottom block.
-ma=-kx
(6.00kgx9.8m/s^2)/-(8.50kN/m)=x
x=6.92m? and then i added 0.11m
and its still wrong
I also took the F=ma and divided by 1000 to get into(0.0588) kN and then divided it by 8.50kN and then dded .11m and still wrong

I need to find all three block eventually

Last edited: Oct 25, 2012
2. Oct 25, 2012

### AJ Bentley

That can't be answered without the diagram.

3. Oct 25, 2012

### Moninder

This is how it is from top to bottom
Ceiling
-
spring
-
block
-
spring
-
block
-
spring
-
block

4. Oct 25, 2012

### Staff: Mentor

What was your answer from doing the calculation this way, and how do you know it was wrong?

5. Oct 25, 2012

### Moninder

For the bottom spring I got 0.1169m
Middle Spring 0.3376m
Top spring 0.35075m

I put it in an online program

6. Oct 27, 2012

### Moninder

Still need help

7. Oct 27, 2012

### Staff: Mentor

Note that the spring constant is in kN/m, not N/m.

8. Oct 27, 2012

### Moninder

Yes I know, check my 3 answers i posted, they are with converting the kN

9. Oct 27, 2012

### Staff: Mentor

Looks OK.
Show how you got these last two.

10. Oct 27, 2012

### Moninder

Mg=kx
((6+6)9.8)/(8500N/m) + .22 m =middle spring
Top spring
((6+6+6)9.8)/(8500N/m) + 0.33m

11. Oct 27, 2012

### Staff: Mentor

12. Oct 27, 2012

### Moninder

Yes, i did add in the unstrecthed length?

13. Oct 27, 2012

### Staff: Mentor

For the middle spring you added in twice the unstretched length; for the top spring you added three times the unstretched length.

Each spring has a total length equal to its unstretched length (0.11m for each) plus the amount of stretch (which varies).

14. Oct 27, 2012

### Moninder

Why would i not add in 2 lengths because the middle spring is also carrying the bottom spring

15. Oct 27, 2012

### Staff: Mentor

How does that affect its unstretched length?

16. Oct 27, 2012

### Moninder

Oh now i get it, i should just only add in 1 unstrecthed length because im only calculating for 1 spring

17. Oct 27, 2012

### Staff: Mentor

Right.

18. Oct 27, 2012

Thank you