# 2 hanging masses with springs, find spring constant

1. May 21, 2012

### Hyperfluxe

1. The problem statement, all variables and given/known data
http://i.imgur.com/0gexJ.png

2. Relevant equations
F=kx (Hooke's Law), ƩFy=0

3. The attempt at a solution
I drew a free-body diagram for each block. For block A, I get the equilibrium equation: Fa = (10kg)(9.81ms^-2) = 98.1N = kx_a
x_a = 300mm-250mm = 50mm
k = 1.962kN/m

For block B I get the equilibrium equation of Fb = 98.1 - Wa = kx_b but it seems as I don't need this equation.

Is this correct, and if not, should I draw a "combined" FBD? If so, how? Thanks.

2. May 21, 2012

### Staff: Mentor

Your equation is not obtained from analyzing block A alone, but by treating A and B as a combined object. So you answered your own question, even if you didn't realize it.

If you analyzed A alone, you'd have to include the force from each spring and only the weight of A.

3. May 21, 2012

### Hyperfluxe

Oh I see. From drawing seperate FBD's, I get the equations:
kxa = kxb + mag (block A)
kxb = mbg (block B)
Substitute kxb back into the first equation to get kxa = (ma + mb)g, where xa = 300mm - 250mm, thus k = 1.962kN/m.

Is this correct? Thank you.

4. May 21, 2012

### Staff: Mentor

Perfectly correct. Note that your final equation is equivalent to what you did earlier.

5. May 21, 2012

### Hyperfluxe

Thank you for you time and help =)