Serially joined & vertically hanged springs with a weight in between

[fuzzyorama]

Homework Statement

Three springs, A (spring constant k), B (spring constant 2k) and C (spring constant 4k) are serially joined and are vertically hanged from a ceiling with spring A at the top, spring B in the middle & spring C at the bottom. All three have the same relaxed length of L. The bottom end of spring C is then tied(?) to a table which is tall enough that all three springs are in their relaxed length (the distance between the ceiling & the table's surface is 3L).

If a weight of mass M is put between spring A and spring B, what is

1. the displacement, x of springs B and C,
2. the force, F applied on the spots where springs A and C are tied to (ceiling & table surface)?

Gravitational force is g, no specific numbers. All springs are massless.

Homework Equations

mg = -kx
F = mg

The Attempt at a Solution

Displacement of spring B
mg = 2kx'
x' = mg/2k

Displacement of spring C
mg = -4kx"
2kx' = -4kx"
x" = -x'/2

Wow, I honestly don't know where I'm going with this.

 

[Hootenanny]

Welcome to Physics Forums.

You're on the right lines with your attempt, but you can't consider each spring individually. Since the two lower springs (B&C) are connected you must consider them together, notice how the question only asks for a single displacement (x) for both springs.

So, the first step is to find the effective spring constant for the combined springs B & C.

 

[fuzzyorama]

Thank you. Oh?! OK so

k_BC = (2k*4k)/(2k+4k) = 4k/3

mg = -k_BCx
mg = -(4k/3)*x
and

x = -(3mg/4k)

Is this correct? I kinda have a pretty good feeling about this:) If it is, I'm guessing the force on the table's surface is

F = -k_BCx = -(4k/3)*(-3mg/4k) = mg ?

 

[Hootenanny]

Again, you have the right idea, but it's not quite correct. Take a look http://scienceworld.wolfram.com/physics/SpringsTwoSpringsinSeries.html" for information on how to find the effective spring constant for two springs in series.

 

[fuzzyorama]

Err, I'm sorry.. could you be more specific?

If the effective spring constant for springs B and C is

k_eff = ($$\frac{1}{k_{B}}$$ + $$\frac{1}{k_{C}}$$)$$^{-1}$$

and k_B = 2k, k_C = 4k, then

k_eff = ($$\frac{1}{2k}$$ + $$\frac{1}{4k}$$)$$^{-1}$$ = ($$\frac{2k+4k}{2k*4k}$$)$$^{-1}$$ = $$\frac{8k^{2}}{6k}$$

so eventually it will be k_eff = $$\frac{4k}{3}$$, right?

 

[Hootenanny]

Err, I'm sorry.. could you be more specific?

If the effective spring constant for springs B and C is

k_eff = ($$\frac{1}{k_{B}}$$ + $$\frac{1}{k_{C}}$$)$$^{-1}$$

and k_B = 2k, k_C = 4k, then

k_eff = ($$\frac{1}{2k}$$ + $$\frac{1}{4k}$$)$$^{-1}$$ = ($$\frac{2k+4k}{2k*4k}$$)$$^{-1}$$ = $$\frac{8k^{2}}{6k}$$

so eventually it will be k_eff = $$\frac{4k}{3}$$, right?

I'm terribly sorry, I misread your post - I though that you have 3k/4. Apologies.

mg = -k_BCx

Does spring A not exert a force on the mass ...?

 

[fuzzyorama]

I'm terribly sorry, I misread your post - I though that you have 3k/4. Apologies.

Does spring A not exert a force on the mass ...?

Heh no problem yo
With spring A taken into account, I can write it as Mg + k_Ax_A = -k_BCx
... or can't I?

Because k_A = k, and k_BC = 4k/3,
Mg + kx = -4kx/3
(skipped)

x = -$$\frac{3kMg}{7}$$

I'm getting confused with the + - signs

 

[Hootenanny]

Heh no problem yo
With spring A taken into account, I can write it as Mg + k_Ax_A = -k_BCx
... or can't I?

Because k_A = k, and k_BC = 4k/3,
Mg + kx = -4kx/3
(skipped)

x = -$$\frac{3kMg}{7}$$

You're almost there, but there's a little mistake going from the previous line to the final line.

I'm getting confused with the + - signs

You're signs are fine. If you're unsure you can do a quick sanity check. If you look at you're original expression:

Heh no problem yo
With spring A taken into account, I can write it as Mg + k_Ax_A = -k_BCx

And consider a positive displacement, then the two forces from the springs will be acting in the same direction as the weight of the mass, as it should be. This also shows that your answer should be negative, since if the displacement were positive, all the three forces would be acting in the same direction and no equilibrium state would be possible.

 

[fuzzyorama]

Oh no I failed at the maths again. How about this?

$$x = -\frac{3Mg}{7k}$$

I can latex!

 

[Hootenanny]

Oh no I failed at the maths again. How about this?

$$x = -\frac{3Mg}{7k}$$

I can latex!

Looks good to me