# A question about an assumption on a Very Basic Hooke's Law problem

A question states:

A uniform beam AOB, O being the mid point of AB, mass $$M$$, rests on three identical vertical springs with stiffness constants $$k_1$$, $$k_2$$ and $$k_3$$ at A, O and B respectively. The bases of the springs are fixed to a horizontal platform.

Determine the compression of the springs and their compressional forces in the case:

(i) $$k_1 = k_3 = k$$ and $$k_2 = 2k$$

My attempt at a solution:
I recognised springs were in parallel, and if springs are parallel in a system with a force applied to them, then they all have the same extension,

(correct me with this statement if not entirely accurate)

If considered as one single effective Hookean spring then,

$$k_\text{eff} = k_1 + k_2 + k_3 = k + 2k + k = 4k$$

then effective restoring force, with same extension $$x$$, is equal to the weight

$$-4kx = Mg$$ then $$x = - Mg/4k$$ is the answer

My Question ( last paragraph) :

I only made the assumption they all had the same extension for the sole reason recognised it was parallel, so a force Mg was being applied to all parallel three springs.

The answer states they have the same extension because 'by symmetry beam is horizontal'

Why does the fact the "beam is horizontal by symmetry" explain the fact they all have the same extension?

• The problem wording is a little unclear to me. Is the weight M located at the mid point O? – Bob D Oct 8 '19 at 12:00
• Also, is the spring with the different k from the others located under the mid point? – Bob D Oct 8 '19 at 12:14

## 2 Answers

"The beam is horizontal by symmetry" is an assumption you can make to allow the extension to be the same for each.

Basically, looking at the springs under the board, we can see that the two less stiff springs are on the outside, and that the stiff spring is over the centre of mass.

If we assume the beam cannot bend, then either all the springs compress the same amount, or A B and C all compress different amounts. This would lead to the bar being on a slope.

Now, for this bar to be not moving (in rotation), the sum of moments also has to equal $$0$$. If $$k_1 = k_3$$, then you can see that only when $$\Delta x_1 = \Delta x_3$$ do the moments sum up to $$0$$.

When $$k_1 = k_2$$ and the bar is not horizontal, you will see a non-zero sum of moments in the bar. This would mean the bar is moving, not static. Basically, if you apply that the sum of forces on the bar is $$0$$ and that the sum of moments is $$0$$ (the two conditions required for this to be static), you will see that when the setup is symmetric, the displacement on each spring will be equal.

If $$k_1 \neq k_2$$ in this case, then the bar would not want to be horizontal when in static equilibrium. I suggest playing around with spring constants and distances from the centre for yourself to see how the math agrees with this. You will see that when the springs are the same stiffness and equal distance from the centre of mass (i.e. symmetry), it leads to a horizontal static equilibrium.

Edit: This assumes A O and B are all equally spaced. If that's not given in the question, then I'm not sure why they said it's symmetrical. If A and B aren't the same distance from O, this isn't symmetrical.

Why does the fact the "beam is horizontal by symmetry" explain the fact they all have the same extension?

I'm assuming M is the mass of the beam so that it can be represented as a concentrated load midpoint. The problem doesn't give the locations of the springs, but I think the spring with k different than the other two springs would have to be midpoint and the other two springs equally spaced from the midpoint in order for the moment of each about the midpoint to be equal, thus keeping the beam horizontal. Without these stipulations, I don't think you can assume the beam will remain horizontal and that there will be equal deformations of all the springs.

It should also be noted that this problem statically indeterminate, i.e., not solvable with only the requirements for static equilibrium (sum of moments and forces equal zero). Equations related to the deformations of the springs are also required.

The diagram below is for a problem I once used in a MOM class. The columns are analogous to the springs and the moduli of elasticity analogous to the spring constant.

Hope this helps.