# Determine the compression of the springs, their compressional forces in the case: k_1=k_3=k, k_2=2k

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:
${k}_{1}={k}_{3}=k$ and ${k}_{2}=2k$
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Winston Cooper
Because the stiffness ${k}_{1}={k}_{3}=k$ of the springs off-center are equal, and also their distances to the center, the system is balanced and the beam keeps horizontal. The displacement $\mathrm{\Delta }y$ is the same for all.
If ${F}_{e}$ is the force applied to each of the springs off-center, and ${F}_{m}$ to the spring at the center: ${F}_{e}=k\mathrm{\Delta }y$ and ${F}_{m}=2k\mathrm{\Delta }y$
As we know that $2{F}_{e}+{F}_{m}=Mg$, we have 3 equations and 3 unknown (${F}_{e},{F}_{m},\mathrm{\Delta }y\right)$
$Mg-2{F}_{e}=2k\mathrm{\Delta }y$
$Mg-2{F}_{e}=2k\frac{{F}_{e}}{k}\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}{F}_{e}=\frac{1}{4}Mg$
Resulting that ${F}_{m}=\frac{1}{2}Mg$ and $\mathrm{\Delta }y=\frac{1}{4k}Mg$