A question states:
A uniform beam AOB, O being the mid point of AB, mass M$M$, rests on three identical vertical springs with stiffness constants k1$k_1$, k2$k_2$ and k3 $k_3$ at A, O and B respectively. The bases of the springs are fixed to a horizontal horizontal platform.
Determine the compression of the springs and their compressional forces forces in the case:
(i) k1 = k3 = k$k_1 = k_3 = k$ and k2 = 2k$k_2 = 2k$
My attempt at a solution:
Recognised SpringsI 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)
ifIf considered as one single effective hookeanHookean spring then,
k eff = k1 + k2 + k3 = k + 2k + k = 4k$$k_\text{eff} = k_1 + k_2 + k_3 = k + 2k + k = 4k$$
then effective restoring force, with same extension x$x$, is equal to the weight
-4kx = Mg then x = - Mg/4k = answer
$$-4kx = Mg$$ 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?
