# Very Basic Q on Hooke's Law Problem Pt.2

"A uniform beam AOB, O being the mid point of AB, mass M, rests on three identical

vertical springs with stiffness constants k1, k2 and k3 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:

(ii) k1 = k, k2 = 2k and k3 = 3k "

My Solution (part):

Let

F3 = Restoring force of spring 3

F2 = Restoring force of spring 2

F1 = Restoring force of spring 1


Spring 3 compresses by α Then Spring 2 compresses α+B Then spring 1 compresses by α+2B in order for the straight beam to be inclined, which we visualise when three different springs with spring constants have a rod placed on top of them.

Resolving Vertically: F3 + F2 + F1 = 3k α + 2k ( α+B) + k(α+2B) = k (6α + 4B) = Mg

'Taking moments' Then this is the part I'm stuck on, I do understand the solutions below, yet my Question posed in last graph.

F3 * L/2 cos θ = F1 * L/2 cos θ

F1 = F3

Then from here you can find expression of α in terms of B, or of B in terms of α

Use this, to rewrite the expression k (6α + 4B) = Mg --> to α= Mg/10k

so F3 = 3K*α = 3Mg/10 = F1 Hence F2 = 4Mg/10

My Question: What I don't understand how the written solutions have come to the conclusions to take moments? Is it because there is no net turning force, even without gravity? I also don't understand why the moment of F3 and moment of F1 set to be equal about midpoint O.

• What was the resolution of "Part 1"? Also, you really need to show a diagram. You have not explicitly stated that A and B are at equal distances from O. If in fact that is the case. – Bob D Oct 8 at 20:50