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

If the net moments are not 0 at a point, it means that there would be some component of rotation around that point. For this bar to be stationary, that cannot be the case.

The reason to choose the midpoint as the place to sum moments is just to make it easier. That's because you don't have to consider the forces acting on the point you take the moment, and point O has both the weight of the bar and a spring acting on it, so you can get rid of a term (or two) in your moment equation by taking the moment around point O.

• Hi JMac. If the net moments are not 0 at a point, couldn't the bar could still be stationary (not dynamically rotating) but just not level? – Bob D Oct 8 at 20:46
• @BobD how so? Net moments directly cause rotation, like net forces directly cause acceleration. – JMac Oct 8 at 20:56
• So excluding gravity , there would be no net rotation too? or is this something we would have to assume then, would conclude to take moments? – User1265 Oct 8 at 21:47
• What I'm saying is this is not strictly a statics problem where all members behave as rigid bodies. The springs are not rigid bodies. When they deform the geometry changes. Deformations don't come into the problem in a statics problem. But the springs here will extend or contract, changing the orientation of the beam (how horizontal it is) based on the moments. It's hard to explain without using drawings. If I'm not making any sense to you let me know and I will delete my comments. – Bob D Oct 8 at 21:52
• @BobD It does say the beam "rests" on this. That heavily implies that it's not moving. You can still treat it like a classic statics problem. Given the stiffness of the springs and a rigid bar, there's only one position where the bar will want to remain in that same position (i.e. "rest"). Every other orientation of the bar would have unbalanced moments or forces and cause movement of the bar, which would be oscillations, not rest. – JMac Oct 8 at 22:00