Solving Hooke's Law Problems: Compression & Forces

In summary: Is there any reason why the displacement of the spring is relative to the horizontal platform and not to the mid point of the beam?In summary, the problem involves a uniform beam with a mass M resting on three identical vertical springs with stiffness constants k1, k2, and k3 at different points on the beam. The first scenario assumes that the compression of the springs is the same due to the symmetrical nature of the system. In the second scenario, the compressions form an arithmetic sequence, and the balance of moments equation is used to determine the relationship between the compressions and the forces on the springs. The displacement of the spring is relative to the horizontal platform, not the midpoint of the beam.
  • #1
Sarah0001
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1

Homework Statement


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:

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

2.
The use of Hookes law is only of need here and the rods weight for the first part. The second part additionally requires the use moments which I am confused on as well as the relationship between the relationship between the spring constant and their length of compressions relative to one another.

The Attempt at a Solution


The assumption for the first scenario is that the compression (x) is the same for k2 as it is for k1 and k3. The markscheme makes indication this this is due to the fact that the system of rod and the three springs are symmetrical, hence compression for the middle spring must be the same as the other two.
I can understand this and which I have done.
for part (ii) I am confused the relationship between the length of the compression between the three springs:

Assume spring K3 length compressed by X
K2 " " X + B
K3 " " X+2B

Aswell as the use of moments on this question seen in the uploaded image (solution)

[/B]
 

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  • #2
It is not entirely clear which parts are your work, which parts are as explained to you, and exactly what your question is.
Are you asking for an explanation of the assumption that the compressions form an arithmetic sequence?

There is a slight awkwardness in the question. If we assume that the springs were vertical before compression then the tilting of the beam will mean they are no longer vertical. The geometry gets rather messy. The solution you attached assumes the bases of the springs move closer together so that they can remain vertical.

What is it that you do not understand about the balance of moments equation? It seems straightforward.
 
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  • #3
sorry I did forget to summarise what I was confused on. Its yes (a) why of the assumption that the compressions form an arithmetic sequence?

(b) The part of the equation I'm stuck on is why resolve using cosine theta , and why does it lead the conclusion 2α = 2β.
 
  • #4
Sarah0001 said:
why of the assumption that the compressions form an arithmetic sequence?
Because the beam is straight and the attachments are equally spaced. If the beam is length 2L and slooes at angle θ then the centre of the beam must be L sin(θ) below one end and L sin(θ) above the other.
Sarah0001 said:
why resolve using cosine theta , and why does it lead the conclusion 2α = 2β.
The magnitude of the torque from a force about an axis is the magnitude of the force multipled by the perpendicular distance to the axis (that is, perpendicular to the line of action of the force).
The forces from the springs are vertical, so the perpendicular distances are the horizontal displacements, L cos(θ).

Taking moments about the centre, the two end springs are both at distance L cos(θ), so that cancels, leaving us to conclude that the two forces are equal: k(α+2β)=3k(α).
 
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  • #5
Thank you!
 

Related to Solving Hooke's Law Problems: Compression & Forces

What is Hooke's Law?

Hooke's Law is a principle in physics that states the force needed to compress or stretch an elastic object is directly proportional to the distance it is stretched or compressed.

What is the equation for Hooke's Law?

The equation for Hooke's Law is F = -kx, where F is the force applied, k is the spring constant, and x is the distance the object is stretched or compressed.

How do you solve for the spring constant (k) in Hooke's Law?

To solve for the spring constant, you can rearrange the equation to k = -F/x. This means that the spring constant is equal to the force applied divided by the distance the object is stretched or compressed.

How do you solve for the force (F) in Hooke's Law?

To solve for the force, you can rearrange the equation to F = -kx. This means that the force applied is equal to the spring constant multiplied by the distance the object is stretched or compressed.

What units are used in Hooke's Law?

The units used in Hooke's Law are newtons (N) for force, meters (m) for distance, and newtons per meter (N/m) for spring constant.

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