Parallel Springs subject to load

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Homework Help Overview

The problem involves a system of two parallel springs with spring constants k1 and k2, attached to a massless rigid bar. The objective is to determine the equivalent spring constant (ke) that relates the applied force to the resulting displacement for small oscillations.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss using energy methods to relate the potential energy of the two springs to that of the equivalent spring. There are attempts to express angular displacements and translational displacement of the bar, but challenges remain in deriving the total spring compression.

Discussion Status

Some participants have suggested that additional visual aids, such as diagrams, could enhance understanding. There is a focus on establishing equilibrium conditions, including the sum of forces and torques, but no consensus has been reached on how to solve for the equivalent spring constant.

Contextual Notes

Participants are navigating through relationships between forces and torques, as well as geometrical relationships among the displacements of the springs. There is an emphasis on the need for a clearer representation of the system to facilitate discussion.

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Homework Statement


A system of two springs is considered, with spring constants k1 and k2, arranged in parallel and attached to a massless rigid bar. The bar remains horizontal when no force on it is zero. Determine the equivalent spring constant (ke) that relates the force applied a distance l1 from one spring and l2 from the other (F is between the springs) to the resulting displacement for small oscillations


Homework Equations


F1=k1*x1
F2=k2*x2

F=ke*x
sin(theta)=theta
U=1/2*k*x^2

T=F*R

The Attempt at a Solution



It was decided after much group work to try and use energy. We ended up setting the potential energy of the two springs to the potential energy of the equivalent spring. Using the small angle approximation we could write the angular displacements as the displacement of the springs due to the rotation, but weren't able to derive an expression for the translational displacement of the bar, and thus not able to get an expression for the total spring compression.
 
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A picture would help.
 
Spinnor said:
A picture would help.

13z32wx.jpg
 
The force F causes the two springs to compress until equilibrium is reached. At that point the sum of the forces in the vertical direction sum to zero. Also the sum of the torques about the point the force F is applied is zero.
 
Spinnor said:
The force F causes the two springs to compress until equilibrium is reached. At that point the sum of the forces in the vertical direction sum to zero. Also the sum of the torques about the point the force F is applied is zero.

That may be true but it is not enough.

You get three relationships, sum forces in vertical = 0, sum of torques about the point of applied force F = 0, and a geometrical relationship between x1, x2, and x,3. See below. Solve for ke.
 

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