Parallel Springs subject to load

In summary, the system has three relationships: the sum of forces in the vertical direction is 0, the sum of torques about the point of applied force is 0, and a geometrical relationship between x1, x2, and x3. Solving for ke, we get an equivalent spring constant of 1.5.
  • #1
pige
3
0

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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  • #2
A picture would help.
 
  • #3
Spinnor said:
A picture would help.

13z32wx.jpg
 
  • #4
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.
 
  • #5
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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1. What are parallel springs subject to load?

Parallel springs subject to load refer to a mechanical system in which two or more springs are connected in parallel to support a load. This means that the springs are arranged side by side and each one carries a portion of the load.

2. How do parallel springs subject to load behave?

When a load is applied to a system of parallel springs, each spring will compress or stretch in proportion to the amount of load it is carrying. This means that the total force exerted by the system of springs is equal to the sum of the individual forces exerted by each spring.

3. What is the equation for calculating the spring constant of parallel springs subject to load?

The equation for calculating the spring constant of parallel springs subject to load is k = k1 + k2 + ... + kn, where k is the total spring constant and k1, k2, ... , kn are the individual spring constants.

4. How does the number of springs affect the behavior of parallel springs subject to load?

The number of springs in a parallel system affects the overall stiffness of the system. The more springs that are added, the stiffer the system becomes. This means that a larger number of springs can support a heavier load without excessive deformation.

5. What are the advantages of using parallel springs subject to load?

The use of parallel springs subject to load offers several advantages, including increased load capacity, improved stability, and reduced stress on individual springs. Additionally, the use of multiple springs allows for more precise control over the amount of force exerted on the load.

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