Multiple-spring system problem....

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The discussion revolves around a three-spring system design where the calculations are disputed by the company president, who holds a fundamentally incorrect assumption. The user has provided a diagram outlining the problem and the differing assumptions leading to conflicting results. It is established that in static equilibrium, the forces on the moveable members must balance, leading to the conclusion that the total tension in the extension spring is double the force on one side. Clarification is sought on the correct interpretation of the forces involved in the system. The resolution of this debate is crucial for the product's design accuracy.
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The product I'm working on designing simplifies down to a relatively simple three-spring system, which is easily calculated. Unfortunately our company president is weighing in on the design, and doesn't agree with the calculation, and insists the result is something different (based on a fundamentally wrong assumption). I've outlined the problem with a diagram in the attached PDF. At the bottom there are the two different assumptions which were used to come up with the two different results, any assistance into which theory is correct would be greatly appreciated.

Thanks.
 

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Mike Howard said:
The product I'm working on designing simplifies down to a relatively simple three-spring system, which is easily calculated. Unfortunately our company president is weighing in on the design, and doesn't agree with the calculation, and insists the result is something different (based on a fundamentally wrong assumption). I've outlined the problem with a diagram in the attached PDF. At the bottom there are the two different assumptions which were used to come up with the two different results, any assistance into which theory is correct would be greatly appreciated.

Thanks.
If the system is in static equilibrium, then the forces on the left hand moveable member must be equal and opposite, F1=F2. And similarly for the right hand moveable member, F3=F2. The forces on the fixed member are also equal and opposite, because F1=F3.
 
tech99 said:
If the system is in static equilibrium, then the forces on the left hand moveable member must be equal and opposite, F1=F2. And similarly for the right hand moveable member, F3=F2. The forces on the fixed member are also equal and opposite, because F1=F3.

Thanks. So based on that the total tension in the extension spring (which is the big debate) is 2x F1, as its being acted on and extended at both ends, correct?
 
Mike Howard said:
Thanks. So based on that the total tension in the extension spring (which is the big debate) is 2x F1, as its being acted on and extended at both ends, correct?
The extension spring applies force F2 to each of the moveable members. But the tension in the spring is just F2. Imagine a weight hanging up by a rope. If the tension in the rope is 1N, it applies 1N to the weight and 1N to the support.
 
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