Understanding Hooke's Law: Comparing Force Constants of Linearly Joined Springs

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The discussion focuses on understanding Hooke's Law in the context of two springs arranged in series. The total force constant of two linearly joined springs is not simply the sum of their individual constants, k1 and k2. Instead, the effective force constant for springs in series is given by the equation 1/k_total = 1/k1 + 1/k2. Participants emphasize the importance of analyzing forces and deformations through free body diagrams to clarify the relationship. The conversation concludes with a confirmation that the initial assumption about the total force constant was incorrect.
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Q: How does the force constant of two springs hung linearly compare with the individual force constant??
What would be the equation that relates the total force constant, to the individual force constants, k1 and k2, of two springs joined together linearly??

A: The Total force constants is going to be k1 + k2??
 
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Doctor Lee said:
Q: How does the force constant of two springs hung linearly compare with the individual force constant??
What would be the equation that relates the total force constant, to the individual force constants, k1 and k2, of two springs joined together linearly??

A: The Total force constants is going to be k1 + k2??
You should show some work.

Apply a force W = mg by hanging a mass m at the end of the bottom spring(2) (do it so the springs are stationary after stretching). The bottom end of spring 1 will have stretched x1 from its top end and spring 2 a further distance x2 from spring 2's top end.

Do a free body diagram for the forces on m. What are the upward forces? What are the downward forces on m? What do they sum to?

Do that and you should see what the answer is and why.

AM
 
Draw the springs and the forces. Write the appropriate relationship between forces and deformations.
If by "linearly" you mean something like one after another (series), k is not k1+k2.
 


Thanks a lot guys. I get it now.
 
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