MHB What Is the Equivalent Linear Spring Constant at Steady Deflection?

[Dustinsfl]

The force-deflection relation of a steel helical spring used in an engine is found experimentally as \(F(x) = 200x + 50x^2 + 10x^3\), where the force \((F)\) and deflection \((x)\) are measured in pounds and inches, respectively. If the spring undergoes a steady deflection of \(0.5\) in during the operation of the engine, determine the equivalent linear spring constant of the spring at its steady deflection.

If I knew \(F\), I could simply solve the cubic for the real \(x\) and then use the Newtons 2nd law \(F = k_{eq}x\).

In this problem, I don't have \(F\) instead I have the steady deflection. How do I solve this type of problem?

The answer is \(253.75\) lb per in

 

[topsquark]

The force-deflection relation of a steel helical spring used in an engine is found experimentally as \(F(x) = 200x + 50x^2 + 10x^3\), where the force \((F)\) and deflection \((x)\) are measured in pounds and inches, respectively. If the spring undergoes a steady deflection of \(0.5\) in during the operation of the engine, determine the equivalent linear spring constant of the spring at its steady deflection.

If I knew \(F\), I could simply solve the cubic for the real \(x\) and then use the Newtons 2nd law \(F = k_{eq}x\).

In this problem, I don't have \(F\) instead I have the steady deflection. How do I solve this type of problem?

The answer is \(253.75\) lb per in

What do you mean by "steady deflection?" It sounds like the spring is at a particular point and doesn't move?

Also, F = kx is Hooke's Law, not Newton's.

-Dan

 

[Dustinsfl]

What do you mean by "steady deflection?" It sounds like the spring is at a particular point and doesn't move?

Also, F = kx is Hooke's Law, not Newton's.

-Dan

That is just the terminology the book uses.