Re: leaf spring energy

[David Dockstader <DRDOCK00@UKCC.UKY.EDU>]

On Mon, 13 Apr 1998 10:44:15 -0700 (PDT) James Mclean said:
> real leaf spring.)  Now, F=k'*phi=k*x.  Using the same kind of reasoning as
> above, (double d, F const => double phi) I get F=K*c*phi/d=K*x/d^2.  But
> then the work integral gives E = (K*c^2/2)*phi^2 = (K/2)*(x/d)^2.
Here is where I think you have run amuck.  F=k*x=k'phi only for d constant.
Double d, F constant will not double phi, in fact phi will hardly change at
all.  I'm not sure why you want to write this in terms of phi anyway.  Why
not just use the same equations as for the coil spring?
> --
> --James McLean
> jmclean@chem.ucsd.edu
> post doc
> UC San Diego, Chemistry