🚀 go-pugleaf

On Mar 9, 9:20 am, John Dredla <Jolt...@hotmail.com> wrote:
> those tensoin problems always sum of forces in x and y dimension.   In
> X dimensoin when the engine just hung there, there was no acceleration
> hence no net force, one force equaled another.   Now, in the swinging
> block problem, there is radial acceleration.    But both tensions bear
> part of the burden.
>
> On Sat, 8 Mar 2008 02:35:14 -0800 (PST), TEL <ted_e_ll...@hotmail.com>
> wrote:
>
> >On Mar 7, 6:05 am, John Dredla <Jolt...@hotmail.com> wrote:
> >> find the gravity there  grav = 2h/t^2
>
> >> Circumference =2piR, find R    R=C/2pi
>
> >> mass = (grav)R^2)/G       G is the gravitational constant
>
> >> mass of planet / mass of earth
>
> >> --------------------------------------
>
> >> Find the distance from the pole to the block
> >> Hypontenuse^2 - 1/2 length2, then square root
>
> >>   sq rt  H^2 - L1  = L2
>
> >> Tension in lower called Tl
> >> Tension in upper called Tu
>
> >> Tl = Tu - (Hype/L1)mg
>
> >> second part of problem  solve for V
>
> >> a=v^2/R   F=ma
>
> >> substitute a into second equation
>
> >> F=mV^2/R
>
> >> set it equal to the tensions, don't forget to add given upper tension
> >> with found lower tension
>
> >> Hope this helps
>
> >> On Thu, 6 Mar 2008 23:39:39 -0800 (PST), TEL <ted_e_ll...@hotmail.com>
> >> wrote:
>
> >> >If anyone could help w/ 6.27 and the EC 6.57 it would be greatly
> >> >appreciated (before 10:00am :D).  Again, not sure what im missing.
>
> >> >Thanks
>
> >Could you go through your logic on these 2, how you derived them and
> >such?

Thnx John for all your help :D