🚀 go-pugleaf

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_llama@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?