🚀 go-pugleaf

Diagram three masses, outer two affect the middle one.  they are in a
row, M1, unknown mass Mu, and 3M1.

M1<---------->Mu<--------------------------------------->3M1

Force on Mu from M1 equal to force on Mu from 3M1

distances are x and (1.8-x)

<-----x-------->    <------------------1.8-x-------------->

Force of gravity is GMM/r^2

two forces equal to each other

GM1Mu/x^2    =  G3M1Mu/(1.8-x)^2

Upper GM1Mu cancels with GM1Mu leaving 

1/x^2             =   3/(1.8-x)(1.8-x)


That should get you to the end

~J


On Wed, 5 Mar 2008 22:14:29 -0800 (PST), TEL <ted_e_llama@hotmail.com>
wrote:

>On Mar 5, 4:45�pm, John Dredla <Jolt...@hotmail.com> wrote:
>> 6.23 ends up with a quadratic equation to solve for the radius. this
>> will point you in the right direction. �My my original masses are
>> 3.60m apart, so the mass that would go in the middle is radius 3.60-r
>>
>> ~J
>>
>> On Wed, 5 Mar 2008 02:32:55 -0800 (PST), TEL <ted_e_ll...@hotmail.com>
>> wrote:
>>
>> >Im currently stuck on problem 6.22 and 6.23 from the homework. �Im
>> >guessing I misunderstood what Dr. Pravica said about multiple objects
>> >effects on one another. �Any help in the right direction would be
>> >appreciated.
>
>Maybe you can tell me what im missing.  I have it set up that the
>force between m and 3m is (G*m*3m)/1.8^2 (1.8 was the distance i was
>given) and this force needs to equal the negative of the force of the
>3rd mass, -(G*m1*m3)/r^2.  I set these 2 equal to each other and tried
>to solve which gave me a distance of 1.04m.  Can you tell me where I
>am going wrong?