Question 522528
<pre>
Rule:

Quantity that varies = k·{{{("directly(s)")/("inversely(s)")}}}
</pre>
>>...The force of attraction between two bodies varies...<<
<pre>
That's the quantity that varies.  We'll call it F. That goes on
the left of the equal sign.
</pre>
>>...directly with the product of their masses...<<
<pre>
So the two masses go on top on the right of the equal sign.  We 
will call them m<sub>1</sub> and m<sub>2</sub>.
</pre>
>>...and inversely with the square of the distance between them...<<
<pre>
That's the inversely.  We'll call it d². It goes on the bottom on the 
right. So we have:

              F = k·{{{(m[1]m[2])/d^2}}}

That's the incomplete formula.  We must find k before we have the
complete formula:

</pre>
>>...The force of attraction is 6 pounds when masses of 64 slugs and 108 slugs are 24 feet apart.<<
<pre>
We substitute F=6, m<sub>1</sub>=64, m<sub>2</sub>=108, d=24 

              6 = k·{{{(64*108)/24^2}}}
              6 = k·12
              {{{6/20}}} = k
              0.5 = k

Now we have k.  Now we substitute 0.5 for k in this:

              F = k·{{{(m[1]m[2])/d^2}}}

              F = 0.5·{{{(m[1]m[2])/d^2}}}

And that is the complete formula.  Now that we have the complete
formula, we are ready to use it.  Here is what we are to use it to
find:
</pre>
>>..Find the force of attraction when two masses are 25 slugs each and 60 feet apart..<<
<pre>
We substitute m<sub>1</sub> = 25, m<sub>2</sub> = 25, d = 60

              F = 0.5·{{{(m[1]m[2])/d^2}}}

              F = 0.5·{{{(25*25)/60^2}}}

              F = 0.5·{{{625/3600}}}

              F = 0.5·{{{25/144}}}
       
              F = {{{1/2}}}·{{{25/144}}}

              F = 25/288 = .0868 pounds.

Edwin</pre>