Question 1042973
.
a weight of 75 pounds is placed in a lever 15 feet from the fulcrum. how far from the fulcrum on the other 
side must a weight of 90 pounds be placed in order to give equilibrium?
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~


<pre>
According to the description, it is 1-st kind lever (1-st class lever).

For levers, see <A HREF=https://en.wikipedia.org/wiki/Lever>this</A> Wikipedia article.


See also the lesson <A HREF=https://www.algebra.com/algebra/homework/proportions/lessons/Using-proportions-to-solve-word-problems-in-Physics.lesson>Using proportions to solve word problems in Physics</A> in this site.


The condition for equilibrium for such levers is this equality

{{{F[1]*L[1]}}} = {{{F[2]*L[2]}}},

where {{{F[1]}}}, {{{F[2]}}} are the forces applied and {{{L[1]}}}  and {{{L[2]}}} are their armes.


In our case 75*15 = 90*x, which gives x = {{{75*15/90}}} = 12.5 feet.

<U>Answer</U>.  12.5 feet.
</pre>