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A rod is 8 feet long. A weight of 100 pounds is fastened to one end, a weight of 75 pounds to the other.
At a distance of 2 feet from the 100 pound weight a third weight of 65 pounds is fastened. Where is the balancing point.
a) if the weight of the rod can be neglected? b) if the rod weighs 10 pounds per foot?
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(a)
Let x be the distance (in feet) from the end, where 100 pound is fastened, to the balancing point
(called fulcrum: see Wikipedia article https://en.wikipedia.org/wiki/Lever ).
Then the 100-pound weight has the arm of of x feet;
the 75-pound weight has the arm of (8-x) feet;
the 65-pound weight has the arm of (x-2) feet.
The balance condition is
00*x + 65*(x-2) = 75*(8-x). (1)
Simplify and solve for x.
(b)
If the rod weights 10 pounds per foot, then add 
to the left side of the equation (1) and 
to the right side
of this equation. It is to express that the rod weight applies half of the arm weight at the center of each arm.
You will get new balance equation
00*x + 65*(x-2) + 
= 75*(8-x) + 
. (2)
Simplify and solve for x.
Solved.
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On levers, see my lesson
- Using proportions to solve word problems in Physics
in this site.
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