Re: please tell me how to figure the force applied from an 1800lb. weight

You know the right place to start, so let's continue with straight math.  
As you stated F=MxA.  Let's apply this to your problem.  First 
acceleration; acceleration in this case is a product of gravity.  Assuming 
the dropped object is in fact dropped, and not pushed down, it is 
considered a free falling object.  Such an object has an acceleration of 
9.8 meters per second per second, or about 32 feet per second per second, 
downward (on Earth). Therefore, the force in this problem can be expressed 
as 
   F = 1800lb x 32 feet / sec /sec 
However, the object is dropped from 10 feet, which means that it 
accelerates proportional with the amount of time it spends "in flight."  
As such, we express this variance as 

   F = 1800lb x [ (32 feet/sec/sec) x time ]

In this case, you would have to measure the time that this object is "in 
flight" to accurately calculate acceleration.  Let's say, for the purpose 
of calculation that time (t) is equal to 2 seconds.  The force could then 
be accurately calculated for by

   F = 1800lb x [  (32 feet/sec/sec) x 2)

which goes to 

   F = 1800 lb. x (64 feet/sec)

and then 

   F = 115200 ft lbs. for every second the force is applied

The rest of the calculation depends on peripheral forces like drag and air 
resistance.  The 300-lb. weight on top of the pole really doesn’t matter 
since it doesn’t exert any opposing force on the dropped object.  These 
calculations probably make more sense in metric where you can use units 
like newtons to express force.  

ADMIN NOTE: Norman Parker adds

The answer stated that in order to get foot-pounds from pounds, you multiply 
it by the acceleration of gravity. Thats the way it works with newtons, 
but pounds are already multiplied by gravity. The answerer is thinking 
of newtons = mass x gravity. The American counterpart to that is 
Pounds = slugs x gravity. Secondly the amount of force in the American 
system is in pounds, the term foot-pounds is a measurement of work, and 
does not correctly describe force.
Sincerely,
Norman L. Parker