Telemarker, your question doesn't make any sense. You will still weigh 175lbs when you come to rest. The force applied to your anchor is determined by the distance you fall, the amount of rope out, and the stretchiness of the rope. You have to take into account the stretchiness of the rope because with a rope with no stretch (not physically possible) you would generate an infinite force (obviously not possible).
The question I think you are asking would probably be better stated as, "Assume I tie my rope directly to the anchor, I weigh 80kg, and I fall 1.65m above my anchor with no intermediate pro (total fall distance would be ~10ft) and a rope of average stretchiness (say 8%). What force is applied to the anchor?"
Well, the answer to that question, according to my calculation is 20.2kN (or roughly 4000lbs).
Now here is the part I'm less sure about:
If you were being belayed through the anchor with a perfect pulley between you and the belayer, the force felt by the anchor would be TWICE the above value. However, because biners aren't good pulleys, there is a lot of friction there and the anchor doesn't feel quite TWICE the force. I don't know what a reasonable coefficient of friction for a carabiner is, but Dave's fall force machine gives a total force reading of 32kN, so he is apparently assuming a coefficient of friction of 0.8.
I'm no physicist, so I hope someone more knowledgable than myself will review this post and correct me if I'm wrong.