posted by rory .
A construction supervisor of mass M = 83.7 kg is standing on a board of mass m = 30.5 kg. The board is supported by two saw horses that are a distance of 4.50 m apart. If the man stands a distance x1 = 1.30 m away from saw horse 1 as shown, what is the force that the board exerts on that saw horse?
Set the total moment about saw horse 2 equal to zero. Call the upward force at saw horse 1 "F1".
F1*4.50 = Mg(4.50 - 1.30) + mg(2.25)
Solve for F1. The number 2.25 meters appears because the weight of the board acts through the center of gravity.
I still don't quite understand how you got 2.25 meters...
2.25 m is the location of midpoint of the board, where the board's weight acts. That is measured from saw horse 2. I am assuming that the saw horses are at the ends. If a figure (which you did not provide) says otherwise, use the appropriate distance.
ok so i solved all that out and it said i was wrong and told me this
There are two forces acting on the saw horses. One is due the weight of the board. This is evenly distributed between the two saw horses, because the board is placed symmetrically on them.
The other force acting on the saw horse is due to the weight of the man. If the man was standing exactly in the middle, his weight would also be distributed evenly over both saw horses. However, he is standing off to one side, and so his weight gets distributed proportionally to the distances of the man to the saw horses.
To do the math, it is perhaps easiest to use the top of saw horse 2 as the pivot points for calculating the torques on the board. (They have to add up to zero!).