physics
posted by jess on .
A meter stick is pivoted at its 50cm mark but does not balance because of nonuniformities in its material that cause its center of gravity to be displaced from its geometrical center. However, when masses of 150 and 200 grams are placed at the 10cm and 75cm marks respectively, balance is obtained. The masses are then interchanged and balance is again obtained by shifting the pivot point to the 43cm mark. Find the mass of the meter stick and the location of its center of gravity.
I know there is going to be 2 unknowns and x+7 is the new distance from the pivot to the c of mass of the meter stick, but I am so lost as to how to put this together, please help. thanks

First balance
...150................M.........200......
...................^.....................
....10............50..x..........75......
second balance
...150................M.........200......
..............^...........................
....10.......43.......x..........75......
M=mass
x=distance from left (=graduation) 
diagrams help, i had these almost exactly drawn out, but i do not know how to create the equations to solve for the unknowns. thanks

First balance
...150................M.........200......
...................^.....................
....10.........50.....x..........75......
second balance
...200................M.........150......
..............^...........................
....10....43..........x..........75......
third line: distance from left
M=mass
x=distance from left (=graduation)
For first balance, take moments about pivot point (=50), clockwise = positive, counterwise = negative:
150(5010)+M(x50)+200(7550)=0
For second balance:
200(4310)+M(x43)+150(7543)=0
Solve for M and Mx.
Divide Mx by M to get x.
I get
M=800/7=114.3
x=58.75 cm from zero.