A meter stick is pivoted at its 50-cm mark but does not balance because of non-uniformities 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 10-cm and 75-cm marks respectively, balance is obtained. The masses are then interchanged and balance is again obtained by shifting the pivot point to the 43-cm mark. Find the mass of the meter stick and the location of its center of gravity.

Question

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

Answer 1

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)

Answer 2

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

Answer 3

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(50-10)+M(x-50)+200(75-50)=0
For second balance:
-200(43-10)+M(x-43)+150(75-43)=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.

Answer 4

Hw didi get ur M and Mx