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.

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
...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.

Hw didi get ur M and Mx

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

To solve this problem, we can use the principle of moments. The principle of moments states that for an object to be in rotational equilibrium, the sum of the clockwise moments must be equal to the sum of the counterclockwise moments.

Let's denote the mass of the meter stick as M and the distance of its center of gravity from the pivot point as d.

First, let's consider the first situation where masses of 150g and 200g are placed at the 10cm and 75cm marks, respectively. The clockwise moments are given by the product of the mass and the distance from the pivot point, while the counterclockwise moments are given by the product of the mass and the distance from the pivot point minus the length of the meter stick.

So, for this situation, we have:

(clockwise moments) + (counterclockwise moments) = 0
(150g * 10cm) + (200g * 75cm - M * 100cm) = 0

Now, let's consider the second situation where the masses are interchanged, and balance is obtained by shifting the pivot point to the 43cm mark. In this case, the clockwise moments are given by the product of the mass and the distance from the pivot point, while the counterclockwise moments are given by the product of the mass and the distance from the pivot point minus the length of the meter stick.

So, for this situation, we have:

(clockwise moments) + (counterclockwise moments) = 0
(200g * 10cm - M * 43cm) + (150g * 75cm - M * 100cm) = 0

Now we have two equations with two unknowns: M (mass of the meter stick) and d (distance of the center of gravity from the pivot point). We can solve these equations simultaneously to find the values of M and d.

To solve the equations, you can rearrange them to isolate each variable and then substitute the value obtained for one variable into the other equation. This will allow you to solve for the remaining variable.

Finally, once you have the values for M and d, you can calculate the location of the center of gravity by dividing the mass moment of the meter stick (M * d) by the mass of the meter stick (M).

I hope this helps! Let me know if you have any further questions.