Trying to understand how center of gravity effects a seesaw-like contraption 
In order to solve this problem, I tried assigning a variable to the masses of the four sections of the meterstick and the constant, 1 kg, to the rock
rock = 1 kg, 1/4 of stick = x
From here I thought the mass of the meter stick was 4x.
Turning the problem into an equation, I got: 1kg + x = 3x.
Isolating x, I got 1kg= 2x.
This leads to x= 0.5kg.
If each x is 0.5kg, the mass of all 4 of the x's put together (a.k.a. the mass of the entire meterstick) is 2 kg.
My textbook says that the mass of the meterstick is 1 kg, but that clearly conflicts with this equation.
Am I missing something here?
The book says that the center of gravity of the stick is at the 50 cm mark, and that I can treat the problem like all of the mass of the meterstick is concentrated at the 50 cm mark. From there I should see that the 50 cm mark is the same distance away from the fulcrum as the stone is, which should lead me to believe that both the stick and the stone have the same mass. Namely, 1kg.
I understand that there is no torque in this system, so the book's solution makes sense to me, but my own mathematical solution makes sense to me as well, so I don't really understand where I messed up.
 A: Gravity creates torque on both sides of the bar by pulling them down, and since the bar is balanced on there, the total torque is zero. Your equation doesn't include the masses' distance from the point of rotation, it would only work for a system that has equal distances for all masses, like a pulley balancing two weights on two of its sides, as the torque would be equal to the radius of the pulley times the mg on both sides, everything would cancel out, just leaving the masses.
For this system take the center of each piece of the bar as a point of mass (mass of each piece of the rod), and calculate the torque that the point creates on the bar. The (scalar) equation for torque is $$Distance \times Force$$
And the forces here will be $m \times g$ for the respective masses. Think of every part of the rod being pulled down from its center with force $m_{rod} \times g$
For the left side torque you get: $$(25 cm \times m_{rock}\times g) + (12.5 cm \times  m_{rod}\times g)$$ ($m_{rod}$ is the mass of one section of the bar here, so the total mass would be $4m_{rod}$)
For the right side you get:
$$
(12.5cm \times  m_{rod}\times g) + (37.5cm \times  m_{rod}\times g) + (62.5cm \times  m_{rod}\times g)
$$
And for the total torque to be zero, these two equations must be equal, so when you rewrite them and cancel out the g's you get $$25 \times m_{rock} = 100 \times m_{rod}$$
$$m_{rock} = 4 m_{rod}$$ and the total mass of the rod as mentioned is  $4 m_{rod}$ so the answer is 1 kg!
I wrote these in centimeters to make it clear, but normally you should just call the length of one piece of the bar $2x$ and calculate arbitrarily without plugging in the values, so the equation would have $xmg +2xm_{rock}g = g(xm+3xm+5xm)$ etc. Hope that's all clear!
