A Faulty Balance!

Relevant wiki: Torque - Equilibrium

The picture gives us a pretty good hint about how to proceed. The balance will be level when there are equal torques on both sides. Let's call the mass of the left plate $m_L$ and the mass of the right plate $m_R$ . Even though these two masses aren't equal, we are told that the torques balance when the plates are empty, so we can write:

$\alpha m_L = \beta m_R$

The other balance cases translate into the following:

$\begin{aligned} \alpha (m_L + m) &= \beta (m_R + 9) \\ \alpha (m_L + 16) &= \beta (m_R + m). \\ \end{aligned}$

Subtracting the first equation from the other two gives:

$\begin{aligned} \alpha m &= 9 \beta \\ 16 \alpha &= \beta m. \\ \end{aligned}$

Then dividing these two equations gives:

$\begin{aligned} \frac{m}{16} &= \frac{9}{m} \\ m^2 &= 144 \\ m &= \boxed{12}. \\ \end{aligned}$

Using the ratio method, the scale can be rewritten as m:9 = 16:m. Multiplying the extremes and means, we get m(m) = 144. Get the square root of both sides, and we have m = 12 g.

We know, Load×Load arm = Effort×Effort Arm
Let load arm = x, effort arm = y..
m×x = 9×y.....(1)
16×x = m×y....(2)
Dividing both the equations, we get,
m×m = 16×9
=> m = 4×3= 12

The summation of moment at the joint is equal to zero

in the first figure, let x be the distance of the mass to the joint and L be the distance from each other of the two masses therefore:

mx = 9(L-x) eq. 1

m(L-x) = 16x eq. 2

equate L-x

mx = $\frac{9(16x)}{m}$

m = 12 grams

there is a simple formula to it... actual wt = root of product of given wt =root of 9 x 16 =root of 144 = 12