Question

Measure the distances from the 200 g and 50 g masses to the fulcrum.

Answer 1

let the mass 200 g is placed at distance x1 from fulcrum on one side and on the other side let say 50 g mass is connected at distance x2 from the fulcrum

Now in order to balance the rod we can say that torque must be zero due to both masses about the fulcrum

now we will write the equation as

$200*x_1 = 50 * x_2$

now we will have

$\frac{x_1}{x_2} = \frac{1}{4}$

now we will do the same for two other masses 100 g and 300 g

now the equation will be

$100*x_1 = 300 * x_2$

$\frac{x_1}{x_2} = \frac{3}{1}$

now from above two equations we can say that the distance of two masses are always in the inverse ratio of the two masses

so in general we can say

$\frac{x_1}{x_2} = \frac{m_2}{m_1}$

now if we use the same for both masses 100 g we can say

$100*x_1 = 100* x_2$

$\frac{x_1}{x_2} = \frac{1}{1}$

so our general formula is perfect here

#4

now in this situation we have to balance 50 g and 100 g mass on one side of the scale with 200 g mass on the other side

let 50 g mass is at distance x1 and 100 g mass is at distance x2 from fulcrum on one side while 200 g mass is at distance x3 on the other side

so now by torque balance we can say

$50* x_1 + 100* x_2 = 200 * x_3$

$\frac{x_1}{x_3} + 2*\frac{x_2}{x_3} = 4$

so above is the general formula to balance this type of situation

So here we can not use the previous equation here