A beam of metal is acted upon by two forces (parallel) of 15 N and 20 N at a distance of 35 cm from each other. At what point the beam will be balanced ?

Let the balance be at x cm from 20 N force.

Anticlockwise moment = 20(35-x)

Clockwise moment = 15x

According to principle of moment :

Anticlockwise moment = Clockwise moment

Substituting the values in the formula above, we get,

$20(35-\text{x}) = 15\text{x} \\[0.5em] \Rightarrow (20 \times 35) - 20 \text{x} = 15\text{x} \\[0.5em] \Rightarrow (20 \times 35) = (15 + 20)\text{x} \\[0.5em] \Rightarrow (20 \times 35) = 35 \text{x} \\[0.5em] \Rightarrow \text{x} = \dfrac{(20 \times 35)}{35} = 20 \text{ cm} \\[0.5em]$

Hence, the beam will be balanced at 20 cm