The mass of $0.3 mathrm{~kg}$ is replaced with a $0.4 mathrm{~kg}$. The strings and horizontal rods have negligible mass.

Solve for the figure’s data of Example $8.13$:

As the textbook mentions, the balance point corresponds to zero net torque assigns that

$$
begin{align*}
tau &= m_{1} ~ g ~ r_{1} – m_{2} ~ g ~ r_{2} \
&= 0 \
m_{1} ~ r_{1} &= m_{2} ~ r_{2} \
end{align*}
$$

Rearrange and solve for the value of the mass $m_{1}$:

$$
begin{align*}
m_{1} &= dfrac{ m_{2} ~ r_{2} }{ r_{1} } \
&= dfrac{ left( 0.2 mathrm{~kg} +0.4 mathrm{~kg} + 0.72 mathrm{~kg} + 0.48 mathrm{~kg} right) times 18 mathrm{~cm} }{ 31 mathrm{~cm} } \
&= 1.045 mathrm{~kg}
end{align*}
$$

So, the mass $m_{1} =1.045 mathrm{~kg}$, in this case, is greater than the mass $m_{1}$ in example $8.13$.