Answer
$m_{4}=m=1kg$
Work Step by Step
İf system is in equilibrium then net torque is zero. Therefore,
For penguin (3) and (4) torque is
$\Sigma \overrightarrow {t}=m_{4}\times 3Lg-m_{3}\times Lg=0\Rightarrow m_{3}=3m_{4}=3m$
For penguin (2) and (3) and (4), the torque is:
$\sum \overrightarrow {t}=\left( m_{3}+m_{4}\right) g\times 3L-m_{2}gL=0\Rightarrow m_{2}=3\left( m_{3}+m_{4}\right) =3\left( 3m+m\right) =12m$
For penguin (1),(2),(3),(4), the torque is:
$m_{1}g\times L=\left( m_{2}+m_{3}+m_{4}\right) g\times 3L\Rightarrow m_{1}=3\left( m_{2}+m_{3}+m_{4}\right) =3\left( 12m+3m+m\right) =48m=48kg\Rightarrow m=1kg$
So the mass of penguin (4) is
$m_{4}=m=1kg$
