2 kg
Explanation
to maintain the system in equilibrium the sum of momentums(or torques) in The point of contatc must be equal to zero,so
[tex]M_1+M_2=0[/tex]
the torque(M) is given by
[tex]M=\text{force}\cdot\text{distance}[/tex]
so for M1
[tex]\begin{gathered} M1=\text{force}\cdot\text{distance} \\ \text{force}=\text{ mass }\cdot gravity \\ F=m_1g \\ dis\tan ce=50-25=25\text{ cm=0.25 m} \\ so,\text{ replace} \\ M1=m_1g\cdot0.25\text{ } \\ M1=0.25m_1g \end{gathered}[/tex]
and M2 must be equal, so
[tex]\begin{gathered} M2=1\operatorname{kg}\cdot g\cdot0.5m \\ M2=0.5\text{ g } \end{gathered}[/tex]
now, as M1= M2
set equal and solve for m
[tex]\begin{gathered} M1=M2 \\ 0.25m_1g=0.5g \\ \text{divide both sides by 0.25 g} \\ \frac{0.25m_1g}{0.25g}=\frac{0.5g}{0.25g} \\ m_1=2\text{ kg} \end{gathered}[/tex]
therefore, the mass of the rock is 2 kg
