Final Answer:
To ensure that the mass (m) does not move in the Atwood machine composed of three masses (m), (m₁), and (m₂), the condition (m = m₁ + m₂) must be satisfied.
Step-by-step explanation:
In an Atwood machine, the net force acting on the system determines the motion of the masses. For the system to be in equilibrium, the net force must be zero. Considering the masses \(m\), \(m₁\), and \(m₂\), if \(m\) is such that (m = m₁ + m₂), then the gravitational forces on either side of the pulley will balance each other, resulting in an equilibrium condition where (m) does not move.
The gravitational force on (m) is given by (F₍g, m₎ = m ⋅ g), where \(g\) is the acceleration due to gravity. Similarly, the gravitational forces on \(m₁\) and (m₂) are (F₍g, m₁₎ = m₁ ⋅ g) and (F₍g, m₂₎ = m₂ ⋅ g). In equilibrium, \(F₍g, m₎ = F₍g, m₁₎ + F₍g, m₂₎\), which leads to the equation \(m ⋅ g = m₁ ⋅ g + m₂ ⋅ g\). Canceling (g) from both sides gives (m = m₁ + m₂).
In conclusion, for the system to be in equilibrium and for \(m\) not to move, the mass \(m\) must be equal to the sum of \(m₁\) and \(m₂\). This condition ensures that the gravitational forces acting on either side of the pulley are balanced, resulting in a stable configuration where \(m\) remains stationary.