In this problem, we are tasked with determining the missing masses and tensions in a system of objects connected by ropes and rods, all while ensuring the system is in both linear and rotational equilibrium. The key principles to remember are that the sum of all forces and the sum of all torques must equal zero. We will denote the masses as follows: mass A, mass B (given as 4 kg), and mass C. The gravitational acceleration is simplified to \( g = 10 \, \text{m/s}^2 \).

To begin, we identify the tensions in the system: \( T_A \) for mass A, \( T_B \) for mass B, \( T_C \) for mass C, and \( T_{BC} \) for the combined tension of masses B and C. Since mass B is known, we can calculate the force acting on it as \( m_B g = 4 \, \text{kg} \times 10 \, \text{m/s}^2 = 40 \, \text{N} \). This means that the tension \( T_B \) must also equal 40 N to maintain equilibrium.

Next, we analyze the torques acting on the system. For mass C, we can express the torque balance around a pivot point. The torque due to \( T_B \) must equal the torque due to \( T_C \). This can be expressed mathematically as:

\[ T_B \cdot r_B = T_C \cdot r_C \]

Given that \( T_B = 40 \, \text{N} \) and the distances \( r_B = 1 \, \text{m} \) and \( r_C = 2 \, \text{m} \), we can solve for \( T_C \):

\[ 40 \cdot 1 = T_C \cdot 2 \implies T_C = \frac{40}{2} = 20 \, \text{N} \]

Now that we have \( T_C \), we can find the mass C using the equation \( T_C = m_C g \):

\[ 20 = m_C \cdot 10 \implies m_C = 2 \, \text{kg} \]

With the mass of C known, we can now find the combined tension \( T_{BC} \) acting on both masses B and C:

\[ T_{BC} = T_B + T_C = 40 + 20 = 60 \, \text{N} \]

Next, we need to find the mass A. The tension \( T_A \) must balance the combined weight of masses B and C. Using the relationship of distances and masses, we can express the torque balance for mass A:

\[ T_A \cdot r_A = T_{BC} \cdot r_{BC} \]

Here, \( r_A = 4 \, \text{m} \) and \( r_{BC} = 1 \, \text{m} \), leading to:

\[ T_A \cdot 4 = 60 \cdot 1 \implies T_A = \frac{60}{4} = 15 \, \text{N} \]

Finally, we can find the mass A using the equation \( T_A = m_A g \):

\[ 15 = m_A \cdot 10 \implies m_A = 1.5 \, \text{kg} \]

In summary, the masses and tensions in the system are as follows: mass A is 1.5 kg, mass B is 4 kg, mass C is 2 kg, and the tensions are \( T_A = 15 \, \text{N} \), \( T_B = 40 \, \text{N} \), \( T_C = 20 \, \text{N} \), and \( T_{BC} = 60 \, \text{N} \). Understanding the relationships between forces and torques is crucial for solving such equilibrium problems efficiently.