Two blocks of mass $2.9\,\text{kg}$ and $1.9\,\text{kg}$ are suspended from a rigid support $S$ by two inextensible wires each of length 1 meter, see fig. The upper wire has negligible mass and the lower wire has a uniform mass of $0.2\,\text{kg/m}$. The whole system of blocks wires and support have an upward acceleration of $0.2\,\text{m/s}^{2}$. Acceleration due to gravity is $9.8\,\text{m/s}^{2}$. (i) Find the tension at the mid-point of the lower wire. (ii) Find the tension at the mid-point of the upper wire.

Step 1: Identify the given parameters.

The system parameters are given as follows:

$$m_1 = 2.9\,\text{kg}$$ $$m_2 = 1.9\,\text{kg}$$ $$L = 1\,\text{m}$$ $$\lambda = 0.2\,\text{kg/m}$$ $$a = 0.2\,\text{m/s}^{2}$$ $$g = 9.8\,\text{m/s}^{2}$$

Step 2: Formulate the tension equation for an upward accelerating system.

Applying Newton's second law to a section supporting a mass $m_{\text{below}}$ yields the tension equation:

$$T - m_{\text{below}}g = m_{\text{below}}a$$ $$T = m_{\text{below}}(g + a)$$

Step 3: Find the tension at the mid-point of the lower wire.

The mass below the mid-point consists of the lower block and half of the lower wire.

$$m_{\text{below}} = m_2 + \lambda\frac{L}{2}$$ $$m_{\text{below}} = 1.9 + 0.2 \times 0.5 = 2.0\,\text{kg}$$ $$T_1 = m_{\text{below}}(g + a)$$ $$T_1 = 2.0 \times (9.8 + 0.2)$$ $$T_1 = 20\,\text{N}$$

Step 4: Find the tension at the mid-point of the upper wire.

The massless upper wire supports the entire mass of the system below it.

$$m_{\text{total}} = m_1 + m_2 + \lambda L$$ $$m_{\text{total}} = 2.9 + 1.9 + 0.2 \times 1 = 5.0\,\text{kg}$$ $$T_2 = m_{\text{total}}(g + a)$$ $$T_2 = 5.0 \times (9.8 + 0.2)$$ $$T_2 = 50\,\text{N}$$

Answer: (i) Tension at the mid-point of the lower wire is $20\,\text{N}$, (ii) Tension at the mid-point of the upper wire is $50\,\text{N}$.