Two blocks of mass $$M_1 = 20$$ kg and $$M_2 = 12$$ kg are connected by a metal rod of mass 8 kg. The system is pulled vertically up by applying a force of 480 N as shown. The tension at the mid-point of the rod is:

Total Mass ($$M_{total}$$): $$M_1 + M_2 + M_{rod} = 20\text{ kg} + 12\text{ kg} + 8\text{ kg} = 40\text{ kg}$$.

Total Weight ($$W$$): $$M_{total} \times g = 40\text{ kg} \times 10\text{ m/s}^2 = 400\text{ N}$$.

$$F_{net} = 480\text{ N} - 400\text{ N} = 80\text{ N}$$

Acceleration ($$a$$): $$a = \frac{F_{net}}{M_{total}} = \frac{80\text{ N}}{40\text{ kg}} = 2\text{ m/s}^2 \text{ (upwards)}$$

To find the tension ($$T$$) at the midpoint of the rod, we isolate the part of the system below that point. This lower section includes block $$M_2$$ and the bottom half of the rod.

Mass of the lower section ($$m'$$): $$m' = M_2 + \frac{M_{rod}}{2} = 12\text{ kg} + \frac{8\text{ kg}}{2} = 12 + 4 = 16\text{ kg}$$
$$T - m'g = m'a$$

$$T = m'(g + a)$$

$$T = 16 \times (10 + 2)$$

$$T = 16 \times 12 = 192\text{ N}$$