Question

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 :

• A. $144\,N$
• B. $96\,N$
• C. $240\,N$
• D. $192\,N$

Answer 1

The Correct Option is D

Let's solve the problem step-by-step to find the tension at the mid-point of the rod.

1. First, calculate the total mass of the system. The system consists of two blocks and one rod:
   M_{\text{total}} = M_1 + M_2 + M_{\text{rod}} = 20\, \text{kg} + 12\, \text{kg} + 8\, \text{kg} = 40\, \text{kg}
2. Calculate the acceleration of the system using Newton's second law:
   F = M_{\text{total}} \cdot g + M_{\text{total}} \cdot a
   480 = 40 \cdot 9.8 + 40 \cdot a
   Solve for a:
   480 = 392 + 40a
   40a = 480 - 392 = 88
   a = \frac{88}{40} = 2.2 \, \text{m/s}^2
3. Find the tension at the midpoint of the rod. To find this, consider the rod to be divided into two equal parts. The mass of half of the rod is m_{\text{rod}/2} = 4\, \text{kg}.
4. The part of the system below the midpoint consists of M_2 and half of the rod:
   M_{\text{below}} = M_2 + m_{\text{rod}/2} = 12\, \text{kg} + 4\, \text{kg} = 16\, \text{kg}
5. Apply Newton's second law to the lower part of the system:
   T = M_{\text{below}} \cdot (g + a)
   T = 16 \cdot (9.8 + 2.2)
   T = 16 \cdot 12 = 192\, \text{N}

Therefore, the tension at the mid-point of the rod is 192\, \text{N}.