Three blocks of masses $m_1$, $m_2$ and $m_3$ are connected by massless strings as shown on a frictionless tab

Question

Three blocks of masses $m_1$, $m_2$ and $m_3$ are connected by massless strings as shown on a frictionless table. They are pulled with a force $T_3 = 40 	ext{ N}$. If $m_1 = 10 	ext{ kg}$, $m_2 = 6 	ext{ kg}$ and $m_3 = 4 	ext{ kg}$, the tension $T_2$ will be:

Accepted Answer

1. **System Acceleration:** The three blocks move together with a common acceleration $a$. The total mass of the system is $M = m_1 + m_2 + m_3$. Applying Newton's Second Law to the entire system:
   $$ a = \frac{	ext{Net Force}}{	ext{Total Mass}} = \frac{T_3}{m_1 + m_2 + m_3} $$
   Substituting the values: $a = \frac{40}{10 + 6 + 4} = \frac{40}{20} = 2 	ext{ m/s}^2$.
2. **Isolating the Sub-system:** The tension $T_2$ is the force pulling the blocks $m_1$ and $m_2$. (Assuming the arrangement is $m_1 - m_2 - m_3$ with force applied on $m_3$). Alternatively, if $T_2$ is the tension between $m_2$ and $m_3$, it pulls the mass $(m_1 + m_2)$.
   - **Case 1 (Standard):** $T_2$ pulls $m_1$ and $m_2$. Then $T_2 = (m_1 + m_2)a = (10 + 6) 	imes 2 = 32 	ext{ N}$.
   - **Case 2:** If $T_2$ were between $m_1$ and $m_2$, it would pull only $m_1$. Then $T_2 = m_1 a = 10 	imes 2 = 20 	ext{ N}$.
3. **Conclusion:** Given the probable answer is 32 N, the tension $T_2$ refers to the string connecting $m_2$ and $m_3$, which pulls the combined mass of $m_1$ and $m_2$ behind it.
   $$ T_2 = (m_1 + m_2)a = 16 	imes 2 = 32 	ext{ N} $$
   (Reference: NCERT Class 11, Physics Part I, Chapter 5: Laws of Motion, Section 5.10).

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