A system consists of three masses $m_1$, $m_2$ and $m_3$ connected by a string passing over a pulley P. The ma

Question

A system consists of three masses $m_1$, $m_2$ and $m_3$ connected by a string passing over a pulley P. The mass $m_1$ hangs freely and $m_2$ and $m_3$ are on a rough horizontal table (the coefficient of friction = $\mu$). The pulley is frictionless and of negligible mass. The downward acceleration of mass $m_1$ is: (Assume $m_1=m_2=m_3=m$)

Accepted Answer

1. **System Analysis:** The three blocks move together with a common acceleration $a$. The hanging mass $m_1$ pulls the system downwards, while the friction on masses $m_2$ and $m_3$ opposes the motion.
2. **Forces:**
   - **Driving Force:** Gravitational force on the hanging mass $m_1$: $F_{gravity} = m_1g$.
   - **Resisting Force:** Frictional force on the horizontal masses $m_2$ and $m_3$. Since they are on a rough table with coefficient $\mu$, the kinetic friction is $f_k = \mu N$. Here $N = mg$.
     Total Friction $f = f_2 + f_3 = \mu m_2g + \mu m_3g = \mu g(m_2 + m_3)$.
3. **Equation of Motion (Newton's Second Law):**
   $$F_{net} = M_{total} 	imes a$$
   $$m_1g - \mu g(m_2 + m_3) = (m_1 + m_2 + m_3)a$$
4. **Substitution:** Given $m_1 = m_2 = m_3 = m$.
   $$mg - \mu g(m + m) = (m + m + m)a$$
   $$mg - 2\mu mg = 3ma$$
   $$mg(1 - 2\mu) = 3ma$$
5. **Solving for acceleration ($a$):**
   $$a = \frac{g(1 - 2\mu)}{3}$$

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