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 \(m_1 = m_2 = m_3 = m\))

• A. $\frac{g(1-g\mu)}{9}$
• B. $\frac{2g\mu}{3}$
• C. $\frac{g(1-2\mu)}{3}$
• D. $\frac{g(1-2\mu)}{2}$

Answer 1

The Correct Option is C

To find the downward acceleration of mass \(m_1\), let's analyze the forces acting on the system.

Given:

• Masses: \(m_1 = m_2 = m_3 = m\)
• Coefficient of friction: \(\mu\)
• Gravitational acceleration: \(g\)

Step-by-step Explanation:

1. For mass \(m_1\):
   • The downward force is \(T + ma\) due to gravity.
2. For mass \(m_2\) and \(m_3\) on a rough horizontal table:
   • The tension in the string provides the force. Friction acts in the opposite direction.
   • Net force on \(m_2\) and \(m_3\) = \( T - \mu mg \).

Setting Up Equations:

1. For mass \(m_1\):
   \( m_1g - T = m_1a \)
   \( mg - T = ma \)
   \( T = mg - ma \)
2. For masses \(m_2\) and \(m_3\):
   \( 2T - 2\mu mg = 2ma \)
   \( T - \mu mg = ma \)

Solving the System:

Add the equations for both systems:

\[ \begin{align*} mg - ma + 2T - 2\mu mg &= 3ma \\ mg - ma + (mg - ma) - \mu mg &= 2ma \\ 3mg - 3\mu mg &= 3ma \\ a = \frac{g(1-2\mu)}{3} \end{align*} \]

Conclusion:

The downward acceleration of mass \(m_1\) is \(\frac{g(1-2\mu)}{3}\).