Introduction
In classical mechanics, the Atwood machine consists of two objects of different masses connected by an inextensible string over a frictionless pulley. This is a classic application of Newton's Second Law.
The Problem
Given two masses, $m_1 = 7\text{ kg}$ and $m_2 = 12\text{ kg}$, find the acceleration $a$ of the system and the tension $T$ in the string.
Step-by-Step Solution
1. Free Body Diagrams (FBD)
- For $m_1$: Tension $T$ acts upward, gravity $m_1g$ acts downward. Since $m_2 > m_1$, $m_1$ moves upward. Equation: $T - m_1g = m_1a$ (1)
- For $m_2$: Tension $T$ acts upward, gravity $m_2g$ acts downward. $m_2$ moves downward. Equation: $m_2g - T = m_2a$ (2)
2. Solving for Acceleration ($a$)
Adding equation (1) and (2): $(T - m_1g) + (m_2g - T) = m_1a + m_2a$ $g(m_2 - m_1) = a(m_1 + m_2)$ $a = g \frac{m_2 - m_1}{m_1 + m_2}$
Plugging in values ($g \approx 9.8\text{ m/s}^2$): $a = 9.8 \cdot \frac{12 - 7}{12 + 7} = 9.8 \cdot \frac{5}{19} \approx 2.58\text{ m/s}^2$
3. Solving for Tension ($T$)
Substitute $a$ into equation (1): $T = m_1(g + a) = 7(9.8 + 2.58) = 7(12.38) \approx 86.66\text{ N}$
Conclusion
The system accelerates at $2.58\text{ m/s}^2$ and the string experiences a tension of approximately $86.66\text{ N}$.