Understanding the Atwood Machine
An Atwood machine consists of two objects of different masses connected by a light, inextensible string passing over a pulley. Because the string is inextensible, both masses move with the same magnitude of acceleration, $a$. Since the pulley is light and frictionless, the tension $T$ in the string is uniform throughout.
Given Data
- Mass 1 ($m_1$) = 7 kg
- Mass 2 ($m_2$) = 12 kg
- Acceleration due to gravity ($g$) = 9.8 m/s²
Free-Body Diagram Analysis
To solve this, we define the direction of motion. Since $m_2 > m_1$, the system will accelerate such that $m_2$ moves downward and $m_1$ moves upward.
For $m_1$ (7 kg):
- Forces acting: Tension ($T$) upward, Weight ($m_1g$) downward.
- Equation of motion: $T - m_1g = m_1a$ --- (Eq. 1)
For $m_2$ (12 kg):
- Forces acting: Tension ($T$) upward, Weight ($m_2g$) downward.
- Equation of motion: $m_2g - T = m_2a$ --- (Eq. 2)
Solving for Acceleration ($a$)
Add Eq. 1 and Eq. 2 together to eliminate $T$: $(T - m_1g) + (m_2g - T) = m_1a + m_2a$ $g(m_2 - m_1) = a(m_1 + m_2)$ $a = g \frac{m_2 - m_1}{m_1 + m_2}$
Plugging in the values: $a = 9.8 \times \frac{12 - 7}{12 + 7} = 9.8 \times \frac{5}{19} \approx 2.58 \text{ m/s}^2$
Solving for Tension ($T$)
Use Eq. 1 to solve for $T$: $T = m_1(g + a)$ $T = 7(9.8 + 2.58) = 7(12.38) \approx 86.66 \text{ N}$
Summary
- Acceleration: ~2.58 m/s²
- Tension: ~86.66 N
This method allows you to systematically decompose complex mechanical systems into individual components, making it a cornerstone of classical mechanics.