Understanding the Atwood Machine
An Atwood machine consists of two objects of different masses ($m_1$ and $m_2$) connected by an inextensible string over a pulley. In this problem, we are given:
- Mass 1 ($m_1$): $7 \text{ kg}$
- Mass 2 ($m_2$): $12 \text{ kg}$
- Gravity ($g$): Approximately $9.8 \text{ m/s}^2$
Because $m_2 > m_1$, the heavier mass will accelerate downward, while the lighter mass will accelerate upward with the same magnitude of acceleration $a$. The tension $T$ in the string is uniform throughout.
Step 1: Free Body Diagrams
To solve this, we apply Newton’s Second Law ($F_{net} = ma$) to each mass individually.
For Mass 1 ($7 \text{ kg}$), moving upward:
- The forces acting on it are Tension ($T$) upward and weight ($m_1g$) downward.
- Equation: $T - m_1g = m_1a$ --- (Eq. 1)
For Mass 2 ($12 \text{ kg}$), moving downward:
- The forces are Weight ($m_2g$) downward and Tension ($T$) upward.
- Equation: $m_2g - T = m_2a$ --- (Eq. 2)
Step 2: Calculating Acceleration
To find the acceleration, add Equation 1 and Equation 2 to eliminate $T$:
$(T - m_1g) + (m_2g - T) = m_1a + m_2a$
$g(m_2 - m_1) = a(m_1 + m_2)$
$a = g \cdot \frac{m_2 - m_1}{m_1 + m_2}$
Plugging in the values:
$a = 9.8 \cdot \frac{12 - 7}{12 + 7} = 9.8 \cdot \frac{5}{19} \approx 2.58 \text{ m/s}^2$
Step 3: Calculating Tension
Now, substitute the value of $a$ back into Equation 1 to solve for $T$:
$T = m_1(g + a)$
$T = 7 \cdot (9.8 + 2.58) = 7 \cdot 12.38 = 86.66 \text{ N}$
Summary of Results
- Acceleration of masses: $\approx 2.58 \text{ m/s}^2$
- Tension in the string: $\approx 86.66 \text{ N}$
These values represent the dynamics of the system the instant the masses are released.