Introduction
In classical mechanics, the Atwood machine is a classic setup used to demonstrate the application of Newton's Laws of Motion. It consists of two objects of different masses connected by a massless, inextensible rope passing over a frictionless, massless pulley.
The Physics Problem
We are given:
- Mass of the load ($m_1$) = 15 kg
- Mass of the counterweight ($m_2$) = 28 kg
- Gravity ($g$) = 9.8 m/s²
Since $m_2 > m_1$, the counterweight will accelerate downwards, pulling the 15 kg load upwards with the same magnitude of acceleration $a$. The tension $T$ in the rope is uniform throughout because the pulley is frictionless and massless.
Step-by-Step Solution
1. Free Body Diagram for the Load ($m_1$)
The forces acting on the 15 kg load are gravity ($m_1g$) acting downwards and tension ($T$) acting upwards. Since it accelerates upwards, the net force equation is: $T - m_1g = m_1a \quad \text{(Equation 1)}$
2. Free Body Diagram for the Counterweight ($m_2$)
The forces acting on the 28 kg counterweight are gravity ($m_2g$) acting downwards and tension ($T$) acting upwards. Since it accelerates downwards, the net force equation is: $m_2g - T = m_2a \quad \text{(Equation 2)}$
3. Solving for Acceleration ($a$)
Add Equation 1 and Equation 2 to eliminate $T$: $(T - m_1g) + (m_2g - T) = m_1a + m_2a$ $g(m_2 - m_1) = a(m_1 + m_2)$ $a = g \cdot \frac{m_2 - m_1}{m_2 + m_1}$
Substitute the values: $a = 9.8 \cdot \frac{28 - 15}{28 + 15} = 9.8 \cdot \frac{13}{43} \approx 2.96 \text{ m/s}^2$
4. Solving for Tension ($T$)
Rearrange Equation 1 to find $T$: $T = m_1(g + a)$ $T = 15(9.8 + 2.96) = 15(12.76) \approx 191.4 \text{ N}$
Summary of Results
- Acceleration of the load: 2.96 m/s² (upwards)
- Tension in the rope: 191.4 N