Problem — Atwood Machine (Symbolic Solution)

Problem

Two masses $m_1 > m_2$ are connected by an inextensible rope that passes over an ideal pulley (Atwood machine). Find symbolically the acceleration of the system and the tension in the rope. Comment on the limiting cases $m_1 = m_2$ and $m_2 \to 0$.

Solution

Step 1 — Choice of directions. The rope is inextensible and the pulley is ideal (massless and frictionless): the magnitude of the acceleration $a$ is the same for both masses and the tension $T$ is the same throughout the rope. Since $m_1 > m_2$, mass $m_1$ descends and $m_2$ rises. We take the direction of motion of each mass as positive.

Step 2 — Newton’s second law for each mass.

$m_1:\quad m_1 g - T = m_1 a$

$m_2:\quad T - m_2 g = m_2 a$

Step 3 — Acceleration. Adding the two equations, the tension cancels out:

$(m_1 - m_2)\,g = (m_1 + m_2)\,a$

$\ev{a = \dfrac{m_1 - m_2}{m_1 + m_2}\,g}$

Step 4 — Tension. Substituting $a$ into the second equation, $T = m_2(g + a)$:

$T = m_2 g\left(1 + \frac{m_1 - m_2}{m_1 + m_2}\right) = m_2 g \cdot \frac{2 m_1}{m_1 + m_2}$

$\ev{T = \dfrac{2\,m_1 m_2}{m_1 + m_2}\,g}$

Limiting cases.

• $m_1 = m_2$: this gives $a = 0$ and $T = m_1 g$. The system is balanced, remains in equilibrium (or moves at constant velocity), and the rope simply supports the weight of one mass. Consistent with intuition.
• $m_2 \to 0$: this gives $a \to g$ and $T \to 0$. With only one mass hanging, it falls in free fall, and the rope, having nothing to hold back on the other side, is no longer under tension. This too is physically reasonable.

Links

Topics: Dynamics Concepts: Newton’s second law · Tension Skills: Symbolic setup Objects: Atwood machine