The Atwood machine

Problem

A cat ($6\;\text{kg}$) and a giraffe ($66\;\text{kg}$) are connected by a rope passing over an ideal pulley. Find the acceleration and the tension in the rope.

Atwood machine: the heavy body (giraffe) descends, the light one (cat) rises; the tension $\vec{t}$ is the same on both branches of the rope.

Solution

Data. $m_g = 6\;\text{kg}$ (cat, light), $m_G = 66\;\text{kg}$ (giraffe, heavy).

Constraints. Ideal rope and pulley: same tension $t$ on both branches and same magnitude of acceleration $a$. The heavy body descends, the light one rises.

Light body (cat, rises) — axis pointing upward positive: $-m_g g + t = m_g a \quad\Rightarrow\quad -58{,}9 + t = 6a$ with $m_g g = 6 \cdot 9{,}81 = 58{,}9\;\text{N}$.

Heavy body (giraffe, descends) — same axis pointing upward, acceleration $-a$: $-m_G g + t = -m_G a \quad\Rightarrow\quad -647 + t = -66a$ with $m_G g = 66 \cdot 9{,}81 = 647\;\text{N}$.

Subtracting the first from the second (the tension cancels): $-647 + 58{,}9 = -72a \quad\Rightarrow\quad -588 = -72a \quad\Rightarrow\quad a = 8{,}2\;\text{m/s}^2$

Tension (from the first equation): $t = 58{,}9 + 6 \cdot 8{,}2 = 58{,}9 + 49{,}2 = 108\;\text{N}$

Result. $\ev{a = \frac{m_G - m_g}{m_G + m_g}\,g = 8{,}2\;\text{m/s}^2, \qquad t \approx 108\;\text{N}}$

Linked items

Topics: Dinamica Concepts: Seconda legge di Newton · Tensione · Forza peso Skills: Applicazione delle leggi di Newton · Diagramma di corpo libero Objects: Macchina di Atwood · Puleggia · Fune ideale