Atwood machine (numerical values)

Problem

Atwood machine: two masses $m_1 = 3{,}0\;\text{kg}$ and $m_2 = 4{,}0\;\text{kg}$ hang from the two sides of an ideal pulley (massless, frictionless). Determine the acceleration of the system and the tension in the rope.

Solution

Step 1 — Free body of the two masses. The ideal pulley merely redirects the rope without changing its tension, so the rope pulls both masses with the same tension $T$. Since $m_2 > m_1$, mass $m_2$ descends and $m_1$ rises, with the same magnitude of acceleration $a$ (inextensible rope). We write the second law taking the direction of motion as positive for each mass.

For $m_1$ (rises, driving force $T$, braking weight):

$T - m_1 g = m_1 a$

For $m_2$ (descends, driving weight, braking tension):

$m_2 g - T = m_2 a$

Step 2 — Acceleration. Adding the two equations term by term eliminates $T$:

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

$a = \frac{(m_2 - m_1)\,g}{m_1 + m_2} = \frac{(4{,}0 - 3{,}0)\cdot 9{,}81}{3{,}0 + 4{,}0} = \frac{9{,}81}{7{,}0}$

$\ev{a \approx 1{,}4\;\text{m/s}^2}$

Step 3 — Tension. We substitute $a$ into the equation for $m_1$, or use the symbolic formula $T = \dfrac{2 m_1 m_2 g}{m_1 + m_2}$:

$T = \frac{2 \cdot 3{,}0 \cdot 4{,}0 \cdot 9{,}81}{3{,}0 + 4{,}0} = \frac{235{,}4}{7{,}0}$

$\ev{T \approx 33{,}6\;\text{N}}$

Sanity check. The tension $33{,}6\;\text{N}$ lies between the two weights $P_1 = 29{,}4\;\text{N}$ and $P_2 = 39{,}2\;\text{N}$, as it must: the rope must pull $m_1$ harder than its own weight (to make it rise) and $m_2$ less than its own weight (to make it descend).

Links

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