Q.Two blocks of masses $3\ \text{kg}$ and $5\ \text{kg}$ are connected by a light inextensible string passing over a frictionless, massless pulley, with both blocks hanging freely on either side (an Atwood machine). Find the acceleration of the system and the tension in the string. (Take $g = 9.8\ \text{m/s}^2$.)

Given: $m_1 = 3\ \text{kg}$, $m_2 = 5\ \text{kg}$, connected over a frictionless, massless pulley, $g = 9.8\ \text{m/s}^2$.

Acceleration. The heavier mass $m_2$ falls, and the lighter mass $m_1$ rises, both with the same magnitude of acceleration: $a = \frac{(m_2 - m_1)g}{m_1 + m_2} = \frac{(5-3) \times 9.8}{3+5} = \frac{2 \times 9.8}{8} = \frac{19.6}{8} = 2.45\ \text{m/s}^2$

Tension. Applying Newton's second law to $m_1$ (which accelerates upward, so tension exceeds its weight): $T - m_1 g = m_1 a \Rightarrow T = m_1(g+a) = 3 \times (9.8 + 2.45) = 3 \times 12.25 = 36.75\ \text{N}$ Checking with $m_2$ (which accelerates downward): $m_2 g - T = m_2 a \Rightarrow T = m_2(g-a) = 5 \times (9.8-2.45) = 5 \times 7.35 = 36.75\ \text{N}$, in agreement.

Use the standard Atwood-machine result $a = (m_2-m_1)g/(m_1+m_2)$, then find tension by applying Newton's second law to either mass and checking both give the same value.

• Using $(m_1+m_2)g$ in the numerator instead of $(m_2-m_1)g$, which ignores that the two weights partly cancel.
• Forgetting that the tension is neither mass's own weight alone, but a value strictly between the two weights, $m_1 g < T < m_2 g$.