Two bodies of equal mass m hang on a rope over a pulley. An…
Two bodies of equal mass $m$ hang on a rope over a pulley. An extra mass $\Delta m$ is placed on one of them. Find (1) the acceleration, (2) the rope tension, (3) the force with which the added mass presses on the body it lies on, (4) the force on the pulley's axle. The pulley is massless and frictionless.
Explanation of Method
Atwood machine with an added mass.
Step 1: Acceleration
\[ (m + \Delta m)g - T = (m + \Delta m)a, \]
\[ T - mg = ma. \]
\[a = \frac{\Delta m}{2m + \Delta m}\,g\]
Step 2: Tension
\[T = m\,(g + a) = \frac{2m\,(m + \Delta m)}{2m + \Delta m}\,g\]
Step 3: Force of the added mass on the body
The added mass moves down with acceleration $a$ under its weight and the upward push $N$ of the body below it: $\Delta m\,g - N = \Delta m\,a \;\Longrightarrow\; N = \Delta m\,(g - a)$.
\[N = \frac{2m\,\Delta m}{2m + \Delta m}\,g \ \ (\text{less than its weight } \Delta m\,g)\]
Step 4: Force on the axle
Both rope ends pull the pulley straight down with $T$: $F_{\text{axle}} = 2T = \frac{4m\,(m + \Delta m)}{2m + \Delta m}\,g$.
Check
$F_{\text{axle}} \lt (2m + \Delta m)g$: the axle carries less than the total weight while the system accelerates, since $(2m + \Delta m)^2 - 4m(m + \Delta m) = \Delta m^2 \gt 0\ \checkmark$
Result
\[\boxed{\,a = \frac{\Delta m}{2m + \Delta m}\,g\,}\]
\[\boxed{\,T = m\,(g + a) = \frac{2m\,(m + \Delta m)}{2m + \Delta m}\,g\,}\]
\[\boxed{\,N = \frac{2m\,\Delta m}{2m + \Delta m}\,g \ \ (\text{less than its weight } \Delta m\,g)\,}\]
\[\boxed{\,F_{\text{axle}} = 2T = \frac{4m\,(m + \Delta m)}{2m + \Delta m}\,g\,}\]
Remark
About the original solution. Parts 1, 2 and 4 are correct (its forms of $T$ and $F_{\text{axle}}$ are equivalent to the ones above). In part 3 it gave the full weight $\Delta m\,g$; but the added mass is accelerating downward, so the body under it only has to push up with $\Delta m(g - a)$, and by Newton's third law that is the force the added mass exerts on it.