Atwood's pulley: the force on the axle

Problem

Atwood’s pulley: mind the axle. On the ideal pulley of the previous problem we replace the numerical values with $m = 5\;\mathrm{kg}$ and $M = 10\;\mathrm{kg}$, leaving the system free to move. Question: what is the magnitude of the force that the pulley’s pin must exert upward to remain still? Compare the result with the value $(M+m)g$ that would apply if the system were locked (for example by locking the rope on the pulley): why does the pin bear less weight when the system moves? Adapted from (Povey 2015).

Solution

Acceleration of the Atwood machine. The two masses are connected by the ideal rope over the ideal pulley; $M$ descends, $m$ rises, with the same magnitude of acceleration: $a = \frac{(M-m)g}{M+m} = \frac{(10-5)\cdot 9{,}8}{15} \approx 3{,}3\;\mathrm{m/s^2}$

Rope tension. For the mass that rises (or descends) the same $T$ is obtained; using $m$: $T = m(g+a) = \frac{2Mm\,g}{M+m} = \frac{2\cdot 10\cdot 5\cdot 9{,}8}{15} \approx 65{,}3\;\mathrm{N}$

$\ev{T = \dfrac{2Mm\,g}{M+m} \approx 65{,}3\;\mathrm{N}}$

Force on the axle. The ideal rope, wrapped over the pulley, pulls the axle downward on both sides with the same tension $T$. For the pulley (ideal, of negligible mass) to remain in equilibrium, the pin must push upward with: $F_\text{axle} = 2T = \frac{4Mm\,g}{M+m} = \frac{4\cdot 10\cdot 5\cdot 9{,}8}{15}$

$\ev{F_\text{axle} = \dfrac{4Mm\,g}{M+m} \approx 131\;\mathrm{N}}$

Comparison with the locked system. If the rope were locked on the pulley, nothing would move: the pin would support the entire hanging weight, $(M+m)g = 15\cdot 9{,}8 = 147\;\mathrm{N}$ When instead the system is free, the pin bears only $\approx 131\;\mathrm{N}$, i.e. less than $147\;\mathrm{N}$.

Why the pin bears less weight when the system moves. Because the two masses accelerate: $M$ descends accelerating downward, $m$ rises accelerating upward. By Newton’s second law, the rope must exert a force smaller than the weight of $M$ (to let it accelerate downward) and greater than the weight of $m$ (to make it accelerate upward): the tension is the same $T$ on both sides and takes an intermediate value. The sum of the two weights is no longer balanced solely by the pin: part of the weight-force of $M$ “goes” into accelerating the masses instead of being transmitted to the axle. Formally: $(M+m)g - F_\text{axle} = (M+m)g - \frac{4Mm\,g}{M+m} = \frac{(M-m)^2 g}{M+m} = (M-m)\,a$ This difference, $(M-m)g - \dots$, is exactly the net force that accelerates the system. In the limiting case $M=m$ nothing moves and $F_\text{axle} = (M+m)g$: the pin bears the full weight only when the system is in equilibrium.

Links

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