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Tete edited Sep 3, 2026, 5:02 PM

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			### Statement

			$2.4.13.$ A smooth, uniform rope of length $l$ and mass $m$ is thrown over a small pulley so that it is initially in equilibrium. The rope is slightly displaced and it starts to slide off the pulley. With what force does it act on the pulley at the moment when the length of the rope on one side of it is equal to $\frac{l}{3}$?
		@@ -4,7 +4,7 @@Statement

			### Solution

		−	![For problem $2.4.13$ \|910x540, %](../../img/2.4.13/Savchenko.png)
		+	![For problem $2.4.13$ \|910x540, 80%](../../img/2.4.13/Savchenko.png)

			From the initial moment (left figure) to the moment in question (middle figure), the rope of length $l/6$ moves from the left part to the right part (assuming that the pulley is small). The change in gravitational potential energy is that of the rope of mass $m/6$ moving from the red section to the green section in the middle figure, i.e.

			\[\frac{m}{6}\cdot g\cdot\left(-\frac{l}{6}\right)=-\frac{mgl}{36}.\]

			If $v$ is the speed of the rope, then conservation of energy yields

			\[\frac{1}{2}mv^2-\frac{mgl}{36}=0\qquad\Rightarrow\qquad v^2=\frac{gl}{18}.\]

			In a short time interval $\Delta t$ afterward, the rope of length $v\Delta t$ and mass $\Delta m=mv\Delta t/l$ moves from the left part to the right part. The change in gravitational potential energy is that of the rope of mass $\Delta m$ moving from the red section to the green section in the right figure, i.e.

			\[\Delta m\cdot g\cdot\left(-\frac{l}{3}-v\Delta t\right)\approx-\frac{mgv\Delta t}{3}.\]

			Conservation of energy yields

			\[\frac{1}{2}m(v+\Delta v)^2-\frac{1}{2}mv^2-\frac{mgv\Delta t}{3}=0\qquad\Rightarrow\qquad\Delta v\approx\frac{g\Delta t}{3}.\]

			Consequently, the change in downward momentum of the right part of the rope is

			\[\frac{2m}{3}\cdot\Delta v+\Delta m\cdot v=\frac{5}{18}\cdot mg\Delta t,\]

			and the change in upward momentum of the left part of the rope is

			\[\frac{m}{3}\cdot\Delta v-\Delta m\cdot v=\frac{1}{18}\cdot mg\Delta t.\]

			Thus, the downward impulse on the entire rope is $2mg\Delta t/9$, which means that the total force on the rope is $2mg/9$. Since gravity is $mg$, the upward normal force from the pulley is $7mg/9$, which has the same magnitude as the force that the rope exerts on the pulley.

			#### Answer

			$\frac{7mg}{9}$
		unchanged lines 27

	### Statement		### Statement

	$2.4.13.$ A smooth, uniform rope of length $l$ and mass $m$ is thrown over a small pulley so that it is initially in equilibrium. The rope is slightly displaced and it starts to slide off the pulley. With what force does it act on the pulley at the moment when the length of the rope on one side of it is equal to $\frac{l}{3}$?		$2.4.13.$ A smooth, uniform rope of length $l$ and mass $m$ is thrown over a small pulley so that it is initially in equilibrium. The rope is slightly displaced and it starts to slide off the pulley. With what force does it act on the pulley at the moment when the length of the rope on one side of it is equal to $\frac{l}{3}$?
	@@ -4,7 +4,7 @@Statement

	### Solution		### Solution

	![For problem $2.4.13$ \|910x540, %](../../img/2.4.13/Savchenko.png)		![For problem $2.4.13$ \|910x540, 80%](../../img/2.4.13/Savchenko.png)

	From the initial moment (left figure) to the moment in question (middle figure), the rope of length $l/6$ moves from the left part to the right part (assuming that the pulley is small). The change in gravitational potential energy is that of the rope of mass $m/6$ moving from the red section to the green section in the middle figure, i.e.		From the initial moment (left figure) to the moment in question (middle figure), the rope of length $l/6$ moves from the left part to the right part (assuming that the pulley is small). The change in gravitational potential energy is that of the rope of mass $m/6$ moving from the red section to the green section in the middle figure, i.e.

	\[\frac{m}{6}\cdot g\cdot\left(-\frac{l}{6}\right)=-\frac{mgl}{36}.\]		\[\frac{m}{6}\cdot g\cdot\left(-\frac{l}{6}\right)=-\frac{mgl}{36}.\]

	If $v$ is the speed of the rope, then conservation of energy yields		If $v$ is the speed of the rope, then conservation of energy yields

	\[\frac{1}{2}mv^2-\frac{mgl}{36}=0\qquad\Rightarrow\qquad v^2=\frac{gl}{18}.\]		\[\frac{1}{2}mv^2-\frac{mgl}{36}=0\qquad\Rightarrow\qquad v^2=\frac{gl}{18}.\]

	In a short time interval $\Delta t$ afterward, the rope of length $v\Delta t$ and mass $\Delta m=mv\Delta t/l$ moves from the left part to the right part. The change in gravitational potential energy is that of the rope of mass $\Delta m$ moving from the red section to the green section in the right figure, i.e.		In a short time interval $\Delta t$ afterward, the rope of length $v\Delta t$ and mass $\Delta m=mv\Delta t/l$ moves from the left part to the right part. The change in gravitational potential energy is that of the rope of mass $\Delta m$ moving from the red section to the green section in the right figure, i.e.

	\[\Delta m\cdot g\cdot\left(-\frac{l}{3}-v\Delta t\right)\approx-\frac{mgv\Delta t}{3}.\]		\[\Delta m\cdot g\cdot\left(-\frac{l}{3}-v\Delta t\right)\approx-\frac{mgv\Delta t}{3}.\]

	Conservation of energy yields		Conservation of energy yields

	\[\frac{1}{2}m(v+\Delta v)^2-\frac{1}{2}mv^2-\frac{mgv\Delta t}{3}=0\qquad\Rightarrow\qquad\Delta v\approx\frac{g\Delta t}{3}.\]		\[\frac{1}{2}m(v+\Delta v)^2-\frac{1}{2}mv^2-\frac{mgv\Delta t}{3}=0\qquad\Rightarrow\qquad\Delta v\approx\frac{g\Delta t}{3}.\]

	Consequently, the change in downward momentum of the right part of the rope is		Consequently, the change in downward momentum of the right part of the rope is

	\[\frac{2m}{3}\cdot\Delta v+\Delta m\cdot v=\frac{5}{18}\cdot mg\Delta t,\]		\[\frac{2m}{3}\cdot\Delta v+\Delta m\cdot v=\frac{5}{18}\cdot mg\Delta t,\]

	and the change in upward momentum of the left part of the rope is		and the change in upward momentum of the left part of the rope is

	\[\frac{m}{3}\cdot\Delta v-\Delta m\cdot v=\frac{1}{18}\cdot mg\Delta t.\]		\[\frac{m}{3}\cdot\Delta v-\Delta m\cdot v=\frac{1}{18}\cdot mg\Delta t.\]

	Thus, the downward impulse on the entire rope is $2mg\Delta t/9$, which means that the total force on the rope is $2mg/9$. Since gravity is $mg$, the upward normal force from the pulley is $7mg/9$, which has the same magnitude as the force that the rope exerts on the pulley.		Thus, the downward impulse on the entire rope is $2mg\Delta t/9$, which means that the total force on the rope is $2mg/9$. Since gravity is $mg$, the upward normal force from the pulley is $7mg/9$, which has the same magnitude as the force that the rope exerts on the pulley.

	#### Answer		#### Answer

	$\frac{7mg}{9}$		$\frac{7mg}{9}$
	unchanged lines 27