$2.1.31.$ The cargo system shown in the figure is located on a smooth horizontal table. The lower right weight is pulled along the table with a force of $F$, as indicated in the figure. The coefficient of friction between loads of mass $m_1$ and $m_2$ is equal to $\mu$. Find the acceleration of all loads in the system.
### Solution
Let's renumber the weights as shown in the figure, and direct the $X$ axis to the right.
It is clear that then none of the weights can have negative acceleration.
Let's prove that weights $3$ and $4$ move as a single whole. To do this, let's assume the opposite: let weight $3$ slide on weight $4$. Then a friction force arises between them
$$
F_{fr}= \mu mg
$$
and elastic force arises in the thread
$$
T> \mu mg
$$
In this case, the acceleration of load $2$ would be directed to the left, which is impossible. Therefore, the accelerations of loads $2$, $3$, and $4$ are the same.
Let us denote the acceleration of these loads as $\bar{a}_{1}=\bar{a}_{2}$, and the acceleration of load $1$ as $\bar{a}_{1}$.
Now consider two cases.
Case 1. Let loads $1$ and $2$ be at relative rest and $\bar{a}_{1}=\bar{a}_{2}$.
Let us denote the modulus of the static friction force between them as $F_1$, the modulus of the friction force between loads $3$ and $4$ as $F_2$, and the modulus of the elastic force of the thread as $T$ .
The same result can be obtained in another way. Since the friction between all surfaces is static friction, the system of loads moves as one body with mass $M=2(M+m)$.
Let load $2$ slide on load $1$. Then the friction force acts on load $1$
$$
F_{fr}^{\prime}= \mu mg
$$
and this load receives acceleration $a_{1}=\frac{F- \mu mg}{m}$.
The system of loads $2$, $3$ and $4$ moves as one body, the mass of which is $M_{0}=2m+M$ with acceleration
$$
a_{2}=\frac{\mu mg}{2m+M}
$$
The first case is realized if
$$
@@ -94,17 +94,18 @@Solution
F \geq\frac{2 \mu m (m+M)g}{2m+M}
$$
+
### Альтернативное решение:
+
+
+
#### Answer
For $F\leqslant\frac{2\mu m_{1}g(m_{1}+m_{2})}{m_{2}+2m_{1}}\equiv F_{0}$ we get $a_{1\text{sing}}=a_{1\text{right}}=a_{2\text{right}}=\frac{F}{2(m_{1}+m_{2})}$
For $F\geqslant F_{0}$ we get $a_{2\text{right}}=\frac{F-\mu m_{1}g}{m_{2}}$, $a_{1\text{left}}=a_{1\text{right}}=a_{2\text{left}}=\frac{\mu m_{1}g}{m_{2}+2m_{1}}$
$2.1.31.$ The cargo system shown in the figure is located on a smooth horizontal table. The lower right weight is pulled along the table with a force of $F$, as indicated in the figure. The coefficient of friction between loads of mass $m_1$ and $m_2$ is equal to $\mu$. Find the acceleration of all loads in the system.
$2.1.31.$ The cargo system shown in the figure is located on a smooth horizontal table. The lower right weight is pulled along the table with a force of $F$, as indicated in the figure. The coefficient of friction between loads of mass $m_1$ and $m_2$ is equal to $\mu$. Find the acceleration of all loads in the system.
### Solution
### Solution
Let's renumber the weights as shown in the figure, and direct the $X$ axis to the right.
Let's renumber the weights as shown in the figure, and direct the $X$ axis to the right.
It is clear that then none of the weights can have negative acceleration.
It is clear that then none of the weights can have negative acceleration.
Let's prove that weights $3$ and $4$ move as a single whole. To do this, let's assume the opposite: let weight $3$ slide on weight $4$. Then a friction force arises between them
Let's prove that weights $3$ and $4$ move as a single whole. To do this, let's assume the opposite: let weight $3$ slide on weight $4$. Then a friction force arises between them
$$
$$
F_{fr}= \mu mg
F_{fr}= \mu mg
$$
$$
and elastic force arises in the thread
and elastic force arises in the thread
$$
$$
T> \mu mg
T> \mu mg
$$
$$
In this case, the acceleration of load $2$ would be directed to the left, which is impossible. Therefore, the accelerations of loads $2$, $3$, and $4$ are the same.
In this case, the acceleration of load $2$ would be directed to the left, which is impossible. Therefore, the accelerations of loads $2$, $3$, and $4$ are the same.
Let us denote the acceleration of these loads as $\bar{a}_{1}=\bar{a}_{2}$, and the acceleration of load $1$ as $\bar{a}_{1}$.
Let us denote the acceleration of these loads as $\bar{a}_{1}=\bar{a}_{2}$, and the acceleration of load $1$ as $\bar{a}_{1}$.
Now consider two cases.
Now consider two cases.
Case 1. Let loads $1$ and $2$ be at relative rest and $\bar{a}_{1}=\bar{a}_{2}$.
Case 1. Let loads $1$ and $2$ be at relative rest and $\bar{a}_{1}=\bar{a}_{2}$.
Let us denote the modulus of the static friction force between them as $F_1$, the modulus of the friction force between loads $3$ and $4$ as $F_2$, and the modulus of the elastic force of the thread as $T$ .
Let us denote the modulus of the static friction force between them as $F_1$, the modulus of the friction force between loads $3$ and $4$ as $F_2$, and the modulus of the elastic force of the thread as $T$ .
The same result can be obtained in another way. Since the friction between all surfaces is static friction, the system of loads moves as one body with mass $M=2(M+m)$.
The same result can be obtained in another way. Since the friction between all surfaces is static friction, the system of loads moves as one body with mass $M=2(M+m)$.
Let load $2$ slide on load $1$. Then the friction force acts on load $1$
Let load $2$ slide on load $1$. Then the friction force acts on load $1$
$$
$$
F_{fr}^{\prime}= \mu mg
F_{fr}^{\prime}= \mu mg
$$
$$
and this load receives acceleration $a_{1}=\frac{F- \mu mg}{m}$.
and this load receives acceleration $a_{1}=\frac{F- \mu mg}{m}$.
The system of loads $2$, $3$ and $4$ moves as one body, the mass of which is $M_{0}=2m+M$ with acceleration
The system of loads $2$, $3$ and $4$ moves as one body, the mass of which is $M_{0}=2m+M$ with acceleration
$$
$$
a_{2}=\frac{\mu mg}{2m+M}
a_{2}=\frac{\mu mg}{2m+M}
$$
$$
The first case is realized if
The first case is realized if
$$
$$
@@ -94,17 +94,18 @@Solution
F \geq\frac{2 \mu m (m+M)g}{2m+M}
F \geq\frac{2 \mu m (m+M)g}{2m+M}
$$
$$
### Альтернативное решение:
#### Answer
#### Answer
For $F\leqslant\frac{2\mu m_{1}g(m_{1}+m_{2})}{m_{2}+2m_{1}}\equiv F_{0}$ we get $a_{1\text{sing}}=a_{1\text{right}}=a_{2\text{right}}=\frac{F}{2(m_{1}+m_{2})}$
For $F\leqslant\frac{2\mu m_{1}g(m_{1}+m_{2})}{m_{2}+2m_{1}}\equiv F_{0}$ we get $a_{1\text{sing}}=a_{1\text{right}}=a_{2\text{right}}=\frac{F}{2(m_{1}+m_{2})}$
For $F\geqslant F_{0}$ we get $a_{2\text{right}}=\frac{F-\mu m_{1}g}{m_{2}}$, $a_{1\text{left}}=a_{1\text{right}}=a_{2\text{left}}=\frac{\mu m_{1}g}{m_{2}+2m_{1}}$
For $F\geqslant F_{0}$ we get $a_{2\text{right}}=\frac{F-\mu m_{1}g}{m_{2}}$, $a_{1\text{left}}=a_{1\text{right}}=a_{2\text{left}}=\frac{\mu m_{1}g}{m_{2}+2m_{1}}$
