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| <title>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.</title> | | <title>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.</title> |
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| <h3 id="back-link"><a href="../../#2.1">$\leftarrow$Back</a></h3> | | <h3 id="back-link"><a href="../../#2.1">$\leftarrow$Back</a></h3> |
| | | |
| <h3> Statement </h3> | | <h3> Statement </h3> |
| <p> | | <p> |
| $2.1.31.$ 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. |
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| <figcaption> | | <figcaption> |
| For problem $2.1.31$ | | For problem $2.1.31$ |
| </figcaption> | | </figcaption> |
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| </center> | | </center> |
| <p> | | <p> |
| </p> | | </p> |
| | | |
| <h3>Solution</h3> | | <h3>Solution</h3> |
| <p> | | <p> |
| <p> | | <p> |
| 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. |
| </p> | | </p> |
| <br> | | <br> |
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| <center> | | <center> |
| <figure> | | <figure> |
| <img src="https://savchenkosolutions.com/2/2.1.31/sol.png" | | <img src="https://savchenkosolutions.com/2/2.1.31/sol.png" |
| loading="lazy" width="350" /> | | loading="lazy" width="350" /> |
| <figcaption> | | <figcaption> |
| Forces acting on the weights | | Forces acting on the weights |
| </figcaption> | | </figcaption> |
| </figure> | | </figure> |
| </center> | | </center> |
| | | |
| <p> | | <p> |
| 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. |
| </p> | | </p> |
| <p> | | <p> |
| 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. |
| </p> | | </p> |
| <p> | | <p> |
| 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 $\bar{a}_{1}$. |
| </p> | | </p> |
| <p> | | <p> |
| Now consider two cases. | | Now consider two cases. |
| </p> | | </p> |
| | | |
| <p> | | <p> |
| 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 $\bar{a}_{1}=\bar{a}_{2}$. |
| </p> | | </p> |
| <p> | | <p> |
| 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 $T$ . </p> | | 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 $T$ . </p> |
| | | |
| <p> | | <p> |
| Then: for cargo $1$ | | Then: for cargo $1$ |
| $$ F-F_{1}=Ma_{1} $$ | | $$ F-F_{1}=Ma_{1} $$ |
| for cargo $2$ | | for cargo $2$ |
| $$ F_{1}-T=ma_{2} $$ | | $$ F_{1}-T=ma_{2} $$ |
| for cargo $3$ | | for cargo $3$ |
| $$ T-F_{2}=ma_{2} $$ | | $$ T-F_{2}=ma_{2} $$ |
| for cargo $4$ | | for cargo $4$ |
| $$ F_{2}=Ma_{2} $$ | | $$ F_{2}=Ma_{2} $$ |
| Solving this system of equations, we obtain: | | Solving this system of equations, we obtain: |
| $$ F_{1}=\frac{2m+M}{2(M+m)}F,a_{1}=a_{2}=\frac{F}{2(M+m)} $$ | | $$ F_{1}=\frac{2m+M}{2(M+m)}F,a_{1}=a_{2}=\frac{F}{2(M+m)} $$ |
| 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 $M=2(M+m)$. | | 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 $M=2(M+m)$. |
| </p> | | </p> |
| <p> | | <p> |
| Therefore | | Therefore |
| $$ \bar{F}=m \bar{a}_{1},\bar{a}_{1}=\bar{a}_{2}= \frac{\bar{F}}{2(M+m)} $$ | | $$ \bar{F}=m \bar{a}_{1},\bar{a}_{1}=\bar{a}_{2}= \frac{\bar{F}}{2(M+m)} $$ |
| Case $2$. | | Case $2$. |
| </p> | | </p> |
| <p> | | <p> |
| 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 $a_{1}=\frac{F- \mu mg}{m}$. |
| </p> | | </p> |
| <p> | | <p> |
| 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 |
| $$ F \geq \frac{2 \mu m (m+M)g}{2m+M} $$ | | $$ F \geq \frac{2 \mu m (m+M)g}{2m+M} $$ |
| </p> | | </p> |
| </p> | | </p> |
| | | |
| <h4>Answer</h4> | | <h4>Answer</h4> |
| <p> | | <p> |
| <p> | | <p> |
| 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 $a_{1\text{sing}}=a_{1\text{right}}=a_{2\text{right}}=\frac{F}{2(m_{1}+m_{2})}$ |
| </p> | | </p> |
| <p> | | <p> |
| For $F\geqslant F_{0}$ we get $a_{2\text{right}}=\frac{F-\mu m_{1}g}{m_{2}}$, $a_{1\text{left}}=a_{1\text{right}}=a_{2\text{left}}=\frac{\mu m_{1}g}{m_{2}+2m_{1}}$ </p> | | For $F\geqslant F_{0}$ we get $a_{2\text{right}}=\frac{F-\mu m_{1}g}{m_{2}}$, $a_{1\text{left}}=a_{1\text{right}}=a_{2\text{left}}=\frac{\mu m_{1}g}{m_{2}+2m_{1}}$ </p> |
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