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| <h3> Statement </h3> | | <h3> Statement </h3> |
| <p> | | <p> |
| $2.1.25.$ A thread spanned over a block with a fixed axis is passed through a slot. At the ends of the thread are suspended loads, the mass of which is $m_1$ and $m_2$. Determine the acceleration of loads if a constant frictional force $F_{tr}$ acts on the thread from the side of the slot when it moves. | | $2.1.25.$ A thread spanned over a block with a fixed axis is passed through a slot. At the ends of the thread are suspended loads, the mass of which is $m_1$ and $m_2$. Determine the acceleration of loads if a constant frictional force $F_{tr}$ acts on the thread from the side of the slot when it moves. |
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| For problem $2.1.25$ | | For problem $2.1.25$ |
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| <h3>Solution</h3> | | <h3>Solution</h3> |
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| loading="lazy" width="130" /> | | loading="lazy" width="130" /> |
| <figcaption> | | <figcaption> |
| Forces acting on loads | | Forces acting on loads |
| </figcaption> | | </figcaption> |
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| <p> | | <p> |
| 1. Due to the weightlessness and non-stretchability of the thread, as well as the ideal properties of the block (no losses and low weight), the problem can be solved in the following approximation) | | 1. Due to the weightlessness and non-stretchability of the thread, as well as the ideal properties of the block (no losses and low weight), the problem can be solved in the following approximation) |
| $$ a_1=a_2=a $$ | | $$ a_1=a_2=a $$ |
| | | |
| $$ T_1=T_2=T $$ | | $$ T_1=T_2=T $$ |
| 2. The equations of motion of loads in projection onto the vertical axis in this case are written as follows: | | 2. The equations of motion of loads in projection onto the vertical axis in this case are written as follows: |
| $$ \left\{\begin{matrix} | | $$ \left\{\begin{matrix} |
| m_1a = m_2g - T \\ | | m_1a = m_2g - T \\ |
| m_2a =T-m_2g-F_{fr} | | m_2a =T-m_2g-F_{fr} |
| \end{matrix}\right. $$ | | \end{matrix}\right. $$ |
| 3. Solving the equations together, we obtain | | 3. Solving the equations together, we obtain |
| $$ a=\frac{(m_1-m_2)g-F_{fr}}{m_1+m_2} $$ | | $$ a=\frac{(m_1-m_2)g-F_{fr}}{m_1+m_2} $$ |
| 4. Substituting the acceleration value into the first equation of the system allows us to determine the tension of the thread | | 4. Substituting the acceleration value into the first equation of the system allows us to determine the tension of the thread |
| $$ \boxed{T=m_1\frac{2m_2g+F_{fr}}{m_1+m_2}} $$ | | $$ \boxed{T=m_1\frac{2m_2g+F_{fr}}{m_1+m_2}} $$ |
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| <h4>Answer</h4> | | <h4>Answer</h4> |
| <p> | | <p> |
| $$T=m_1\frac{2m_2g+F_{fr}}{m_1+m_2}$$ | | $$T=m_1\frac{2m_2g+F_{fr}}{m_1+m_2}$$ |
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