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| <h3 id="back-link"><a href="../../#2.2">$\leftarrow$Back</a></h3> | | <h3 id="back-link"><a href="../../#2.2">$\leftarrow$Back</a></h3> |
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| <h3> Statement </h3> | | <h3> Statement </h3> |
| | | |
| <p> | | <p> |
| $2.2.24$ | | $2.2.24$ |
| A monkey of mass $m$ is balanced by a counterweight on block $A$. Block $A$ is balanced by a weight of $2m$ on block $B$. The system is stationary. How will the load move if the monkey starts to evenly select the rope at a speed $u$ relative to itself? Ignore the mass of blocks and friction | | A monkey of mass $m$ is balanced by a counterweight on block $A$. Block $A$ is balanced by a weight of $2m$ on block $B$. The system is stationary. How will the load move if the monkey starts to evenly select the rope at a speed $u$ relative to itself? Ignore the mass of blocks and friction |
| </p> | | </p> |
| <center> | | <center> |
| <figure> | | <figure> |
| <img src="statement.png" | | <img src="statement.png" |
| loading="lazy" width="120" /> | | loading="lazy" width="120" /> |
| <figcaption> | | <figcaption> |
| For problem 2.2.24 | | For problem 2.2.24 |
| </figcaption> | | </figcaption> |
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| <h3>Solution</h3> | | <h3>Solution</h3> |
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| <figure> | | <figure> |
| <img src="draw.png" | | <img src="draw.png" |
| loading="lazy" width="180" /> | | loading="lazy" width="180" /> |
| <figcaption> | | <figcaption> |
| Forces acting on the system | | Forces acting on the system |
| </figcaption> | | </figcaption> |
| </figure> | | </figure> |
| </center> | | </center> |
| <p> | | <p> |
| Newton's second law of motion for y-axis: | | Newton's second law of motion for y-axis: |
| $$ | | $$ |
| \begin{cases} | | \begin{cases} |
| 2T - 2mg = 2ma_{1y} \\ | | 2T - 2mg = 2ma_{1y} \\ |
| T - mg = ma_{2y} \\ | | T - mg = ma_{2y} \\ |
| T - mg = ma_{3y} | | T - mg = ma_{3y} |
| \end{cases} | | \end{cases} |
| $$ | | $$ |
| From these three equations, it's evident that: | | From these three equations, it's evident that: |
| $$ | | $$ |
| a_1 = a_{2y} = a_{3y}$$ | | a_1 = a_{2y} = a_{3y}$$ |
| $$\quad \int_0^{v_{1y}} dv_{1y} = \int_0^{v_{2y}} dv_{2y} = \int_0^{v_{3y}} dv_{3y}$$ | | $$\quad \int_0^{v_{1y}} dv_{1y} = \int_0^{v_{2y}} dv_{2y} = \int_0^{v_{3y}} dv_{3y}$$ |
| $$v_{1y} = v_{2y} = v_{3y} | | $$v_{1y} = v_{2y} = v_{3y} |
| $$ | | $$ |
| The lengths of two ropes: | | The lengths of two ropes: |
| $$ | | $$ |
| \begin{cases} | | \begin{cases} |
| L_1 = 2y_b - y_1 - y_a \\ | | L_1 = 2y_b - y_1 - y_a \\ |
| L_2 = 2y_A - y_2 - y_k | | L_2 = 2y_A - y_2 - y_k |
| \end{cases} | | \end{cases} |
| $$ | | $$ |
| Taking the derivative of the rope lengths with respect to time, we obtain these relationships: | | Taking the derivative of the rope lengths with respect to time, we obtain these relationships: |
| $$ | | $$ |
| \begin{cases} | | \begin{cases} |
| 0 = 2v_{by} - v_{1y} - v_{Ay} \\ | | 0 = 2v_{by} - v_{1y} - v_{Ay} \\ |
| 0 = 2v_{Ay} - v_{2y} - v_{ky} | | 0 = 2v_{Ay} - v_{2y} - v_{ky} |
| \end{cases} | | \end{cases} |
| $$ | | $$ |
| From velocity-addition formula: | | From velocity-addition formula: |
| $$ | | $$ |
| \vec{v} = \vec{v}^\prime + \vec{u}$$ | | \vec{v} = \vec{v}^\prime + \vec{u}$$ |
| $ | | $ |
| v_{ky} = v_{3y} + u_y$ from the conditions of the problem | | v_{ky} = v_{3y} + u_y$ from the conditions of the problem |
| </p> | | </p> |
| <p> | | <p> |
| Eventually: | | Eventually: |
| $$ | | $$ |
| v_{1y} = -v_Ay \quad (v_{by} = 0)$$ | | v_{1y} = -v_Ay \quad (v_{by} = 0)$$ |
| $$ | | $$ |
| 2v_{1y} + v_{2y} + v_{ky} = 0$$ | | 2v_{1y} + v_{2y} + v_{ky} = 0$$ |
| $$ | | $$ |
| 2v_{1y} + v_{2y} + v_{3y} + u_y = 0 | | 2v_{1y} + v_{2y} + v_{3y} + u_y = 0 |
| $$ | | $$ |
| $$ | | $$ |
| 4v_{1y} = -u_y$$ | | 4v_{1y} = -u_y$$ |
| $$ | | $$ |
| v_{1y} = -\frac{u_y}{4} | | v_{1y} = -\frac{u_y}{4} |
| $$ | | $$ |
| </p> | | </p> |
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| <p style="text-align: right; font-style: italic; font-size: 16;">physicshub</p> | | <p style="text-align: right; font-style: italic; font-size: 16;">physicshub</p> |
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