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| + | <meta name="author" content="Aliaksandr Melnichenka"> | ||
| + | <meta name="date" content="2023-10" scheme="YYYY-MM"> | ||
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| + | <title>The body with wedges installed in its cutouts is located between two parallel walls as shown in the figure. Find the limit angle at the top of the wedges at which the body can move to the right and cannot move to the left. The coefficients of friction of the wedges against the walls and body are equal to \mu_1 and \mu_2, respectively.</title> | ||
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| + | <header style="text-align:center;"> | ||
| + | <h2>Solutions of Savchenko Problems in Physics</h2> | ||
| + | <p class="author"> | ||
| + | Aliaksandr Melnichenka <br/> | ||
| + | October 2023 | ||
| + | </p> | ||
| + | </header> | ||
| + | |||
| + | <h3 id="back-link"><a href="../../#2.8">$\leftarrow$Back</a></h3> | ||
| + | |||
| + | <h3> Statement </h3> | ||
| + | <p> | ||
| + | $2.8.14^*.$ The body with wedges installed in its cutouts is located between two parallel walls as shown in the figure. Find the limit angle at the top of the wedges at which the body can move to the right and cannot move to the left. The coefficients of friction of the wedges against the walls and body are equal to $\mu_1$ and $\mu_2$, respectively. | ||
| + | </p> | ||
| + | <center> | ||
| + | <figure> | ||
| + | <img src="2.8.14.png" | ||
| + | loading="lazy" width="200" /> | ||
| + | <figcaption> | ||
| + | For problem $2.8.14^*$ | ||
| + | </figcaption> | ||
| + | </figure> | ||
| + | </center> | ||
| + | <p> | ||
| + | </p> | ||
| + | |||
| + | <h3>Solution</h3> | ||
| + | <p> | ||
| + | </p> | ||
| + | <center> | ||
| + | <figure> | ||
| + | <img src="2.8.14_1.png" | ||
| + | loading="lazy" width="250" /> | ||
| + | <figcaption> | ||
| + | Forces acting on the wedge | ||
| + | </figcaption> | ||
| + | </figure> | ||
| + | </center> | ||
| + | <p> | ||
| + | |||
| + | State of forces equilibrium for the wedge for the horizontal axis: | ||
| + | $$N_3+\mu_1N_1=\left(\mu_2\cos\alpha+\sin\alpha\right)N_2$$ | ||
| + | In the set mode, the force that will act on the left wall will be equal to zero ($N_3=0$) | ||
| + | $$\frac{N_1}{N_2}=\frac{\mu_2\cos\alpha+\sin\alpha}{\mu_1}\quad(1)$$ | ||
| + | Similarly, for the vertical axis | ||
| + | $$N_1+\mu_2N_2\sin\alpha=N_2\cos\alpha$$ | ||
| + | $$\frac{N_1}{N_2}=\cos\alpha-\mu_2\sin\alpha\quad(2)$$ | ||
| + | Substituing $(2)$ into $(1)$ | ||
| + | $$\mu_1\cos\alpha-\mu_1\mu_2\sin\alpha=\mu_2\cos\alpha+\sin\alpha$$ | ||
| + | Divide the left and right parts by $\cos\alpha$ | ||
| + | $$\left(1+\mu_1\mu_2\right)\tan\alpha=\mu_1-\mu_2$$ | ||
| + | Whence the required angle | ||
| + | $$\boxed{\tan\alpha=\frac{{\mu}_{1}-\mu_{2}}{1+\mu_{1}\mu_{2}}}$$ | ||
| + | </p> | ||
| + | |||
| + | <h4>Answer</h4> | ||
| + | <p> | ||
| + | $$\operatorname{tg}\alpha=({\mu}_{1}-\mu_{2})/(1+\mu_{1}\mu_{2})$$ | ||
| + | </p> | ||
| + | <p style="text-align: right; font-style: italic; font-size: 14;"> | ||
| + | Almaskhan Arsen<br> | ||
| + | </p> | ||
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| <meta charset="utf-8"> | |||
| <meta name="viewport" content="width=device-width, initial-scale=1.0"> | |||
| <meta http-equiv="content-language" content="en"> | |||
| <meta name="keywords" content="Savchenko Problems in Physics, Savchenko solutions, physics problems, physics olympiad preparation, IPhO, Jaan Kalda"> | |||
| <meta name="description" content="The body with wedges installed in its cutouts is located between two parallel walls as shown in the figure. Find the limit angle at the top of the wedges at which the body can move to the right and cannot move to the left. The coefficients of friction of the wedges against the walls and body are equal to \mu_1 and \mu_2, respectively."> | |||
| <meta name="author" content="Aliaksandr Melnichenka"> | |||
| <meta name="date" content="2023-10" scheme="YYYY-MM"> | |||
| <meta property="og:title" content="The body with wedges installed in its cutouts is located between two parallel walls as shown in the figure. Find the limit angle at the top of the wedges at which the body can move to the right and cannot move to the left. The coefficients of friction of the wedges against the walls and body are equal to \mu_1 and \mu_2, respectively."> | |||
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| </head> | |||
| <body style=""> | |||
| <header style="text-align:center;"> | |||
| <h2>Solutions of Savchenko Problems in Physics</h2> | |||
| <p class="author"> | |||
| Aliaksandr Melnichenka <br/> | |||
| October 2023 | |||
| </p> | |||
| </header> | |||
| <h3 id="back-link"><a href="../../#2.8">$\leftarrow$Back</a></h3> | |||
| <h3> Statement </h3> | |||
| <p> | |||
| $2.8.14^*.$ The body with wedges installed in its cutouts is located between two parallel walls as shown in the figure. Find the limit angle at the top of the wedges at which the body can move to the right and cannot move to the left. The coefficients of friction of the wedges against the walls and body are equal to $\mu_1$ and $\mu_2$, respectively. | |||
| </p> | |||
| <center> | |||
| <figure> | |||
| <img src="2.8.14.png" | |||
| loading="lazy" width="200" /> | |||
| <figcaption> | |||
| For problem $2.8.14^*$ | |||
| </figcaption> | |||
| </figure> | |||
| </center> | |||
| <p> | |||
| </p> | |||
| <h3>Solution</h3> | |||
| <p> | |||
| </p> | |||
| <center> | |||
| <figure> | |||
| <img src="2.8.14_1.png" | |||
| loading="lazy" width="250" /> | |||
| <figcaption> | |||
| Forces acting on the wedge | |||
| </figcaption> | |||
| </figure> | |||
| </center> | |||
| <p> | |||
| State of forces equilibrium for the wedge for the horizontal axis: | |||
| $$N_3+\mu_1N_1=\left(\mu_2\cos\alpha+\sin\alpha\right)N_2$$ | |||
| In the set mode, the force that will act on the left wall will be equal to zero ($N_3=0$) | |||
| $$\frac{N_1}{N_2}=\frac{\mu_2\cos\alpha+\sin\alpha}{\mu_1}\quad(1)$$ | |||
| Similarly, for the vertical axis | |||
| $$N_1+\mu_2N_2\sin\alpha=N_2\cos\alpha$$ | |||
| $$\frac{N_1}{N_2}=\cos\alpha-\mu_2\sin\alpha\quad(2)$$ | |||
| Substituing $(2)$ into $(1)$ | |||
| $$\mu_1\cos\alpha-\mu_1\mu_2\sin\alpha=\mu_2\cos\alpha+\sin\alpha$$ | |||
| Divide the left and right parts by $\cos\alpha$ | |||
| $$\left(1+\mu_1\mu_2\right)\tan\alpha=\mu_1-\mu_2$$ | |||
| Whence the required angle | |||
| $$\boxed{\tan\alpha=\frac{{\mu}_{1}-\mu_{2}}{1+\mu_{1}\mu_{2}}}$$ | |||
| </p> | |||
| <h4>Answer</h4> | |||
| <p> | |||
| $$\operatorname{tg}\alpha=({\mu}_{1}-\mu_{2})/(1+\mu_{1}\mu_{2})$$ | |||
| </p> | |||
| <p style="text-align: right; font-style: italic; font-size: 14;"> | |||
| Almaskhan Arsen<br> | |||
| </p> | |||
| <footer class="row container"> | |||
| <br> | |||
| <p> | |||
| <small> © <strong>Savchenko Solutions</strong>, 2023-2024 <br></small> | |||
| </p> | |||
| <p> | |||
| <small>All rights belong to the authors. <br> Commercial use of materials - with the written permission of the authors. <br> alex@savchenkosolutions.com <br></small> | |||
| </p> | |||
| </footer> | |||
| </body> | |||
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