<meta name="keywords" content="Savchenko Problems in Physics, Savchenko solutions, physics problems, physics olympiad preparation, IPhO, Jaan Kalda">
<meta name="description" content="Through three holes made in a desktop pass threads united by a common knot. At the free ending of each thread is hanged equal loads. Determine the angles between threads. There is no friction.">
<meta property="og:title" content="Through three holes made in a desktop pass threads united by a common knot. At the free ending of each thread is hanged equal loads. Determine the angles between threads. There is no friction.">
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@@ -14,9 +14,9 @@
<meta property="og:description" content="Through three holes made in a desktop pass threads united by a common knot. At the free ending of each thread is hanged equal loads. Determine the angles between threads. There is no friction.">
<title>Through three holes made in a desktop pass threads united by a common knot. At the free ending of each thread is hanged equal loads. Determine the angles between threads. There is no friction.</title>
Through three holes made in a desktop pass threads united by a common knot. At the free ending of each thread is hanged equal loads. Determine the angles between threads. There is no friction.
</p>
<center>
<figure>
<img src="statement.png"
loading="lazy" alt="2.8.4" width="200" />
<figcaption>
For problem 2.8.4
</figcaption>
</figure>
</center>
<h3>Solution</h3>
<p>
Tension in each thread is the same ($T=mg$). Moreover, note that the absolute value of the sum of two any tensions must be equal to the third one which is also $T$ and be directed in opposed sense for maintain the equilibrium. So, consider for example the angle $\gamma$
</p>
<center>
<figure>
<img src="draw.png"
loading="lazy" alt="2.8.4" width="300" />
<figcaption>
Top view
</figcaption>
</figure>
</center>
<p>
$$T_R = T = 2T\cos{\frac{\gamma}{2}}$$
Hence
$$\cos{\frac{\gamma}{2}} = 0.5$$
$$\gamma = \frac{2\pi}{3} = 120^\circ$$
But, since the above condition of the sum of two any tensions always is true, the three angles must be equal.
<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>
<meta name="keywords" content="Savchenko Problems in Physics, Savchenko solutions, physics problems, physics olympiad preparation, IPhO, Jaan Kalda">
<meta name="keywords" content="Savchenko Problems in Physics, Savchenko solutions, physics problems, physics olympiad preparation, IPhO, Jaan Kalda">
<meta name="description" content="Through three holes made in a desktop pass threads united by a common knot. At the free ending of each thread is hanged equal loads. Determine the angles between threads. There is no friction.">
<meta name="description" content="Through three holes made in a desktop pass threads united by a common knot. At the free ending of each thread is hanged equal loads. Determine the angles between threads. There is no friction.">
<meta property="og:title" content="Through three holes made in a desktop pass threads united by a common knot. At the free ending of each thread is hanged equal loads. Determine the angles between threads. There is no friction.">
<meta property="og:title" content="Through three holes made in a desktop pass threads united by a common knot. At the free ending of each thread is hanged equal loads. Determine the angles between threads. There is no friction.">
<meta property="og:image" content="img/logo.png">
<meta property="og:image" content="img/logo.png">
@@ -14,9 +14,9 @@
<meta property="og:description" content="Through three holes made in a desktop pass threads united by a common knot. At the free ending of each thread is hanged equal loads. Determine the angles between threads. There is no friction.">
<meta property="og:description" content="Through three holes made in a desktop pass threads united by a common knot. At the free ending of each thread is hanged equal loads. Determine the angles between threads. There is no friction.">
<title>Through three holes made in a desktop pass threads united by a common knot. At the free ending of each thread is hanged equal loads. Determine the angles between threads. There is no friction.</title>
<title>Through three holes made in a desktop pass threads united by a common knot. At the free ending of each thread is hanged equal loads. Determine the angles between threads. There is no friction.</title>
Through three holes made in a desktop pass threads united by a common knot. At the free ending of each thread is hanged equal loads. Determine the angles between threads. There is no friction.
Through three holes made in a desktop pass threads united by a common knot. At the free ending of each thread is hanged equal loads. Determine the angles between threads. There is no friction.
</p>
</p>
<center>
<center>
<figure>
<figure>
<img src="statement.png"
<img src="statement.png"
loading="lazy" alt="2.8.4" width="200" />
loading="lazy" alt="2.8.4" width="200" />
<figcaption>
<figcaption>
For problem 2.8.4
For problem 2.8.4
</figcaption>
</figcaption>
</figure>
</figure>
</center>
</center>
<h3>Solution</h3>
<h3>Solution</h3>
<p>
<p>
Tension in each thread is the same ($T=mg$). Moreover, note that the absolute value of the sum of two any tensions must be equal to the third one which is also $T$ and be directed in opposed sense for maintain the equilibrium. So, consider for example the angle $\gamma$
Tension in each thread is the same ($T=mg$). Moreover, note that the absolute value of the sum of two any tensions must be equal to the third one which is also $T$ and be directed in opposed sense for maintain the equilibrium. So, consider for example the angle $\gamma$
</p>
</p>
<center>
<center>
<figure>
<figure>
<img src="draw.png"
<img src="draw.png"
loading="lazy" alt="2.8.4" width="300" />
loading="lazy" alt="2.8.4" width="300" />
<figcaption>
<figcaption>
Top view
Top view
</figcaption>
</figcaption>
</figure>
</figure>
</center>
</center>
<p>
<p>
$$T_R = T = 2T\cos{\frac{\gamma}{2}}$$
$$T_R = T = 2T\cos{\frac{\gamma}{2}}$$
Hence
Hence
$$\cos{\frac{\gamma}{2}} = 0.5$$
$$\cos{\frac{\gamma}{2}} = 0.5$$
$$\gamma = \frac{2\pi}{3} = 120^\circ$$
$$\gamma = \frac{2\pi}{3} = 120^\circ$$
But, since the above condition of the sum of two any tensions always is true, the three angles must be equal.
But, since the above condition of the sum of two any tensions always is true, the three angles must be equal.
<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>
<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>
