<meta name="description" content="A website with solutions to physics problems from Savchenko Textbook">
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<meta name="description" content="The largest dataset of solutions of 'Savchenko. Problems in Physics'. Savchenko’s Problems in General Physics is widely used to prepare for olympiads and it is a useful tool to
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master and sharpen your skills and techniques in comptetitive problem solving. Some of these problems were a source
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of inspiration for Jaan Kalda’s handouts and to some NBPhO problems. You may find problems from old IPhO
Determine the force of the tension of the thread connecting two balls of volume 10 cm$^3$ each, if the upper ball floats half-immersed in water. The lower ball is three times heavier than the upper one.
</p>
<center>
<figure>
<img src="statement.png"
loading="lazy" alt="4.2.12" width="200" />
<figcaption>
For problem 4.2.12
</figcaption>
</figure>
</center>
<h3>Solution</h3>
<p>
Applying Newton's Second Law...
</p>
<center>
<figure>
<img src="draw.png"
loading="lazy" alt="4.2.12" width="200" />
<figcaption>
Force analysis
</figcaption>
</figure>
</center>
<p>
For upper ball
$$m_{\uparrow}~g+T = F_{\uparrow}$$
$$m_{\uparrow}~g+T = \rho_0~g~\frac{V}{2} \;(1)$$
For lower ball
$$m_{\downarrow}~g = T+F_{\downarrow}$$
$$m_{\downarrow}~g = T+\rho_0~g~V \;(2)$$
Separating the gravity's force term from $(1)$
$$m_{\uparrow}~g = \rho_0~g~\frac{V}{2}-T \;(3)$$
Dividing $(2)$ by $(3)$, and taking account that $m_{\downarrow}=3~m_{\uparrow}$
$$\frac{\rho_0~g~V+T}{\rho_0~g~\frac{V}{2}-T}=3$$
and separating $T$
$$T = \frac{\rho_0~g~V}{8}$$
Taking $\rho_0 = 1~\rm{g/cm^3} = 10^{-3}~\rm{kg/cm^3}$ (water's density) and $g=9.8~\rm{m/s^2}$ (acceleration due to gravity at Earth's surface),
<small>All rights belong to the authors. <br> Commercial use of materials - with the written permission of the authors. <br> astrosander01@gmail.com <br></small>
<meta name="keywords" content="Savchenko Problems in Physics, Savchenko solutions, physics problems, physics olympiad preparation, IPhO, Jaan Kalda">
<meta name="description" content="A website with solutions to physics problems from Savchenko Textbook">
<meta name="description" content="The largest dataset of solutions of 'Savchenko. Problems in Physics'. Savchenko’s Problems in General Physics is widely used to prepare for olympiads and it is a useful tool to
master and sharpen your skills and techniques in comptetitive problem solving. Some of these problems were a source
of inspiration for Jaan Kalda’s handouts and to some NBPhO problems. You may find problems from old IPhO
Determine the force of the tension of the thread connecting two balls of volume 10 cm$^3$ each, if the upper ball floats half-immersed in water. The lower ball is three times heavier than the upper one.
Determine the force of the tension of the thread connecting two balls of volume 10 cm$^3$ each, if the upper ball floats half-immersed in water. The lower ball is three times heavier than the upper one.
</p>
</p>
<center>
<center>
<figure>
<figure>
<img src="statement.png"
<img src="statement.png"
loading="lazy" alt="4.2.12" width="200" />
loading="lazy" alt="4.2.12" width="200" />
<figcaption>
<figcaption>
For problem 4.2.12
For problem 4.2.12
</figcaption>
</figcaption>
</figure>
</figure>
</center>
</center>
<h3>Solution</h3>
<h3>Solution</h3>
<p>
<p>
Applying Newton's Second Law...
Applying Newton's Second Law...
</p>
</p>
<center>
<center>
<figure>
<figure>
<img src="draw.png"
<img src="draw.png"
loading="lazy" alt="4.2.12" width="200" />
loading="lazy" alt="4.2.12" width="200" />
<figcaption>
<figcaption>
Force analysis
Force analysis
</figcaption>
</figcaption>
</figure>
</figure>
</center>
</center>
<p>
<p>
For upper ball
For upper ball
$$m_{\uparrow}~g+T = F_{\uparrow}$$
$$m_{\uparrow}~g+T = F_{\uparrow}$$
$$m_{\uparrow}~g+T = \rho_0~g~\frac{V}{2} \;(1)$$
$$m_{\uparrow}~g+T = \rho_0~g~\frac{V}{2} \;(1)$$
For lower ball
For lower ball
$$m_{\downarrow}~g = T+F_{\downarrow}$$
$$m_{\downarrow}~g = T+F_{\downarrow}$$
$$m_{\downarrow}~g = T+\rho_0~g~V \;(2)$$
$$m_{\downarrow}~g = T+\rho_0~g~V \;(2)$$
Separating the gravity's force term from $(1)$
Separating the gravity's force term from $(1)$
$$m_{\uparrow}~g = \rho_0~g~\frac{V}{2}-T \;(3)$$
$$m_{\uparrow}~g = \rho_0~g~\frac{V}{2}-T \;(3)$$
Dividing $(2)$ by $(3)$, and taking account that $m_{\downarrow}=3~m_{\uparrow}$
Dividing $(2)$ by $(3)$, and taking account that $m_{\downarrow}=3~m_{\uparrow}$
$$\frac{\rho_0~g~V+T}{\rho_0~g~\frac{V}{2}-T}=3$$
$$\frac{\rho_0~g~V+T}{\rho_0~g~\frac{V}{2}-T}=3$$
and separating $T$
and separating $T$
$$T = \frac{\rho_0~g~V}{8}$$
$$T = \frac{\rho_0~g~V}{8}$$
Taking $\rho_0 = 1~\rm{g/cm^3} = 10^{-3}~\rm{kg/cm^3}$ (water's density) and $g=9.8~\rm{m/s^2}$ (acceleration due to gravity at Earth's surface),
Taking $\rho_0 = 1~\rm{g/cm^3} = 10^{-3}~\rm{kg/cm^3}$ (water's density) and $g=9.8~\rm{m/s^2}$ (acceleration due to gravity at Earth's surface),
<small>All rights belong to the authors. <br> Commercial use of materials - with the written permission of the authors. <br> astrosander01@gmail.com <br></small>
<small>All rights belong to the authors. <br> Commercial use of materials - with the written permission of the authors. <br> astrosander01@gmail.com <br></small>
