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<title>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.</title>
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>
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<meta name="description" content="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.">
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of inspiration for Jaan Kalda’s handouts and to some NBPhO problems. You may find problems from old IPhO
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<meta property="og:description" content="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.">
<title>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.</title>
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>
