With what force a heavy stick presses on the bottom of the reservoir, if a hollow ball of radius r, rigidly connected to the stick, is half immersed in the liquid? The density of the liquid is
ho, the length of the stick is l.
<h3 id="back-link"><a href="/#4.2">$\leftarrow$Back</a></h3>
<h3> Statement </h3>
<p>
$4.2.11$
With what force a heavy stick presses on the bottom of the reservoir, if a hollow ball of radius $r$, rigidly connected to the stick, is half immersed in the liquid? The density of the liquid is $\rho$, the length of the stick is $l$.
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For problem 4.2.11
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<h3>Solution</h3>
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Let's consider the following figure
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Force analysis
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The system is on equilibrium in above position, so total torque is zero ($\sum \vec{\tau} = \vec{0}$). Considering as a pivot point the center of mass of the stick (located at stick's half),
$$F_A\left(\frac{l}{2}+r\right)\sin{\left(\frac{\pi}{2}-\alpha\right)}+mg\left(\frac{l}{2}+r\right)\sin{\left(\frac{\pi}{2}+\alpha\right)}+N\frac{l}{2}\sin{\left(\frac{\pi}{2}+\alpha\right)}=0$$
$$F_A\left(\frac{l}{2}+r\right)\cos{\alpha}-mg\left(\frac{l}{2}+r\right)\cos{\alpha}-N\frac{l}{2}\cos{\alpha}=0$$
$$F_A\left(\frac{l}{2}+r\right)-mg\left(\frac{l}{2}+r\right)-N\frac{l}{2}=0$$
$$\rho\~g\~\frac{V}{2}\left(\frac{l}{2}+r\right)-mg\left(\frac{l}{2}+r\right)-N\frac{l}{2}=0$$
Separating $N$, and considering that ball's mass can be expressed as $m=\rho_0 V$ and ball's volume is $V=\frac{4}{3}\pi r^3$
$$N = \frac{2\pi r^3g}{3}\left(1+2\frac{r}{l}\right)(\rho -2\rho_0)$$
But, since the ball is a hollow sphere, it can be considered massless, so $\rho_0 = 0$, then
$$\boxed{N = \frac{2\pi r^3\rho g}{3}\left(1+2\frac{r}{l}\right)}$$
Another way:
Let's suppose the pivot point at the contact of stick with the bottom. Applying Newton Second Law for rotation ($\sum \vec{\tau} = \vec{0}$) and considering the massless ball
$$F_A\left(l+r\right)\sin{\left(\frac{\pi}{2}-\alpha\right)}+m_s\~g\frac{l}{2}\sin{\left(\frac{\pi}{2}+\alpha\right)}=0$$
$$F_A\left(l+r\right)\cos{\alpha}-m_s\~g\frac{l}{2}\cos{\alpha}=0$$
Hence, the mass of the stick is
$$m_s = \rho\~V\left(1+\frac{r}{l}\right) \quad(1)$$
Applying Newton Second Law
$$F_A+N = m_s\~g$$
$$\rho g \frac{V}{2} + N = m_s g \quad(2)$$
From $(1)$, $(2)$ and taking $V=\frac{4\pi r^3}{3}$, the above answer is obtained again...
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<p style="text-align: right; font-style: italic; font-size: 14;">
BSc. Luis Daniel Fernández Quintana<br>
Physics Department (FCNE)<br>
Universidad de Oriente, Cuba<br>
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