$4.1.17.$ A liquid is poured into a hemispherical bell, the edges of which fit tightly
+
against the surface of a table, through a hole at the top. When the liquid
+
reaches the hole, it lifts the bell and begins to flow from under it. Find the
+
mass of the bell if its inner radius is $R$ and the density of the liquid is $\rho$.
### Solution
At the moment of detachment, the forces acting on the bell are: its weight $W_c$, the force exerted by the fluid on the bell $F_{f-b}$, and the atmospheric force $F_{\text{atm}}$. We do not consider the contact force with the ground, since it becomes zero once detachment occurs. The balance of the interactions, noting that all forces are vertical, is such that
$$
@@ -10,29 +13,36 @@Solution
W_b = F_{f-b} - F_{\text{atm}}
$$
−
The force exerted by the atmosphere can be expressed in terms of the atmospheric pressure $p_0$ and the effective area $\pi R^2$ as $F_{\text{atm}} = p_0 \pi R^2$
+
The force exerted by the atmosphere can be expressed in terms of the atmospheric pressure $p_0$ and the effective area $\pi R^2$ as $F_{\text{atm}} = p_0 \pi R^2$:
$$
−
W_b = F_{f-b} - p_0 \pi R^2 \tag{1}
+
W_b = F_{f-b} - p_0 \pi R^2.\tag{1}
$$
To determine $F_{f-b}$, we will use an indirect method. Instead of focusing on the bell, we shift our attention to the fluid, which remains at rest until the bell is lifted. The forces acting on it are: its weight $W_f$, the contact force $F_{b-f}$ exerted by the bell on the fluid, and the normal force $N$ exerted by the ground. Thus,
$$
−
N = W_f + F_{b-f}
+
N = W_f + F_{b-f}.
$$
Taking into account Newton's third law and the fact that we are considering absolute values of the forces, $F_{f-b} = F_{b-f}$, therefore
$$
−
F_{f-b} = N - W_f
+
F_{f-b} = N - W_f.
$$
The force exerted by the ground on the fluid (equal in magnitude to the force the fluid exerts on the ground) is given by the pressure $p = p_0 + \rho g R$ and the contact area $\pi R^2$:
$$
N = p \pi R^2 = (p_0 + \rho g R)\, \pi R^2 \tag{2}
$$
+
+
\begin{equation}
+
\begin{split}
+
N &= p \pi R^2 \\\
+
&= (p_0 + \rho g R)\, \pi R^2
+
\end{split} \tag{2}
+
\end{equation}
The weight of the fluid is given by its mass $m_f = \rho V_f$ and its volume $V_f = \frac{2}{3}\pi R^3$:
$$
W_f = m_fg = \rho V_F g = \frac{2}{3}\rho g \pi R^3
$$
Thus, relation (2) can be rewritten as
$$
F_{f-b} = (p_0 + \rho g R)\pi R^2 - \frac{2}{3}\rho g \pi R^3
= \left(p_0 + \rho g R - \frac{2}{3}\rho g R\right)\pi R^2
$4.1.17.$ A liquid is poured into a hemispherical bell, the edges of which fit tightly
against the surface of a table, through a hole at the top. When the liquid
reaches the hole, it lifts the bell and begins to flow from under it. Find the
mass of the bell if its inner radius is $R$ and the density of the liquid is $\rho$.
### Solution
### Solution
At the moment of detachment, the forces acting on the bell are: its weight $W_c$, the force exerted by the fluid on the bell $F_{f-b}$, and the atmospheric force $F_{\text{atm}}$. We do not consider the contact force with the ground, since it becomes zero once detachment occurs. The balance of the interactions, noting that all forces are vertical, is such that
At the moment of detachment, the forces acting on the bell are: its weight $W_c$, the force exerted by the fluid on the bell $F_{f-b}$, and the atmospheric force $F_{\text{atm}}$. We do not consider the contact force with the ground, since it becomes zero once detachment occurs. The balance of the interactions, noting that all forces are vertical, is such that
$$
$$
@@ -10,29 +13,36 @@Solution
W_b = F_{f-b} - F_{\text{atm}}
W_b = F_{f-b} - F_{\text{atm}}
$$
$$
The force exerted by the atmosphere can be expressed in terms of the atmospheric pressure $p_0$ and the effective area $\pi R^2$ as $F_{\text{atm}} = p_0 \pi R^2$
The force exerted by the atmosphere can be expressed in terms of the atmospheric pressure $p_0$ and the effective area $\pi R^2$ as $F_{\text{atm}} = p_0 \pi R^2$:
$$
$$
W_b = F_{f-b} - p_0 \pi R^2 \tag{1}
W_b = F_{f-b} - p_0 \pi R^2.\tag{1}
$$
$$
To determine $F_{f-b}$, we will use an indirect method. Instead of focusing on the bell, we shift our attention to the fluid, which remains at rest until the bell is lifted. The forces acting on it are: its weight $W_f$, the contact force $F_{b-f}$ exerted by the bell on the fluid, and the normal force $N$ exerted by the ground. Thus,
To determine $F_{f-b}$, we will use an indirect method. Instead of focusing on the bell, we shift our attention to the fluid, which remains at rest until the bell is lifted. The forces acting on it are: its weight $W_f$, the contact force $F_{b-f}$ exerted by the bell on the fluid, and the normal force $N$ exerted by the ground. Thus,
$$
$$
N = W_f + F_{b-f}
N = W_f + F_{b-f}.
$$
$$
Taking into account Newton's third law and the fact that we are considering absolute values of the forces, $F_{f-b} = F_{b-f}$, therefore
Taking into account Newton's third law and the fact that we are considering absolute values of the forces, $F_{f-b} = F_{b-f}$, therefore
$$
$$
F_{f-b} = N - W_f
F_{f-b} = N - W_f.
$$
$$
The force exerted by the ground on the fluid (equal in magnitude to the force the fluid exerts on the ground) is given by the pressure $p = p_0 + \rho g R$ and the contact area $\pi R^2$:
The force exerted by the ground on the fluid (equal in magnitude to the force the fluid exerts on the ground) is given by the pressure $p = p_0 + \rho g R$ and the contact area $\pi R^2$:
$$
$$
N = p \pi R^2 = (p_0 + \rho g R)\, \pi R^2 \tag{2}
N = p \pi R^2 = (p_0 + \rho g R)\, \pi R^2 \tag{2}
$$
$$
\begin{equation}
\begin{split}
N &= p \pi R^2 \\\
&= (p_0 + \rho g R)\, \pi R^2
\end{split} \tag{2}
\end{equation}
The weight of the fluid is given by its mass $m_f = \rho V_f$ and its volume $V_f = \frac{2}{3}\pi R^3$:
The weight of the fluid is given by its mass $m_f = \rho V_f$ and its volume $V_f = \frac{2}{3}\pi R^3$:
$$
$$
W_f = m_fg = \rho V_F g = \frac{2}{3}\rho g \pi R^3
W_f = m_fg = \rho V_F g = \frac{2}{3}\rho g \pi R^3
$$
$$
Thus, relation (2) can be rewritten as
Thus, relation (2) can be rewritten as
$$
$$
F_{f-b} = (p_0 + \rho g R)\pi R^2 - \frac{2}{3}\rho g \pi R^3
F_{f-b} = (p_0 + \rho g R)\pi R^2 - \frac{2}{3}\rho g \pi R^3
= \left(p_0 + \rho g R - \frac{2}{3}\rho g R\right)\pi R^2
= \left(p_0 + \rho g R - \frac{2}{3}\rho g R\right)\pi R^2
