Solved Problem on Static Equilibrium

A hemisphere with weight W rests on a flat horizontal plane. At the A of the diameter AB is applied a force F, the hemisphere inclines until the AB makes with the horizontal plane an angle α. Calculate, this angle knowing that the center of gravity of the hemisphere lies at a distance from the center equal to 3/8 of the radius.

Problem data:

Weight of the hemisphere: W;
Force applied: F;
Radius of the hemisphere: d_OA=R;
Position of the center of gravity: \( d_{CG}=\frac{3}{8}R \);

Problem diagram:

It is given in the problem that the hemisphere is inclined from an angle α, the OA segment \( \overline{OA} \) (Figure 1-A) makes with the horizontal plane that angle α, segment \( \overline{OC} \) is vertical and makes an angle β with the surface of the hemisphere

\[ \begin{gather} \alpha +\beta =90°\\[5pt] \beta =90°-\alpha \end{gather} \]

Segment \( \overline{OD} \), which passes through the center of gravity of the hemisphere, is perpendicular to the surface of the hemisphere, makes an angle of 90°

\[ \begin{gather} \gamma +\beta =90°\\[5pt] \gamma =90°-\beta \end{gather} \]

substituting the value of β found above

\[ \begin{gather} \gamma =90°-(90°-\alpha)\\[5pt] \gamma=90°-90°+\alpha \\[5pt] \gamma =\alpha \end{gather} \]

The problem tells us that the center of gravity of the semisphere is located \( \frac{3}{8} \) units from point O (Figure 1-B), the distance between the vertical segment passing by O and the segment passing through the center of gravity, where the gravitational force is applied, is r₁, we see in the frame highlighted in red

\[ \begin{gather} \sin \alpha =\frac{\text{opposite side}}{\text{hypotenuse}}=\frac{r_{1}} {\frac{3}{8}R}\\[5pt] r_{1}=\frac{3}{8}R\,\sin \alpha \tag{I} \end{gather} \]

Figura 1

The distance between the vertical segment that passes by O and the segment passing by the end of the hemisphere, where the force \( \vec{F} \), is applied, is r₂ (Figure 1-B)

\[ \begin{gather} \cos \alpha =\frac{\text{adjacent side}}{\text {hypotenuse}}=\frac{r_{2}}{R}\\[5pt] r_{2}=R\,\cos \alpha \tag{II} \end{gather} \]

As the gravitational force and the force applied at the end of the hemisphere are vertical, we can represent the system as a horizontal bar, given by the projection of the surface of the hemisphere, supported at the point O (Figure 1-C).
We assume point O as the origin of the reference frame, we will assume the clockwise as positive.

Solution

The torque of a force is given by

\[ \begin{gather} \bbox[#99CCFF,10px] {{\Large \tau}=Fd} \tag{III} \end{gather} \]

"Forgetting" the force \( \vec{F} \) and considering only the gravitational force \( \vec{F}_{g} \) makes the bar rotate in the opposite direction of the chosen orientation, the torque of this force will be negative, using the expression (III)

\[ \begin{gather} {\Large \tau}_{F_{g}}=-F_{g}r_{1} \tag{IV} \end{gather} \]

substituting the expression (I) into expression (IV)

\[ \begin{gather} {\Large \tau}_{F_{g}}=-\frac{{3}}{8}F_{g}\,R\,\sin \alpha \tag{V} \end{gather} \]

"Forgetting" gravitational force \( {\vec F}_{g} \) and considering only the force \( \vec{F} \), it rotates in the same sense of the chosen orientation, the torque of this force will be positive, using the expression (III)

\[ \begin{gather} {\Large \tau}_{F}=Fr_{2} \tag{VI} \end{gather} \]

substituting the expression (II) into expression (VI)

\[ \begin{gather} {\Large \tau}_{F}=FR\,\cos \alpha \tag{VII} \end{gather} \]

Using the condition that the sum of the moments is equal to zero

\[ \begin{gather} \bbox[#99CCFF,10px] {\sum {\Large \tau}=0} \tag{VIII} \end{gather} \]

substituting expressions (V) and (VII) in the condition (VIII)

\[ \begin{gather} {\Large \tau}_{F_{g}}+{\Large \tau}_{F}=0\\[5pt] -{\frac{3}{8}}F_{g}\,R\,\sin \alpha+F\,R\,\cos \alpha =0\\[5pt] F\,\cancel{R}\,\cos \alpha=\frac{3}{8}\,F_{g}\,\cancel{R}\,\sin \alpha\\[5pt] \frac{\sin \alpha }{\cos \alpha }=\frac{8}{3}\frac{F}{F_{g}} \end{gather} \]

From the Trigonometry \( \tan\alpha =\dfrac{\sin \alpha }{\cos \alpha } \)

\[ \begin{gather} \tan\alpha =\frac{8}{3}\frac{F}{F_{g}} \end{gather} \]

\[ \begin{gather} \bbox[#FFCCCC,10px] {\alpha =\arctan\,\frac{8}{3}\frac{F}{F_{g}}} \end{gather} \]

Fisicaexe - Physics Solved Problems by Elcio Brandani Mondadori is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License .