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:

The problem tells us that the center of gravity of the hemisphere is located at \( \frac{3}{8} \) units from the point O (Figure 1-A).

Figure 1

We consider the gravitational force \( {\vec F}_{g} \) concentrated at this point and the force \( \vec{F} \) applied to point A at a distance equal to the radius of the hemisphere (Figure 1-B). When the force \( \vec{F} \) is applied the hemisphere is inclined until the diameter AB makes an angle α relative to the horizontal.
We assume point O as the origin of the reference frame, we assume the clockwise direction as positive.

Solution

The torque of a force is given by

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

"Forgetting" the force \( \vec{F} \), the gravitational force \( {\vec F}_{g} \) can be decomposed into two components, one component parallel to the radius passing through the center of gravity \( {\vec F}_{gP} \) and another component perpendicular, or normal \( {\vec F}_{gN} \). The angle between the horizontal and the line segment AB is equal to α, the angle between the line segment AB and the vertical passing through O we label β, then the sum of α and β should be 90°, are complementary angles, then β should be \( \alpha +\beta =90°\Rightarrow \beta =90°-\alpha \).

Figure 2

Since the normal component of the gravitational force is in the same direction of the segment AB and the gravitational force is vertical, the angle between them is also β, which are corresponding angles. Only the normal component contributes to the hemisphere rotate around the point O, as this component rotates counterclockwise, the torque of this force will be negative, applying the expression (I)

\[ \begin{gather} {\Large \tau}_{F_{gN}}=-F_{gN}d_{CG} \tag{II} \end{gather} \]

The normal component of the gravitational force is given by

\[ \begin{gather} F_{gN}=F\cos \beta \\[5pt] F_{gN}=F\cos (90°-\alpha ) \end{gather} \]

From the Trigonometry \( \cos (a-b)=\cos a\,\cos b+\sin a\,\sin b \)

\[ \cos (a-b)=\cos a\,\cos b+\sin a\,\sin b \]

\[ \begin{gather} F_{gN}=F_{g}(\cos 90°\cos \alpha +\sin 90°\sin \alpha)\\[5pt] F_{gN}=F_{g}(0\times \cos \alpha +1\times \sin \alpha) \\[5pt] F_{gN}=F_{g}\sin \alpha \tag{III} \end{gather} \]

substituting the expression (III) into expression (II)

\[ \begin{gather} {\Large \tau}_{\,F_{gN}}=-F_{g}\,\sin \alpha \,d_{CG} \tag{IV} \end{gather} \]

"Forgetting" the gravitational force \( {\vec F}_{g} \), the force \( \vec{F} \) can be decomposed into two components, one component parallel to the radius passing through the center O \( {\vec F}_{P} \) and another component perpendicular or normal component \( {\vec F}_{N} \). The angle between the horizontal and the line segment AB is equal to α, the angle between the line segment AB and the vertical through A we label β, then the sum of α and β should be 90°, are complementary angles, then β should be \( \alpha +\beta =90°\Rightarrow \beta =90°-\alpha \).

Figure 3

Only the normal component contributes to the hemisphere rotate around the point O, as this component rotates in the same direction of orientation, the torque of this force will be positive, applying expression (I)

\[ \begin{gather} {\Large \tau}_{F_{N}}=F_{N}\,d_{OA} \tag{V} \end{gather} \]

The normal component of the force is given by

\[ \begin{gather} F_{N}=F\sin \beta\\[5pt] F_{N}=F\sin (90°-\alpha) \end{gather} \]

From the Trigonometry \( \sin (a-b)=\sin a\,\cos b-\sin b\,\cos a \)

\( \sin (a-b)=\sin a\,\cos b-\sin b\,\cos a \)

\[ \begin{gather} F_{N}=F\,(\sin 90°\cos \alpha -\sin \alpha \cos 90°)\\[5pt] F_{N}=F\,(1\times\cos \alpha -\sin \alpha \times 0) \\[5pt] F_{N}=F\,\cos \alpha \tag{VI} \end{gather} \]

substituting the expression (VI) into expression (V)

\[ \begin{gather} {\Large \tau}_{F_{N}}=F\cos \alpha \,d_{OA} \tag{VII} \end{gather} \]

Using the condition that the sum of the torques is zero

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

substituting the expressions (IV) and (VII) into the condition (VIII)

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

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

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

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

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