Solution to Question 2
1. Coordinate Geometry & Tilt
The system pivots about the line connecting the feet of legs $B$ and $D$. Let this axis be the origin.
Due to the shorter leg $A$, the table tilts by an angle $\theta$. From the geometry of the setup (viewed from the side as shown in the diagram), the tangent of the tilt angle is the height difference divided by the horizontal half-diagonal $r = l/\sqrt{2}$:
$$ \tan \theta = \frac{\Delta h}{r} = \frac{\sqrt{2} \Delta h}{l} $$
2. Lever Arms Calculation
Crucially, both the table’s Center of Mass (CM) and the added mass $m_0$ are located at height $h$ above the feet. The tilt affects their horizontal lever arms as follows:
- Destabilizing Arm ($x_{cm}$): The CM starts above the pivot. Tilting shifts it horizontally by:
$$ x_{cm} = h \sin \theta $$
- Restoring Arm ($x_{mass}$): The mass $m_0$ is at horizontal distance $r$ from the center. However, the tilt also moves the “top” of the table closer to the pivot line. The effective horizontal distance is:
$$ x_{mass} = r \cos \theta – h \sin \theta $$
3. Torque Balance
Equating the torques about the pivot axis:
$$ \tau_{restoring} = \tau_{destabilizing} $$
$$ m_0 g (r \cos \theta – h \sin \theta) = M g (h \sin \theta) $$
Dividing by $g \cos \theta$ (to introduce $\tan \theta$):
$$ m_0 (r – h \tan \theta) = M h \tan \theta $$
4. Solving for Mass $M$
Rearranging the equation:
$$ M = m_0 \left( \frac{r – h \tan \theta}{h \tan \theta} \right) = m_0 \left( \frac{r}{h \tan \theta} – 1 \right) $$
Substituting $\tan \theta = \frac{\Delta h}{r}$:
$$ M = m_0 \left( \frac{r}{h (\Delta h / r)} – 1 \right) = m_0 \left( \frac{r^2}{h \Delta h} – 1 \right) $$
Since $r = l/\sqrt{2}$, we have $r^2 = l^2/2$. Thus:
$$ M = m_0 \left( \frac{l^2}{2 h \Delta h} – 1 \right) $$
5. Numerical Calculation
Substituting the values: $m_0 = 0.2$ kg, $l = 1.0$ m, $h = 1.0$ m, $\Delta h = 0.01$ m.
$$ M = 0.2 \left( \frac{1.0^2}{2(1.0)(0.01)} – 1 \right) $$
$$ M = 0.2 \left( \frac{1}{0.02} – 1 \right) $$
$$ M = 0.2 (50 – 1) = 0.2 (49) $$
$$ M = 9.8 \text{ kg} $$
The system pivots about the line connecting the feet of legs $B$ and $D$. Let this axis be the origin. Due to the shorter leg $A$, the table tilts by an angle $\theta$. From the geometry of the setup (viewed from the side as shown in the diagram), the tangent of the tilt angle is the height difference divided by the horizontal half-diagonal $r = l/\sqrt{2}$: $$ \tan \theta = \frac{\Delta h}{r} = \frac{\sqrt{2} \Delta h}{l} $$
Crucially, both the table’s Center of Mass (CM) and the added mass $m_0$ are located at height $h$ above the feet. The tilt affects their horizontal lever arms as follows:
- Destabilizing Arm ($x_{cm}$): The CM starts above the pivot. Tilting shifts it horizontally by: $$ x_{cm} = h \sin \theta $$
- Restoring Arm ($x_{mass}$): The mass $m_0$ is at horizontal distance $r$ from the center. However, the tilt also moves the “top” of the table closer to the pivot line. The effective horizontal distance is: $$ x_{mass} = r \cos \theta – h \sin \theta $$
Equating the torques about the pivot axis: $$ \tau_{restoring} = \tau_{destabilizing} $$ $$ m_0 g (r \cos \theta – h \sin \theta) = M g (h \sin \theta) $$ Dividing by $g \cos \theta$ (to introduce $\tan \theta$): $$ m_0 (r – h \tan \theta) = M h \tan \theta $$
Rearranging the equation: $$ M = m_0 \left( \frac{r – h \tan \theta}{h \tan \theta} \right) = m_0 \left( \frac{r}{h \tan \theta} – 1 \right) $$ Substituting $\tan \theta = \frac{\Delta h}{r}$: $$ M = m_0 \left( \frac{r}{h (\Delta h / r)} – 1 \right) = m_0 \left( \frac{r^2}{h \Delta h} – 1 \right) $$ Since $r = l/\sqrt{2}$, we have $r^2 = l^2/2$. Thus: $$ M = m_0 \left( \frac{l^2}{2 h \Delta h} – 1 \right) $$
Substituting the values: $m_0 = 0.2$ kg, $l = 1.0$ m, $h = 1.0$ m, $\Delta h = 0.01$ m. $$ M = 0.2 \left( \frac{1.0^2}{2(1.0)(0.01)} – 1 \right) $$ $$ M = 0.2 \left( \frac{1}{0.02} – 1 \right) $$ $$ M = 0.2 (50 – 1) = 0.2 (49) $$