Solution for Question 10
Consider a geometrically similar lamina of characteristic length $r$ (e.g., its linear dimension). The mass of the lamina $m$ is proportional to its area, which scales as $r^2$ (assuming uniform density and constant thickness). $$ m \propto r^2 $$ The friction force $f$ acting on the lamina is proportional to the normal force ($mg$). $$ f = \mu N = \mu mg \propto \mu (r^2) g $$ The torque $\tau$ is the product of this force and the lever arm (distance from the axis), which also scales as $r$. $$ \tau \approx f \times r \propto (\mu m g) \times r \propto (\mu r^2 g) \times r $$ Thus, the torque required to overcome friction scales with the cube of the linear dimension: $$ \tau \propto r^3 $$ This implies that for any two similar triangles, the ratio of the torques required to rotate them about homologous vertices is equal to the ratio of their hypotenuses cubed.
The right-angled triangle $ABC$ (hypotenuse $c$) is rotated about vertex $C$. We decompose $ABC$ into two smaller right-angled triangles by dropping an altitude from $C$ to the hypotenuse $AB$ at point $D$.
- Triangle $ACD$: Hypotenuse $b = c \cos \theta$. It is similar to $ABC$ but rotated about the vertex corresponding to angle $B$.
- Triangle $BCD$: Hypotenuse $a = c \sin \theta$. It is similar to $ABC$ but rotated about the vertex corresponding to angle $A$.
Using the $r^3$ scaling law derived in Step 1:
1. Torque for $ACD$ about $C$: Scales with hypotenuse $b$. Homologous to rotation about $B$.
$$ \tau_{ACD} = \tau_B \left(\frac{b}{c}\right)^3 = \tau_B \cos^3 \theta $$
2. Torque for $BCD$ about $C$: Scales with hypotenuse $a$. Homologous to rotation about $A$.
$$ \tau_{BCD} = \tau_A \left(\frac{a}{c}\right)^3 = \tau_A \sin^3 \theta $$
Total torque about C:
$$ \tau_C = \tau_A \sin^3 \theta + \tau_B \cos^3 \theta $$
The minimum force is applied at the maximum distance from the axis.
- For axis $A$: Force $F_A$ at distance $c$. $\tau_A = F_A c$.
- For axis $B$: Force $F_B$ at distance $c$. $\tau_B = F_B c$.
- For axis $C$: Force $F_C$ is applied at vertex $A$ (distance $b = c \cos \theta$) or $B$ (distance $a = c \sin \theta$). Assuming $\theta < 45^\circ$, $b > a$, so maximum lever arm is $c \cos \theta$. Thus, $\tau_C = F_C (c \cos \theta)$.