Physics Solution Q22

Solution to Question 22

(a) Working Principle: By rotating the object while pulling, the ant forces the friction to act mainly against the rotation. The component of friction opposing the translation becomes significantly smaller than the total static friction, allowing the ant to move the heavy object with less force.

(b) Calculation: The ant applies a force $F$ perpendicular to the straw at one end. This causes the straw to pivot about an instantaneous center of rotation (ICR) located at a distance $x$ from the end where the force is applied. Friction acts perpendicular to the rod, opposing the local velocity of the rod segments.

Step 1: Force and Torque Equilibrium

Let $\lambda = \frac{\mu mg}{L}$ be the friction force per unit length. The pivot $P$ divides the rod into lengths $x$ and $L-x$.
For translational equilibrium (sum of vertical forces):

$$ F + \int_x^L \lambda dl – \int_0^x \lambda dl = 0 $$

Since friction on the right segment ($L-x$) acts UP (supporting F) and friction on the left segment ($x$) acts DOWN (opposing F):

$$ F + \lambda(L-x) – \lambda x = 0 \implies F = \lambda(2x – L) \quad \dots(1) $$

If F is minimal, it balances the net friction. Let’s look at the torque balance about the pivot P.
Torque due to $F$ = $Fx$.
Resistive torque due to friction = $\int_0^x \lambda y dy + \int_0^{L-x} \lambda y dy = \frac{\lambda x^2}{2} + \frac{\lambda (L-x)^2}{2}$.
Equilibrium:

$$ F x = \frac{\lambda}{2} [ x^2 + (L-x)^2 ] \quad \dots(2) $$

Step 2: Solving the System

From (1), $F = \frac{\mu mg}{L}(2x – L)$. (Assuming $2x > L$, i.e., pivot is far from the ant).
Substitute $F$ into (2):
$$ \lambda(2x – L) x = \frac{\lambda}{2} (2x^2 – 2Lx + L^2) $$ $$ 4x^2 – 2Lx = 2x^2 – 2Lx + L^2 $$ $$ 2x^2 = L^2 \implies x = \frac{L}{\sqrt{2}} $$

Step 3: Calculating Force

Now substitute $x = L/\sqrt{2}$ back into the Force equation:
$$ F = \frac{\mu mg}{L} \left( 2\frac{L}{\sqrt{2}} – L \right) $$ $$ F = \mu mg (\sqrt{2} – 1) $$
$$ F = \mu mg (\sqrt{2} – 1) \approx 0.414 \mu mg $$