1. Corrected Physics Diagram

The diagram below illustrates the forces. Note that the velocity $v_0$ is down the slope, while the friction impulse that balances gravity is generated only when the pads move slowly relative to the box.

2. Mathematical Derivation

Step 1: Impulse Balance
Since the box moves at a constant velocity $v_0$, the total impulse from gravity (acting down) must equal the total impulse from friction (acting up) over one time period $T = 2\pi/\omega$.

$$ \int_0^T F_{gravity} dt = \int_0^T F_{friction} dt $$ $$ (mg \sin \theta) \cdot T = (\mu mg) \cdot \Delta t_{active} $$

Using small angle approximation $\sin \theta \approx \theta$, we get:

$$ \theta T = \mu \Delta t_{active} \quad \Rightarrow \quad \theta \frac{2\pi}{\omega} = \mu \Delta t_{active} $$

Step 2: Calculate Active Time ($\Delta t_{active}$)
Net friction is non-zero (and acts UP) only when both pads move down relative to the ground. This requires the relative velocity $v_{rel}$ to be smaller than the box velocity $v_0$:

$$ |A\omega \cos(\omega t)| < v_0 $$

Since $v_0$ is very small compared to the max speed $A\omega$, this happens briefly twice per cycle (when $\cos(\omega t) \approx 0$). We can linearize the motion near these points. The total phase angle $\Phi$ per cycle where this condition holds is:

$$ \Phi = 4 \times \frac{v_0}{A\omega} $$

Therefore, the active time is:

$$ \Delta t_{active} = \frac{\Phi}{\omega} = \frac{4 v_0}{A \omega^2} $$

Step 3: Solve for $v_0$
Substitute $\Delta t_{active}$ back into the impulse equation:

$$ mg \theta \frac{2\pi}{\omega} = \mu mg \left( \frac{4 v_0}{A \omega^2} \right) $$

Cancel $mg$ and simplify:

$$ \theta \pi = \mu \frac{2 v_0}{A \omega} $$ $$ v_0 = \frac{\pi A \omega \theta}{2 \mu} $$

3. Calculation

Given values:

• Amplitude $A = 0.25 \text{ mm} = 0.25 \times 10^{-3} \text{ m}$
• Angular Frequency $\omega = 72 \text{ rad/s}$
• Friction Coefficient $\mu = 0.60$
• Angle $\theta = 0.5^\circ$ (Must convert to radians!)

Convert angle:

$$ \theta_{rad} = 0.5 \times \frac{\pi}{180} = \frac{\pi}{360} $$

Substitute into formula:

$$ v_0 = \frac{\pi (0.25 \times 10^{-3}) (72)}{2 (0.60)} \times \frac{\pi}{360} $$ $$ v_0 = \frac{\pi^2 (0.25)(72)}{1.2 \times 360} \times 10^{-3} $$ $$ v_0 = \frac{9.87 \times 18}{432} \text{ mm/s} $$ $$ v_0 \approx 0.41 \text{ mm/s} $$