NLM BYU 29

Let the force exerted by the spring on the block be $F$. By Newton’s third law, the block exerts an equal and opposite force $F$ on the catapult structure. This reaction force acts downwards and to the left at an angle $\theta$ with the horizontal.

The catapult has a negligible mass compared to the forces involved ($mg \ll F$). Therefore, for the catapult to remain in equilibrium (not slip), the horizontal and vertical forces must balance.

Forces on the Catapult:

• Horizontal component of reaction force: $F_H = F \cos\theta$ (acting left).
• Vertical component of reaction force: $F_V = F \sin\theta$ (acting down).
• Normal reaction from ground: $N = F_V = F \sin\theta$ (since mass is negligible).
• Friction force from ground: $f$ (acting right).

For the catapult to stay stationary, the static friction must balance the horizontal thrust: $$f = F \cos\theta$$ The condition for no slipping is $f \le \mu N$. Substituting the values: $$F \cos\theta \le \mu (F \sin\theta)$$ $$\cos\theta \le \mu \sin\theta$$ $$\cot\theta \le \mu \quad \text{or} \quad \tan\theta \ge \frac{1}{\mu}$$

The acceleration of the block of mass $m$ is provided by the horizontal component of the spring force. $$a_x = \frac{F \cos\theta}{m}$$ To maximize acceleration, we need to maximize $\cos\theta$. Since $\cos\theta$ increases as $\theta$ decreases (for $0 < \theta < 90^\circ$), we want the smallest possible angle $\theta$.

From the slipping constraint $\tan\theta \ge \frac{1}{\mu}$, the minimum angle $\theta_{\min}$ occurs when: $$\tan\theta = \frac{1}{\mu}$$ Using the trigonometric identity $\cos\theta = \frac{1}{\sqrt{1+\tan^2\theta}}$: $$\cos\theta_{\max} = \frac{1}{\sqrt{1 + (1/\mu)^2}} = \frac{1}{\sqrt{\frac{\mu^2+1}{\mu^2}}} = \frac{\mu}{\sqrt{1+\mu^2}}$$