Solution to Question 15
1. System Dynamics
Let $m$ be the mass of each cylinder. The total mass is $3m$. Since the cylinders remain in contact, they all move with a common acceleration $a$. $$F = (3m)a \implies a = \frac{F}{3m}$$ We need to analyze the contact forces (normal forces) to find the range of $F$. Let:
- $N_1$: Normal force between A (bottom-left) and C (top).
- $N_2$: Normal force between B (bottom-right) and C (top).
- $N_3$: Normal force between A and B (horizontal contact).
2. Equations of Motion for Each Cylinder
Cylinder C (Top):
Vertical equilibrium (no vertical acceleration):
$$(N_1 + N_2) \sin 60^\circ = mg$$
$$N_1 + N_2 = \frac{mg}{\sin 60^\circ} = \frac{2mg}{\sqrt{3}} \quad \dots(1)$$
Horizontal motion:
$$(N_1 – N_2) \cos 60^\circ = ma$$
$$N_1 – N_2 = \frac{ma}{\cos 60^\circ} = 2ma \quad \dots(2)$$
Solving (1) and (2) for $N_1$ and $N_2$: Adding: $2N_1 = \frac{2mg}{\sqrt{3}} + 2ma \implies N_1 = \frac{mg}{\sqrt{3}} + ma$ Subtracting: $2N_2 = \frac{2mg}{\sqrt{3}} – 2ma \implies N_2 = \frac{mg}{\sqrt{3}} – ma$
Cylinder B (Bottom-Right):
Horizontal motion:
$$N_3 + N_2 \cos 60^\circ = ma$$
$$N_3 = ma – \frac{N_2}{2}$$
Substitute $N_2$:
$$N_3 = ma – \frac{1}{2}\left( \frac{mg}{\sqrt{3}} – ma \right) = \frac{3ma}{2} – \frac{mg}{2\sqrt{3}}$$
3. Contact Conditions
For the cylinders to remain in contact, all normal forces must be non-negative ($N \ge 0$).
Condition 1: Contact between B and C ($N_2 \ge 0$) This prevents C from “falling off” the back. $$\frac{mg}{\sqrt{3}} – ma \ge 0$$ $$a \le \frac{g}{\sqrt{3}}$$ Since $F = 3ma$, the maximum force is: $$F_{max} = 3m \left( \frac{g}{\sqrt{3}} \right) = \sqrt{3}mg$$
Condition 2: Contact between A and B ($N_3 \ge 0$) This prevents A from separating from B. $$\frac{3ma}{2} – \frac{mg}{2\sqrt{3}} \ge 0$$ $$3ma \ge \frac{mg}{\sqrt{3}}$$ $$a \ge \frac{g}{3\sqrt{3}}$$ The minimum force is: $$F_{min} = 3m \left( \frac{g}{3\sqrt{3}} \right) = \frac{mg}{\sqrt{3}}$$
4. Calculation
Given: $m = 10\sqrt{3} \text{ kg}$, $g = 10 \text{ m/s}^2$.
Minimum Force: $$F_{min} = \frac{(10\sqrt{3})(10)}{\sqrt{3}} = 100 \text{ N}$$
Maximum Force: $$F_{max} = \sqrt{3}(10\sqrt{3})(10) = 300 \text{ N}$$