Questions: An adventurous archaeologist crosses between two rock cliffs by slowly going hand over hand along a rope stretched between the cliffs. He stops to rest at the middle of the rope. The rope will break if the tension in it exceeds 2.50 x 10^4 N. Our hero's mass is 70.0 kg. What is the smallest value the angle θ can have if the rope is not to break? Express your answer in degrees.

Solution Steps

• Mass of the hero, $m = 70.0 \, \text{kg}$
• Maximum tension the rope can withstand, $T_{\text{max}} = 2.50 \times 10^4 \, \text{N}$
• Gravitational acceleration, $g = 9.8 \, \text{m/s}^2$

The weight $W$ of the hero is given by: $$W = m \cdot g$$ $$W = 70.0 \, \text{kg} \times 9.8 \, \text{m/s}^2$$ $$W = 686 \, \text{N}$$

At the midpoint, the tension in the rope has two components due to symmetry. Each side of the rope supports half of the weight of the hero. The vertical component of the tension $T$ must balance the weight of the hero: $$2T \sin(\theta) = W$$ $$T \sin(\theta) = \frac{W}{2}$$ $$T \sin(\theta) = \frac{686 \, \text{N}}{2}$$ $$T \sin(\theta) = 343 \, \text{N}$$

The tension $T$ must not exceed the maximum tension $T_{\text{max}}$: $$T \leq T_{\text{max}}$$ $$T = \frac{343 \, \text{N}}{\sin(\theta)}$$ $$\frac{343 \, \text{N}}{\sin(\theta)} \leq 2.50 \times 10^4 \, \text{N}$$ $$\sin(\theta) \geq \frac{343 \, \text{N}}{2.50 \times 10^4 \, \text{N}}$$ $$\sin(\theta) \geq 0.01372$$

$$\theta_{\text{min}} = \sin^{-1}(0.01372)$$ $$\theta_{\text{min}} \approx 0.785^\circ$$

Final Answer