An adventurous archaeologist crosses between two rock cliffs, with a raging river far below, 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 3.00 x 10^4 N, and our hero’s mass is 81.6 kg. a. If the angle that the rope makes with the cliff is 75°, find the tension in the rope. b. What is the smallest value the angle can have if the rope is not to break
The Correct Answer and Explanation is:
Given:
- Mass of the archaeologist, m=81.6 kgm = 81.6 \, \text{kg}m=81.6kg
- Gravitational acceleration, g=9.8 m/s2g = 9.8 \, \text{m/s}^2g=9.8m/s2
- Maximum tension the rope can withstand: Tmax=3.00×104 NT_{\text{max}} = 3.00 \times 10^4 \, \text{N}Tmax=3.00×104N
- The archaeologist is at the center of the rope
- Rope makes an angle θ=75∘\theta = 75^\circθ=75∘ with the horizontal
Part (a): Find the tension in the rope when θ=75∘\theta = 75^\circθ=75∘
At the midpoint, the system is symmetrical. The vertical component of the tension from both sides must support the archaeologist’s weight.
Let TTT be the tension in the rope on one side. The vertical component of tension is TsinθT \sin \thetaTsinθ, and since there are two sides, total upward force is 2Tsinθ2T \sin \theta2Tsinθ. This must balance the weight:2Tsinθ=mg2T \sin \theta = mg2Tsinθ=mgT=mg2sinθT = \frac{mg}{2 \sin \theta}T=2sinθmg
Substituting values:T=81.6×9.82sin75∘T = \frac{81.6 \times 9.8}{2 \sin 75^\circ}T=2sin75∘81.6×9.8T=799.682×0.9659T = \frac{799.68}{2 \times 0.9659}T=2×0.9659799.68T≈799.681.9318≈413.9 NT \approx \frac{799.68}{1.9318} \approx 413.9 \, \text{N}T≈1.9318799.68≈413.9N
Answer (a): The tension in the rope is approximately 413.9 N
Part (b): Find the smallest angle before the rope breaks
Rearrange the tension formula and solve for θ\thetaθ:Tmax=mg2sinθT_{\text{max}} = \frac{mg}{2 \sin \theta}Tmax=2sinθmgsinθ=mg2Tmax\sin \theta = \frac{mg}{2T_{\text{max}}}sinθ=2Tmaxmgsinθ=81.6×9.82×3.00×104\sin \theta = \frac{81.6 \times 9.8}{2 \times 3.00 \times 10^4}sinθ=2×3.00×10481.6×9.8sinθ=799.6860000≈0.01333\sin \theta = \frac{799.68}{60000} \approx 0.01333sinθ=60000799.68≈0.01333θ=arcsin(0.01333)≈0.764∘\theta = \arcsin(0.01333) \approx 0.764^\circθ=arcsin(0.01333)≈0.764∘
Answer (b): The smallest angle the rope can have without breaking is approximately 0.764 degrees
Explanation
When a person hangs motionless at the midpoint of a rope suspended between two cliffs, their weight is supported by the vertical components of tension in the rope on both sides. Since the person is at the center, the forces from both sides are symmetrical, making the analysis simpler. The vertical force due to gravity must be balanced by the combined vertical components of the rope tension.
The gravitational force acting on the person is calculated using mgmgmg, where mmm is the person’s mass and ggg is the gravitational acceleration. This downward force is balanced by two upward forces — the vertical components of the rope tension on either side. Each side contributes a vertical force of TsinθT \sin \thetaTsinθ, so the total vertical force is 2Tsinθ2T \sin \theta2Tsinθ. By setting this equal to the weight, the expression for tension can be derived.
Once the actual tension is computed at a specific angle, the next step involves determining the critical angle at which the tension would reach the maximum limit before the rope breaks. Solving the equation for θ\thetaθ when tension equals its maximum safe value gives the smallest possible angle that still supports the archaeologist safely.
A very small angle results in a nearly horizontal rope, which dramatically increases the tension because more force is needed from each side to hold the weight. Therefore, the closer the angle gets to zero, the more likely the rope will snap. In this case, the rope can withstand even extremely shallow angles, but the calculated lower bound ensures the archaeologist’s safety.
