Question

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 (figure 1). the rope will break if the tension in it exceeds 2.85×10⁴ n, and our hero’s mass is 90.4 kg. you may want to review (page). for help with math skills, you may want to review. part b what is the smallest value the angle θ can have if the rope is not to break? express your answer in degrees. view available hint(s)

Explanation:

Step1: Find the weight of the archaeologist

The weight \( W \) of an object is given by \( W = mg \), where \( m = 90.4 \, \text{kg} \) and \( g = 9.8 \, \text{m/s}^2 \).
\[
W = 90.4 \times 9.8 = 885.92 \, \text{N}
\]

Step2: Analyze the forces on the rope

When the archaeologist is at the middle of the rope, the tension \( T \) in each half of the rope has a vertical component. Let the angle of each rope with the horizontal be \( \theta \). The vertical components of the tension in both ropes balance the weight of the archaeologist. So, \( 2T\sin\theta = W \). We know the maximum tension \( T_{\text{max}} = 2.85\times 10^{4} \, \text{N} \). We need to find the minimum \( \theta \) (since smaller \( \theta \) means larger tension for the same weight, so the minimum \( \theta \) corresponds to the maximum tension before breaking).
Rearranging the formula \( 2T\sin\theta = W \) for \( \sin\theta \):
\[
\sin\theta=\frac{W}{2T}
\]
Substitute \( W = 885.92 \, \text{N} \) and \( T = 2.85\times 10^{4} \, \text{N} \):
\[
\sin\theta=\frac{885.92}{2\times2.85\times 10^{4}}=\frac{885.92}{57000}\approx0.01554
\]

Step3: Calculate the angle \( \theta \)

To find \( \theta \), we take the inverse sine (arcsin) of \( 0.01554 \):
\[
\theta=\arcsin(0.01554)\approx0.89^{\circ}
\]

Answer:

\( \approx 0.89^{\circ} \)