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. if the angle between the rope and the horizontal is θ = 11.6°, find the tension in the rope. express your answer in newtons. view available hint(s) input box n

Explanation:

Step1: Calculate 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 in the vertical direction

Let \( T \) be the tension in the rope. Since there are two segments of the rope (symmetrically placed), the vertical component of each tension is \( T\sin\theta \). The sum of the vertical components of the tension must balance the weight of the archaeologist. So, \( 2T\sin\theta = W \).

Step3: Solve for tension \( T \)

We can rearrange the formula from Step 2 to solve for \( T \): \( T=\frac{W}{2\sin\theta} \).
We know \( W = 885.92 \, \text{N} \) and \( \theta = 11.6^\circ \). First, calculate \( \sin(11.6^\circ) \approx \sin(11.6^\circ) \approx 0.2007 \).
Then, \( T=\frac{885.92}{2\times0.2007}=\frac{885.92}{0.4014}\approx 2207 \, \text{N} \) (rounded to a reasonable number of significant figures)

Answer:

\( \approx 2210 \, \text{N} \) (or more precisely calculated value around 2207 N, can be adjusted based on more precise \( \sin \) value calculation)