**Problem Statement:**A 104.8 N tightrope walker stands at the center of the rope such that each half of the rope makes an angle of 8 degrees with the horizontal. What is the tension in the rope?**Explanation:**To solve this problem, consider the forces acting on the tightrope walker. The walker applies their entire weight downward (104.8 N), which is balanced by the tension in the two segments of the rope. Because the rope is symmetrical, each half of the rope supports half the walker’s weight.**Steps to Calculate Tension:**1. Resolve the forces horizontally and vertically.2. Use the vertical component of the tension to balance the walker’s weight.3. If \( T \) is the tension in the rope, then the vertical component for each half of the rope is \( T \cdot \sin(8^\circ) \).4. The sum of the vertical components from both halves of the rope equals the weight of the walker: \[ 2 \cdot T \cdot \sin(8^\circ) = 104.8 \, \text{N} \]5. Solve for \( T \): \[ T = \frac{104.8}{2 \cdot \sin(8^\circ)} \] \[ T \approx \frac{104.8}{0.278} \] \[ T \approx 188.49 \, \text{N} \]The tension in the rope is approximately 188.49 N.

Given data The weight of 104.8the tightrope is given as T = 104.8 N. The angle made by the rope…

Answered: A 104.8N tightrope walker stands at the…

A 104.8N tightrope walker stands at the center of the rope such that each half of the rope makes an angle of 8 degrees with the horizontal. What is the tension in the rope?

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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