### Problem StatementGiven the following conditions:- Angle α = 35°- Angle β = 55°- Tension in cable 1 is 150 NDetermine:1. The tension in string 2.2. The mass M.### Diagram ExplanationThe diagram illustrates a mass \( M \) suspended from two strings, 1 and 2, which are attached to a horizontal surface at different angles. - String 1 makes an angle α = 35° with the horizontal.- String 2 makes an angle β = 55° with the horizontal.The forces at play include the tension in each string and the weight of the mass \( M \).### Equations and ResolutionTo solve this problem, we can use the equilibrium conditions. The mass is at rest, meaning that the sum of forces in both the x- and y-directions must be zero. 1. **Resolve tension components:** - \( T_1 \), the tension in string 1. - \( T_2 \), the tension in string 2.2. **Equilibrium in the horizontal (x) direction:** \( T_1 \cos(\alpha) = T_2 \cos(\beta) \) Given \( T_1 = 150\,N \): \( 150 \cos(35^\circ) = T_2 \cos(55^\circ) \) Solve for \( T_2 \): \( T_2 = \frac{150 \cos(35^\circ)}{\cos(55^\circ)} \)3. **Equilibrium in the vertical (y) direction:** The sum of the vertical components of the tensions must equal the weight of the mass \( M \): \( T_1 \sin(\alpha) + T_2 \sin(\beta) = M \cdot g \) Given \( g = 9.8\,m/s^2 \): \( 150 \sin(35^\circ) + T_2 \sin(55^\circ) = M \cdot 9.8 \)4. **Calculate mass \( M \):** From the tension calculations, substitute \( T_2 \) back into the vertical direction equation to solve for \( M \): \( M = \frac{150 \sin(35^\

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Answered: If a = 350°, ß= 55° and tension in…

Science

Physics

If a = 350°, ß= 55° and tension in cable 1 is 150 N, determine the tension in string 2 and mass M.

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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