Please help with this homework. Thanks ### Problem Statement:**Title:** Analysis of Forces in a 2-Wire Suspension System**Objective:** Consider a 55-N weight suspended by two wires as shown in the accompanying diagram. If the magnitude of vector \( F_1 \) is 35 N, find angle \( \alpha \) and the magnitude of vector \( F_2 \).**Diagram Explanation:** - A 55-N weight is hanging from a point where two wires meet.- The wire on the left side creates an angle \( \alpha \) with the horizontal.- The wire on the right side creates an angle of 62° with the horizontal.- The left wire is under a tension force of \( F_1 = 35 \text{ N} \). This force vector is directed along the wire, upwards to the left.- The right wire is under a tension force designated as \( F_2 \). This force vector is directed along the wire, upwards to the right.- The downward force due to the 55-N weight is represented vertically.### Steps to Solve:1. **Sum of Vertical Components:** - The vertical components of \( F_1 \) and \( F_2 \) must add up to the weight (55 N) to maintain equilibrium. - \( F_{1y} + F_{2y} = 55 \text{ N} \)2. **Sum of Horizontal Components:** - The horizontal components of \( F_1 \) and \( F_2 \) must balance each other out for equilibrium. - \( F_{1x} = F_{2x} \)3. **Breaking Down the Forces:** - For \( F_1 \): \[ F_{1y} = F_1 \sin(\alpha) = 35 \sin(\alpha) \] \[ F_{1x} = F_1 \cos(\alpha) = 35 \cos(\alpha) \] - For \( F_2 \): \[ F_{2y} = F_2 \sin(62^\circ) \] \[ F_{2x} = F_2 \cos(62^\circ) \]4. **Equations for Equilibrium:** \[ 35 \sin(\alpha) + F_2 \sin(62^\circ

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Answered: Consider a 55-N weight suspended by two…

Consider a 55-N weight suspended by two wires as shown in the accompanying figure. If the magnitude of vector F₁ is 35 N, find angle a and the magnitude of vector F2. Su 55 620 F₂

Physics for Scientists and Engineers: Foundations and Connections

1st Edition

ISBN:9781133939146

Author:Katz, Debora M.

Publisher:Katz, Debora M.

Chapter6: Applications Of Newton’s Laws Of Motion

Section: Chapter Questions

Problem 40PQ

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