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
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
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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![### 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](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F41a1af84-f79e-495e-9bf6-e155c46aa5b8%2Fa73d8446-3145-4536-8c1b-7c74bb7ba5d4%2Fjz40wo_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### 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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