The springs BA and BC each have a stiffness of 500 N/m and an unstretched length of 3 m. Determine the horizontal force F applied to the cord which is attached to the small ring B so that the displacement of the ring from the wall is d = 1.5 m. OC 135 N 158 N 177 N 241 N F k = 500 N/m B k = 500 N/m d 6 m

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Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
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### Problem Statement:
Two springs, BA and BC, each have a stiffness of 500 N/m and an unstretched length of 3 meters. Determine the horizontal force \( F \) applied to the cord, which is attached to the small ring B, so that the displacement of the ring from the wall is \( d = 1.5 \) meters.

### Diagram Explanation:
The diagram displays a mechanical setup involving:
- A hand applying a horizontal force \( F \) pulling a cord attached to a ring \( B \).
- Two springs, \( k = 500 \, \text{N/m} \) each, connected from points \( A \) and \( C \) to the ring \( B \).
- The vertical distance between points \( A \) and \( C \) along the wall is 6 meters.
- The ring \( B \) is displaced 1.5 meters horizontally from the wall.

### Analysis:
We are to identify from the given options the value of force \( F \) that creates the displacement \( d \).

### Given Options:
- 135 N
- 158 N
- 177 N
- 241 N

### Mechanical Analysis:
Given:
- Spring constant, \( k = 500 \, \text{N/m} \)
- Unstretched length, \( L = 3 \) meters
- Vertical distance between A and C, \( h = 6 \) meters
- Horizontal displacement \( d = 1.5 \) meters

### Calculations:
The first step is to find the length of the springs when the ring \( B \) is displaced.

Let's denote the stretched length of the spring as \( l_{new} \).

Using the Pythagorean theorem for triangles \( ABA \) and \( BCB \):

\[ l_{new} = \sqrt{(6/2)^2 + 1.5^2} = \sqrt{3^2 + 1.5^2} = \sqrt{9 + 2.25} = \sqrt{11.25} \approx 3.35 \, \text{m} \]

Extension \( e \) of the spring:
\[ e = l_{new} - 3 = 3.35 - 3 = 0.35 \, \text{m} \]

Force exerted by each spring:
\[ F_s = k \cdot
Transcribed Image Text:### Problem Statement: Two springs, BA and BC, each have a stiffness of 500 N/m and an unstretched length of 3 meters. Determine the horizontal force \( F \) applied to the cord, which is attached to the small ring B, so that the displacement of the ring from the wall is \( d = 1.5 \) meters. ### Diagram Explanation: The diagram displays a mechanical setup involving: - A hand applying a horizontal force \( F \) pulling a cord attached to a ring \( B \). - Two springs, \( k = 500 \, \text{N/m} \) each, connected from points \( A \) and \( C \) to the ring \( B \). - The vertical distance between points \( A \) and \( C \) along the wall is 6 meters. - The ring \( B \) is displaced 1.5 meters horizontally from the wall. ### Analysis: We are to identify from the given options the value of force \( F \) that creates the displacement \( d \). ### Given Options: - 135 N - 158 N - 177 N - 241 N ### Mechanical Analysis: Given: - Spring constant, \( k = 500 \, \text{N/m} \) - Unstretched length, \( L = 3 \) meters - Vertical distance between A and C, \( h = 6 \) meters - Horizontal displacement \( d = 1.5 \) meters ### Calculations: The first step is to find the length of the springs when the ring \( B \) is displaced. Let's denote the stretched length of the spring as \( l_{new} \). Using the Pythagorean theorem for triangles \( ABA \) and \( BCB \): \[ l_{new} = \sqrt{(6/2)^2 + 1.5^2} = \sqrt{3^2 + 1.5^2} = \sqrt{9 + 2.25} = \sqrt{11.25} \approx 3.35 \, \text{m} \] Extension \( e \) of the spring: \[ e = l_{new} - 3 = 3.35 - 3 = 0.35 \, \text{m} \] Force exerted by each spring: \[ F_s = k \cdot
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