### 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

Answered: Image /qna-images/answer/339206cf-d5a3-41f2-9411-f7d69e88c066.jpg

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

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

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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Newton’s Third Law of Motion

The horse pulls the cart and the cart pulls the horse in response as per Newton third law. So the big question is, “Don’t the two forces cancel each other? How does the cart accelerate?”

Question

### 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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