An object of unknown weight is suspended as shown. The tension in rope 1 is 31 N, and the tension in rope 2 is 25 N. What is the weight of the suspended object? 370 Select one: M O a. 40 N b. 39 N O c. 50 N O d. 22 N O e. 36 N Of. 33 N O g. 56 N Clear my choice
An object of unknown weight is suspended as shown. The tension in rope 1 is 31 N, and the tension in rope 2 is 25 N. What is the weight of the suspended object? 370 Select one: M O a. 40 N b. 39 N O c. 50 N O d. 22 N O e. 36 N Of. 33 N O g. 56 N Clear my choice
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)...
Related questions
Question
Solve it for me pls
![### Physics Problem Explanation: Suspended Object
#### Problem Statement
An object of unknown weight is suspended as shown. The tension in rope 1 is 31 N, and the tension in rope 2 is 25 N. What is the weight of the suspended object?
#### Diagram Description
The provided diagram shows an object of mass \( M \) suspended by two ropes.
- The left rope (Rope 1) forms a 37° angle with the horizontal and has a tension of 31 N.
- The right rope (Rope 2) is vertical and has a tension of 25 N.
#### Multiple Choice Options
Select one of the following options:
- a. 40 N
- b. 39 N
- c. 50 N
- d. 22 N
- e. 36 N
- f. 33 N
- g. 56 N
(Highlighted answer: **b. 39 N**)
#### Detailed Analysis
The system is in equilibrium, meaning the sum of all forces acting on the object must be zero.
1. **Vertical Forces:**
- The tension force in Rope 2 is acting directly upwards: \( T_2 = 25 \, \text{N} \).
- The vertical component of the tension in Rope 1 is \( T_{1y} = T_1 \sin(\theta) \).
- The weight of the object \( W \) acts downwards.
Sum of the vertical forces:
\[
T_{1y} + T_2 = W
\]
\[
31 \, \text{N} \sin(37^\circ) + 25 \, \text{N} = W
\]
\[
31 \, \text{N} \times 0.6018 + 25 \, \text{N} = W
\]
\[
18.656 \, \text{N} + 25 \, \text{N} = W
\]
\[
W = 43.656 \, \text{N}
\]
2. **Horizontal Forces:**
- The horizontal force is balanced which means the horizontal component of the tension in Rope 1 (\( T_{1x} \)) should be zero since there are no other horizontal forces](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F472845c7-72fc-4ada-a49a-97c8d680e8c6%2Fb1104ac9-24cd-48e3-b4dc-63f794a2da13%2F5rnikeh_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Physics Problem Explanation: Suspended Object
#### Problem Statement
An object of unknown weight is suspended as shown. The tension in rope 1 is 31 N, and the tension in rope 2 is 25 N. What is the weight of the suspended object?
#### Diagram Description
The provided diagram shows an object of mass \( M \) suspended by two ropes.
- The left rope (Rope 1) forms a 37° angle with the horizontal and has a tension of 31 N.
- The right rope (Rope 2) is vertical and has a tension of 25 N.
#### Multiple Choice Options
Select one of the following options:
- a. 40 N
- b. 39 N
- c. 50 N
- d. 22 N
- e. 36 N
- f. 33 N
- g. 56 N
(Highlighted answer: **b. 39 N**)
#### Detailed Analysis
The system is in equilibrium, meaning the sum of all forces acting on the object must be zero.
1. **Vertical Forces:**
- The tension force in Rope 2 is acting directly upwards: \( T_2 = 25 \, \text{N} \).
- The vertical component of the tension in Rope 1 is \( T_{1y} = T_1 \sin(\theta) \).
- The weight of the object \( W \) acts downwards.
Sum of the vertical forces:
\[
T_{1y} + T_2 = W
\]
\[
31 \, \text{N} \sin(37^\circ) + 25 \, \text{N} = W
\]
\[
31 \, \text{N} \times 0.6018 + 25 \, \text{N} = W
\]
\[
18.656 \, \text{N} + 25 \, \text{N} = W
\]
\[
W = 43.656 \, \text{N}
\]
2. **Horizontal Forces:**
- The horizontal force is balanced which means the horizontal component of the tension in Rope 1 (\( T_{1x} \)) should be zero since there are no other horizontal forces
Expert Solution
This question has been solved!
Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.
This is a popular solution!
Trending now
This is a popular solution!
Step by step
Solved in 2 steps with 2 images
Knowledge Booster
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, physics and related others by exploring similar questions and additional content below.Recommended textbooks for you
College Physics
Physics
ISBN:
9781305952300
Author:
Raymond A. Serway, Chris Vuille
Publisher:
Cengage Learning
University Physics (14th Edition)
Physics
ISBN:
9780133969290
Author:
Hugh D. Young, Roger A. Freedman
Publisher:
PEARSON
Introduction To Quantum Mechanics
Physics
ISBN:
9781107189638
Author:
Griffiths, David J., Schroeter, Darrell F.
Publisher:
Cambridge University Press
College Physics
Physics
ISBN:
9781305952300
Author:
Raymond A. Serway, Chris Vuille
Publisher:
Cengage Learning
University Physics (14th Edition)
Physics
ISBN:
9780133969290
Author:
Hugh D. Young, Roger A. Freedman
Publisher:
PEARSON
Introduction To Quantum Mechanics
Physics
ISBN:
9781107189638
Author:
Griffiths, David J., Schroeter, Darrell F.
Publisher:
Cambridge University Press
Physics for Scientists and Engineers
Physics
ISBN:
9781337553278
Author:
Raymond A. Serway, John W. Jewett
Publisher:
Cengage Learning
Lecture- Tutorials for Introductory Astronomy
Physics
ISBN:
9780321820464
Author:
Edward E. Prather, Tim P. Slater, Jeff P. Adams, Gina Brissenden
Publisher:
Addison-Wesley
College Physics: A Strategic Approach (4th Editio…
Physics
ISBN:
9780134609034
Author:
Randall D. Knight (Professor Emeritus), Brian Jones, Stuart Field
Publisher:
PEARSON