Homework Help is Here – Start Your Trial Now!
Science
Physics
Question 3-a, Numerical: A crane is used to carry a heavy container of 1000 kg as shown in the figure. ↑a If the strings can withstand a maximum tension of 14800 N, what maximum acceleration can the crane have (measured in m/s2) before the string breaks? Newl Devious

Question 3-a, Numerical: A crane is used to carry a heavy container of 1000 kg as shown in the figure. ↑a If the strings can withstand a maximum tension of 14800 N, what maximum acceleration can the crane have (measured in m/s2) before the string breaks? Newl Devious

Glencoe Physics: Principles and Problems, Student Edition
Glencoe Physics: Principles and Problems, Student Edition
1st Edition
ISBN:9780078807213
Author:Paul W. Zitzewitz
Publisher:Paul W. Zitzewitz
Chapter4: Forces In One Dimension
Section: Chapter Questions
Problem 90A
See similar textbooks
icon
Related questions
Question
### Question 3-a. Numerical: A crane is used to carry a heavy container of 1000 kg as shown in the figure. ![Crane lifting a heavy container](image-link.jpg) If the strings can withstand a maximum tension of 14800 N, what maximum acceleration can the crane have (measured in m/s²) before the string breaks? **Answer Box:** (Place your numerical answer here) ### Explanation of Diagram: The diagram shows a crane lifting a container. The container is connected to the crane via a string that can withstand a maximum tension of 14800 N. The objective is to determine the maximum acceleration the crane can sustain without breaking the string. ### Solving the Problem: To solve this problem, we need to use Newton's second law of motion. \[ T = m(a + g) \] Where: - \( T \) is the tension in the string, - \( m \) is the mass of the container, - \( a \) is the acceleration of the crane, - \( g \) is the acceleration due to gravity (approximately \( 9.8 \, \text{m/s}^2 \)). Given: - \( T = 14800 \, \text{N} \), - \( m = 1000 \, \text{kg} \), - \( g = 9.8 \, \text{m/s}^2 \). Plug these values into the equation and solve for \( a \): \[ 14800 = 1000(a + 9.8) \] \[ 14800 = 1000a + 9800 \] \[ 14800 - 9800 = 1000a \] \[ 5000 = 1000a \] \[ a = \frac{5000}{1000} \] \[ a = 5 \, \text{m/s}^2 \] Therefore, the maximum acceleration the crane can have before the string breaks is \( 5 \, \text{m/s}^2 \).
Transcribed Image Text:### Question 3-a. Numerical: A crane is used to carry a heavy container of 1000 kg as shown in the figure. ![Crane lifting a heavy container](image-link.jpg) If the strings can withstand a maximum tension of 14800 N, what maximum acceleration can the crane have (measured in m/s²) before the string breaks? **Answer Box:** (Place your numerical answer here) ### Explanation of Diagram: The diagram shows a crane lifting a container. The container is connected to the crane via a string that can withstand a maximum tension of 14800 N. The objective is to determine the maximum acceleration the crane can sustain without breaking the string. ### Solving the Problem: To solve this problem, we need to use Newton's second law of motion. \[ T = m(a + g) \] Where: - \( T \) is the tension in the string, - \( m \) is the mass of the container, - \( a \) is the acceleration of the crane, - \( g \) is the acceleration due to gravity (approximately \( 9.8 \, \text{m/s}^2 \)). Given: - \( T = 14800 \, \text{N} \), - \( m = 1000 \, \text{kg} \), - \( g = 9.8 \, \text{m/s}^2 \). Plug these values into the equation and solve for \( a \): \[ 14800 = 1000(a + 9.8) \] \[ 14800 = 1000a + 9800 \] \[ 14800 - 9800 = 1000a \] \[ 5000 = 1000a \] \[ a = \frac{5000}{1000} \] \[ a = 5 \, \text{m/s}^2 \] Therefore, the maximum acceleration the crane can have before the string breaks is \( 5 \, \text{m/s}^2 \).
### Question 3-a: Numerical #### Problem Statement: A crane is used to carry a heavy container of 1000 kg as shown in the figure. ![Image of a crane lifting a container](insert-image-link-here) #### Task: If the crane has an upward acceleration \( a = 1.60 \, \text{m/s}^2 \), find the tension \( T \) in its strings measured in Newtons (N). #### Explanation: In the image, we see a crane lifting a container using its lifting mechanism. The container is hanging by a string or cable which experiences tension. The following forces act on the container: 1. Gravitational Force (\( F_g \)): \[ F_g = m \cdot g \] Where \( m = 1000 \, \text{kg} \) is the mass of the container and \( g = 9.81 \, \text{m/s}^2 \) is the acceleration due to gravity. 2. Tension (\( T \)) in the string is counteracting the gravitational force and providing the upward acceleration of the container: \[ T - F_g = m \cdot a \] #### Calculation: 1. Calculate the gravitational force (\( F_g \)): \[ F_g = 1000 \, \text{kg} \times 9.81 \, \text{m/s}^2 = 9810 \, \text{N} \] 2. Calculate the net force providing the upward acceleration (\( F_{\text{net}} \)): \[ F_{\text{net}} = m \cdot a = 1000 \, \text{kg} \times 1.60 \, \text{m/s}^2 = 1600 \, \text{N} \] 3. Since the tension has to counteract both the gravitational force and provide the upward acceleration, the tension \( T \) in the string is: \[ T = F_g + F_{\text{net}} \] \[ T = 9810 \, \text{N} + 1600 \, \text{N} = 11410 \, \text{N} \] #### Answer: The tension \( T \) in the strings is \( 11410 \, \text{N} \). --- ### Graphical Explanation The diagram shows a crane with an extended arm lifting a
Transcribed Image Text:### Question 3-a: Numerical #### Problem Statement: A crane is used to carry a heavy container of 1000 kg as shown in the figure. ![Image of a crane lifting a container](insert-image-link-here) #### Task: If the crane has an upward acceleration \( a = 1.60 \, \text{m/s}^2 \), find the tension \( T \) in its strings measured in Newtons (N). #### Explanation: In the image, we see a crane lifting a container using its lifting mechanism. The container is hanging by a string or cable which experiences tension. The following forces act on the container: 1. Gravitational Force (\( F_g \)): \[ F_g = m \cdot g \] Where \( m = 1000 \, \text{kg} \) is the mass of the container and \( g = 9.81 \, \text{m/s}^2 \) is the acceleration due to gravity. 2. Tension (\( T \)) in the string is counteracting the gravitational force and providing the upward acceleration of the container: \[ T - F_g = m \cdot a \] #### Calculation: 1. Calculate the gravitational force (\( F_g \)): \[ F_g = 1000 \, \text{kg} \times 9.81 \, \text{m/s}^2 = 9810 \, \text{N} \] 2. Calculate the net force providing the upward acceleration (\( F_{\text{net}} \)): \[ F_{\text{net}} = m \cdot a = 1000 \, \text{kg} \times 1.60 \, \text{m/s}^2 = 1600 \, \text{N} \] 3. Since the tension has to counteract both the gravitational force and provide the upward acceleration, the tension \( T \) in the string is: \[ T = F_g + F_{\text{net}} \] \[ T = 9810 \, \text{N} + 1600 \, \text{N} = 11410 \, \text{N} \] #### Answer: The tension \( T \) in the strings is \( 11410 \, \text{N} \). --- ### Graphical Explanation The diagram shows a crane with an extended arm lifting a
Expert Solution
trending now

Trending now

This is a popular solution!

steps

Step by step

Solved in 2 steps with 1 images

SEE SOLUTIONCheck out a sample Q&A here
Blurred answer
Knowledge Booster
Third law of motion
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.
Similar questions
  • SEE MORE QUESTIONS
Recommended textbooks for you
Glencoe Physics: Principles and Problems, Student…
Glencoe Physics: Principles and Problems, Student…
Physics
ISBN:
9780078807213
Author:
Paul W. Zitzewitz
Publisher:
Glencoe/McGraw-Hill
SEE MORE TEXTBOOKS