A worker develops a tension T in the cable as he attempts to move the 48-kg cart up the 21° incline. Determine the resulting acceleration a of the cart if (a) T = 135 N and (b) T = 222 N. Neglect all friction, except that at the worker's feet. The acceleration a is positive if up the slope, negative if down the slope. 48 kg 15 21
A worker develops a tension T in the cable as he attempts to move the 48-kg cart up the 21° incline. Determine the resulting acceleration a of the cart if (a) T = 135 N and (b) T = 222 N. Neglect all friction, except that at the worker's feet. The acceleration a is positive if up the slope, negative if down the slope. 48 kg 15 21
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
Related questions
Question
![**Problem Statement:**
A worker develops a tension \( T \) in the cable as he attempts to move the 48-kg cart up the 21° incline. Determine the resulting acceleration \( a \) of the cart if
(a) \( T = 135 \, \text{N} \) and
(b) \( T = 222 \, \text{N} \). Neglect all friction, except that at the worker's feet. The acceleration \( a \) is positive if up the slope, negative if down the slope.
**Diagram:**
The diagram shows a worker pulling a cart up an inclined plane at an angle of 21° with the horizontal. The cart has a mass of 48 kg and there is a pulley system assisting in the pulling motion. A tension \( T \) is developed in the cable.
**Analysis:**
To solve for the acceleration \( a \), we'll use Newton's second law in the direction parallel to the incline, taking into account the forces involved:
1. The component of gravitational force acting down the incline:
\[
F_{\text{gravity}} = m \cdot g \cdot \sin(\theta)
\]
where \( m = 48 \, \text{kg} \), \( g = 9.81 \, \text{m/s}^2 \), and \( \theta = 21^\circ \).
2. The tension in the cable acts up the incline: \( T \).
3. The net force \( F_{\text{net}} \) equals mass times acceleration \( a \):
\[
F_{\text{net}} = T - m \cdot g \cdot \sin(\theta) = m \cdot a
\]
Solve for \( a \):
\[
a = \frac{T - m \cdot g \cdot \sin(\theta)}{m}
\]
**Calculations:**
For part (a) when \( T = 135 \, \text{N} \):
1. Calculate the component of the gravitational force:
\[
F_{\text{gravity}} = 48 \cdot 9.81 \cdot \sin(21^\circ)
\approx 48 \cdot 9.81 \cdot 0.3584
\approx 168.69 \, \](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc5b82ebd-ba0e-48e3-9657-63e0693b3346%2Fe1603c64-2663-48fa-b182-e525a24362c0%2Fsfqnc7b_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
A worker develops a tension \( T \) in the cable as he attempts to move the 48-kg cart up the 21° incline. Determine the resulting acceleration \( a \) of the cart if
(a) \( T = 135 \, \text{N} \) and
(b) \( T = 222 \, \text{N} \). Neglect all friction, except that at the worker's feet. The acceleration \( a \) is positive if up the slope, negative if down the slope.
**Diagram:**
The diagram shows a worker pulling a cart up an inclined plane at an angle of 21° with the horizontal. The cart has a mass of 48 kg and there is a pulley system assisting in the pulling motion. A tension \( T \) is developed in the cable.
**Analysis:**
To solve for the acceleration \( a \), we'll use Newton's second law in the direction parallel to the incline, taking into account the forces involved:
1. The component of gravitational force acting down the incline:
\[
F_{\text{gravity}} = m \cdot g \cdot \sin(\theta)
\]
where \( m = 48 \, \text{kg} \), \( g = 9.81 \, \text{m/s}^2 \), and \( \theta = 21^\circ \).
2. The tension in the cable acts up the incline: \( T \).
3. The net force \( F_{\text{net}} \) equals mass times acceleration \( a \):
\[
F_{\text{net}} = T - m \cdot g \cdot \sin(\theta) = m \cdot a
\]
Solve for \( a \):
\[
a = \frac{T - m \cdot g \cdot \sin(\theta)}{m}
\]
**Calculations:**
For part (a) when \( T = 135 \, \text{N} \):
1. Calculate the component of the gravitational force:
\[
F_{\text{gravity}} = 48 \cdot 9.81 \cdot \sin(21^\circ)
\approx 48 \cdot 9.81 \cdot 0.3584
\approx 168.69 \, \
Expert Solution
This question has been solved!
Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.
Step by step
Solved in 3 steps with 3 images
Knowledge Booster
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, mechanical-engineering and related others by exploring similar questions and additional content below.Recommended textbooks for you
Elements Of Electromagnetics
Mechanical Engineering
ISBN:
9780190698614
Author:
Sadiku, Matthew N. O.
Publisher:
Oxford University Press
Mechanics of Materials (10th Edition)
Mechanical Engineering
ISBN:
9780134319650
Author:
Russell C. Hibbeler
Publisher:
PEARSON
Thermodynamics: An Engineering Approach
Mechanical Engineering
ISBN:
9781259822674
Author:
Yunus A. Cengel Dr., Michael A. Boles
Publisher:
McGraw-Hill Education
Elements Of Electromagnetics
Mechanical Engineering
ISBN:
9780190698614
Author:
Sadiku, Matthew N. O.
Publisher:
Oxford University Press
Mechanics of Materials (10th Edition)
Mechanical Engineering
ISBN:
9780134319650
Author:
Russell C. Hibbeler
Publisher:
PEARSON
Thermodynamics: An Engineering Approach
Mechanical Engineering
ISBN:
9781259822674
Author:
Yunus A. Cengel Dr., Michael A. Boles
Publisher:
McGraw-Hill Education
Control Systems Engineering
Mechanical Engineering
ISBN:
9781118170519
Author:
Norman S. Nise
Publisher:
WILEY
Mechanics of Materials (MindTap Course List)
Mechanical Engineering
ISBN:
9781337093347
Author:
Barry J. Goodno, James M. Gere
Publisher:
Cengage Learning
Engineering Mechanics: Statics
Mechanical Engineering
ISBN:
9781118807330
Author:
James L. Meriam, L. G. Kraige, J. N. Bolton
Publisher:
WILEY