Homework Help is Here – Start Your Trial Now!
Engineering
Mechanical Engineering
Find the value of the input torque on the crank-shaft mechanism in following Figure. The centers of gravity of link 2, 3 and 4 are at G₂, G3 and G4 respectively. The masses and moments of inertia of the links are: m₂=2 kg, m3 =0.5 kg, m4 0.5 kg, IG2 = 0.001 kg m², IG3 = 0.01 kg m² and IG4 = 0.002 kg m². It is known that: O₂A = 3 cm, AB = 7 cm, AG3 = 2 cm, 0₂ = 60°, ₂ =-20 rad/s², a₂ = -100 rad/s² and that external force P4 has a magnitude of 98 N. consider 03 =338.2° and the following accelerations for links 3 and 4: ag3 = -250.26^i- 849:45^j cm/s², 174:44 rad/ s², aG4 = -25.45^i cm/s², a4 = 0. A = α3 =

Find the value of the input torque on the crank-shaft mechanism in following Figure. The centers of gravity of link 2, 3 and 4 are at G₂, G3 and G4 respectively. The masses and moments of inertia of the links are: m₂=2 kg, m3 =0.5 kg, m4 0.5 kg, IG2 = 0.001 kg m², IG3 = 0.01 kg m² and IG4 = 0.002 kg m². It is known that: O₂A = 3 cm, AB = 7 cm, AG3 = 2 cm, 0₂ = 60°, ₂ =-20 rad/s², a₂ = -100 rad/s² and that external force P4 has a magnitude of 98 N. consider 03 =338.2° and the following accelerations for links 3 and 4: ag3 = -250.26^i- 849:45^j cm/s², 174:44 rad/ s², aG4 = -25.45^i cm/s², a4 = 0. A = α3 =

Elements Of Electromagnetics
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
See similar textbooks
icon
Related questions
Question

Solve both show all steps  and solutions 

**Problem Statement:** Find the value of the input torque on the crank-shaft mechanism in the following figure. The centers of gravity of links 2, 3, and 4 are at \( G_2, G_3, \) and \( G_4 \) respectively. The masses and moments of inertia of the links are: \[ m_2 = 2 \text{ kg}, \, m_3 = 0.5 \text{ kg}, \, m_4 = 0.5 \text{ kg}, \, I_{G_2} = 0.001 \, \text{kg m}^2, \, I_{G_3} = 0.01 \, \text{kg m}^2, \, \text{and} \, I_{G_4} = 0.002 \, \text{kg m}^2. \] It is known that: \[ O_2A = 3 \text{ cm}, \, AB = 7 \text{ cm}, \, AG_3 = 2 \text{ cm}, \, \theta_2 = 60^\circ, \, \omega_2 = -20 \, \text{rad/s}^2, \, \alpha_2 = -100 \, \text{rad/s}^2 \] and that external force \( P_4 \) has a magnitude of 98 N. Consider \( \theta_3 = 338.2^\circ \) and the following accelerations for links 3 and 4: \[ a_{G_3} = -250.26\hat{i} - 849.45\hat{j} \, \text{cm/s}^2, \, \alpha_3 = 174.44 \, \text{rad/s}^2, \, a_{G_4} = -25.45\hat{i} \, \text{cm/s}^2, \, \alpha_4 = 0. \] **Diagram Description:** The diagram depicts a crank-shaft mechanism with links marked as 2, 3, and 4. The point \( O_2 \) serves as the pivot point with \( G_2 \) coinciding with it. \( A \) is a pivot on
Transcribed Image Text:**Problem Statement:** Find the value of the input torque on the crank-shaft mechanism in the following figure. The centers of gravity of links 2, 3, and 4 are at \( G_2, G_3, \) and \( G_4 \) respectively. The masses and moments of inertia of the links are: \[ m_2 = 2 \text{ kg}, \, m_3 = 0.5 \text{ kg}, \, m_4 = 0.5 \text{ kg}, \, I_{G_2} = 0.001 \, \text{kg m}^2, \, I_{G_3} = 0.01 \, \text{kg m}^2, \, \text{and} \, I_{G_4} = 0.002 \, \text{kg m}^2. \] It is known that: \[ O_2A = 3 \text{ cm}, \, AB = 7 \text{ cm}, \, AG_3 = 2 \text{ cm}, \, \theta_2 = 60^\circ, \, \omega_2 = -20 \, \text{rad/s}^2, \, \alpha_2 = -100 \, \text{rad/s}^2 \] and that external force \( P_4 \) has a magnitude of 98 N. Consider \( \theta_3 = 338.2^\circ \) and the following accelerations for links 3 and 4: \[ a_{G_3} = -250.26\hat{i} - 849.45\hat{j} \, \text{cm/s}^2, \, \alpha_3 = 174.44 \, \text{rad/s}^2, \, a_{G_4} = -25.45\hat{i} \, \text{cm/s}^2, \, \alpha_4 = 0. \] **Diagram Description:** The diagram depicts a crank-shaft mechanism with links marked as 2, 3, and 4. The point \( O_2 \) serves as the pivot point with \( G_2 \) coinciding with it. \( A \) is a pivot on
To carry out dynamic force analysis of the four-bar mechanism shown in the figure, it is required to find the inertial radius of the links. **Where:** - \( w2 = 20 \text{ rad/s (cw)}, a2 = 160 \text{ rad/s}^2 \text{ (cw)} \) - \( OA = 250 \text{mm}, OG2 = 110 \text{mm}, AB = 300 \text{mm} \) - \( AG3 = 150 \text{mm}, BC = 300 \text{mm}, CG4 = 140 \text{mm} \) - \( OC = 550 \text{mm}, \angle AOC = 60^\circ \) **The masses & mass moment of inertia of the various members are:** | Link | Mass, m | MMI (IG, \(\text{Kgm}^2\)) | |------|---------|-----------------------------| | 2 | 20.7 kg | 0.01872 | | 3 | 9.66 kg | 0.01105 | | 4 | 23.47 kg| 0.0277 | **Diagrams:** 1. **Diagram (a):** - Shows the four-bar mechanism with labeled points (1, 2, 3, 4, A, B), angles (\(\alpha_3\), \(\alpha_4\)), and forces (\(F_{i2}\), \(F_{i3}\), \(F_{i4}\)). - It also indicates the scale: 1 cm = 10 cm. 2. **Acceleration Polygon:** - Displays the acceleration polygon with points labeled (a, b, a', b', a'', b''). - Scale: 1 cm = 20 m/sec\(^2\). **Note:** The listed scales are not perfectly correct. Consider the following values to find out the correct scale of the acceleration polygon. - \( V_A = 250 \times 20; 5 \text{ m/s} \) - \( V_B = 4 \text{ m/s}, V_{BA} = 4.75 \text{ m/s} \) - \( a'_A = 250 \times 20^2; 100 \text{ m/s
Transcribed Image Text:To carry out dynamic force analysis of the four-bar mechanism shown in the figure, it is required to find the inertial radius of the links. **Where:** - \( w2 = 20 \text{ rad/s (cw)}, a2 = 160 \text{ rad/s}^2 \text{ (cw)} \) - \( OA = 250 \text{mm}, OG2 = 110 \text{mm}, AB = 300 \text{mm} \) - \( AG3 = 150 \text{mm}, BC = 300 \text{mm}, CG4 = 140 \text{mm} \) - \( OC = 550 \text{mm}, \angle AOC = 60^\circ \) **The masses & mass moment of inertia of the various members are:** | Link | Mass, m | MMI (IG, \(\text{Kgm}^2\)) | |------|---------|-----------------------------| | 2 | 20.7 kg | 0.01872 | | 3 | 9.66 kg | 0.01105 | | 4 | 23.47 kg| 0.0277 | **Diagrams:** 1. **Diagram (a):** - Shows the four-bar mechanism with labeled points (1, 2, 3, 4, A, B), angles (\(\alpha_3\), \(\alpha_4\)), and forces (\(F_{i2}\), \(F_{i3}\), \(F_{i4}\)). - It also indicates the scale: 1 cm = 10 cm. 2. **Acceleration Polygon:** - Displays the acceleration polygon with points labeled (a, b, a', b', a'', b''). - Scale: 1 cm = 20 m/sec\(^2\). **Note:** The listed scales are not perfectly correct. Consider the following values to find out the correct scale of the acceleration polygon. - \( V_A = 250 \times 20; 5 \text{ m/s} \) - \( V_B = 4 \text{ m/s}, V_{BA} = 4.75 \text{ m/s} \) - \( a'_A = 250 \times 20^2; 100 \text{ m/s
Expert Solution
steps

Step by step

Solved in 3 steps with 5 images

SEE SOLUTIONCheck out a sample Q&A here
Blurred answer
Follow-up Questions
Read through expert solutions to related follow-up questions below.
Follow-up Question

Please solve both show all steps and solutions please

Solution
Bartleby Expert
SEE SOLUTION
Recommended textbooks for you
Elements Of Electromagnetics
Elements Of Electromagnetics
Mechanical Engineering
ISBN:
9780190698614
Author:
Sadiku, Matthew N. O.
Publisher:
Oxford University Press
Mechanics of Materials (10th Edition)
Mechanics of Materials (10th Edition)
Mechanical Engineering
ISBN:
9780134319650
Author:
Russell C. Hibbeler
Publisher:
PEARSON
Thermodynamics: An Engineering Approach
Thermodynamics: An Engineering Approach
Mechanical Engineering
ISBN:
9781259822674
Author:
Yunus A. Cengel Dr., Michael A. Boles
Publisher:
McGraw-Hill Education
Control Systems Engineering
Control Systems Engineering
Mechanical Engineering
ISBN:
9781118170519
Author:
Norman S. Nise
Publisher:
WILEY
Mechanics of Materials (MindTap Course List)
Mechanics of Materials (MindTap Course List)
Mechanical Engineering
ISBN:
9781337093347
Author:
Barry J. Goodno, James M. Gere
Publisher:
Cengage Learning
Engineering Mechanics: Statics
Engineering Mechanics: Statics
Mechanical Engineering
ISBN:
9781118807330
Author:
James L. Meriam, L. G. Kraige, J. N. Bolton
Publisher:
WILEY
SEE MORE TEXTBOOKS