Homework Help is Here – Start Your Trial Now!
Engineering
Civil Engineering
Multiple Choice Questions 1. Select the correct equations of equilibrium, E F = 0 and E F, = 0 for the forces shown in the figure. 200 N 60

Multiple Choice Questions 1. Select the correct equations of equilibrium, E F = 0 and E F, = 0 for the forces shown in the figure. 200 N 60

Structural Analysis
Structural Analysis
6th Edition
ISBN:9781337630931
Author:KASSIMALI, Aslam.
Publisher:KASSIMALI, Aslam.
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
See similar textbooks
icon
Related questions
icon
Concept explainers
Question
Can someone show me all the work for all these problems please?
ENGR 2301 12. The force has a component of 20 N directed along the -y axis as shown. Represent the force F as a Cartesian vector. 13. Determine the component of projection of the force F along the pipe AB. F-|-201 – 30j +60k) Ib 4 ft 30 Answers: 1-6: ABBCA, B 7: F2x = -318.2 N, F2y=318.2 N 8: 666 N 9: FR=567 N, 0=38.1° 2 10: R = 300 N, 0 = 36.9° 11: x = 24 ft, y = 18 ft, z = 16 ft 12: F= (7.90i – 20j + 8.42k) N 13: 36 Ib
Transcribed Image Text:ENGR 2301 12. The force has a component of 20 N directed along the -y axis as shown. Represent the force F as a Cartesian vector. 13. Determine the component of projection of the force F along the pipe AB. F-|-201 – 30j +60k) Ib 4 ft 30 Answers: 1-6: ABBCA, B 7: F2x = -318.2 N, F2y=318.2 N 8: 666 N 9: FR=567 N, 0=38.1° 2 10: R = 300 N, 0 = 36.9° 11: x = 24 ft, y = 18 ft, z = 16 ft 12: F= (7.90i – 20j + 8.42k) N 13: 36 Ib
ENGR 2301 CH 3 Exam Review Problems Multiple Choice Questions 1. Select the correct equations of equilibrium, EF = 0 and E F, = 0 for the forces shown in the figure. 200 N 60 A) P() - F sin 60° – 200 sin 15° = 0 and -P (2) - F cos 60° + 200 cos 15° = 0 B) P () - F sin 60° – 200 sin 25° = 0 and -P 2) - F cos 60° + 200 cos 25° = 0 C) P (2) - F sin 60° – 200 sin 25° = 0 and -P (). - F cos 60° + 200 cos 25° = 0 D) P(=) - F sin 60° – 200 sin 15° = 0 and -P - F cos 60° + 200 cos 25° = 0 2. The box has weight W and is held in place on the smooth inclined surface by rope AB. A rotated coordinate system is used as shown in the figure. Draw the free-body diagram of the box and select the correct force equilibrium equation along the x direction. 50 A) -T sin 30° +W sin 50° = 0 B) –T sin 30° +W cos 50° + N sin 50° = 0 C) -T cos 30° + W sin 50° = 0 D) -T cos 30° + W cos 50° + N sin 50° = 0 3. Determine the force in each strut and tell whether it is in tension or compression. Determine the force in each strut and tell whether it is in tension or compression. E Fac W=20 Ib A) Fab = 11.47 Ib T, Fac = 25.0 lb C, Fad = 14.97 Ib C B) Fab = 1.76 Ib T, Fac = 5.00 lb T, Fad = 3.53 Ib C C) Fab = 11.47 Ib C, Fac = 25.0 lb T, Fad = 14.97 Ib C D) Fab = 1.76 Ib C, Fac = 5.00 lb T, Fad = 3.53 Ib C 1
Transcribed Image Text:ENGR 2301 CH 3 Exam Review Problems Multiple Choice Questions 1. Select the correct equations of equilibrium, EF = 0 and E F, = 0 for the forces shown in the figure. 200 N 60 A) P() - F sin 60° – 200 sin 15° = 0 and -P (2) - F cos 60° + 200 cos 15° = 0 B) P () - F sin 60° – 200 sin 25° = 0 and -P 2) - F cos 60° + 200 cos 25° = 0 C) P (2) - F sin 60° – 200 sin 25° = 0 and -P (). - F cos 60° + 200 cos 25° = 0 D) P(=) - F sin 60° – 200 sin 15° = 0 and -P - F cos 60° + 200 cos 25° = 0 2. The box has weight W and is held in place on the smooth inclined surface by rope AB. A rotated coordinate system is used as shown in the figure. Draw the free-body diagram of the box and select the correct force equilibrium equation along the x direction. 50 A) -T sin 30° +W sin 50° = 0 B) –T sin 30° +W cos 50° + N sin 50° = 0 C) -T cos 30° + W sin 50° = 0 D) -T cos 30° + W cos 50° + N sin 50° = 0 3. Determine the force in each strut and tell whether it is in tension or compression. Determine the force in each strut and tell whether it is in tension or compression. E Fac W=20 Ib A) Fab = 11.47 Ib T, Fac = 25.0 lb C, Fad = 14.97 Ib C B) Fab = 1.76 Ib T, Fac = 5.00 lb T, Fad = 3.53 Ib C C) Fab = 11.47 Ib C, Fac = 25.0 lb T, Fad = 14.97 Ib C D) Fab = 1.76 Ib C, Fac = 5.00 lb T, Fad = 3.53 Ib C 1
Expert Solution
trending now

Trending now

This is a popular solution!

steps

Step by step

Solved in 2 steps

SEE SOLUTIONCheck out a sample Q&A here
Blurred answer
Knowledge Booster
Management of Labor
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, civil-engineering and related others by exploring similar questions and additional content below.
Recommended textbooks for you
Structural Analysis
Structural Analysis
Civil Engineering
ISBN:
9781337630931
Author:
KASSIMALI, Aslam.
Publisher:
Cengage,
Structural Analysis (10th Edition)
Structural Analysis (10th Edition)
Civil Engineering
ISBN:
9780134610672
Author:
Russell C. Hibbeler
Publisher:
PEARSON
Principles of Foundation Engineering (MindTap Cou…
Principles of Foundation Engineering (MindTap Cou…
Civil Engineering
ISBN:
9781337705028
Author:
Braja M. Das, Nagaratnam Sivakugan
Publisher:
Cengage Learning
Fundamentals of Structural Analysis
Fundamentals of Structural Analysis
Civil Engineering
ISBN:
9780073398006
Author:
Kenneth M. Leet Emeritus, Chia-Ming Uang, Joel Lanning
Publisher:
McGraw-Hill Education
Sustainable Energy
Sustainable Energy
Civil Engineering
ISBN:
9781337551663
Author:
DUNLAP, Richard A.
Publisher:
Cengage,
Traffic and Highway Engineering
Traffic and Highway Engineering
Civil Engineering
ISBN:
9781305156241
Author:
Garber, Nicholas J.
Publisher:
Cengage Learning
SEE MORE TEXTBOOKS