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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
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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
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