Can someone show me all the work for all these problems please? ENGR 230112. The force has a component of 20 N directed along the -y axis as shown. Represent the forceF as a Cartesian vector.13. Determine the component of projection of the force F along the pipe AB.F-|-201 – 30j +60k) Ib4 ft30Answers:1-6: ABBCA, B7: F2x = -318.2 N, F2y=318.2 N8: 666 N9: FR=567 N, 0=38.1° 210: R = 300 N, 0 = 36.9°11: x = 24 ft, y = 18 ft, z = 16 ft12: F= (7.90i – 20j + 8.42k) N13: 36 Ib ENGR 2301CH 3 Exam Review ProblemsMultiple Choice Questions1. Select the correct equations of equilibrium, EF = 0 and E F, = 0 for the forces shown in the figure.200 N60A) P() - F sin 60° – 200 sin 15° = 0 and -P(2) - F cos 60° + 200 cos 15° = 0B) P () - F sin 60° – 200 sin 25° = 0 and -P2) - F cos 60° + 200 cos 25° = 0C) P (2) - F sin 60° – 200 sin 25° = 0 and -P().- F cos 60° + 200 cos 25° = 0D) P(=) - F sin 60° – 200 sin 15° = 0 and -P- F cos 60° + 200 cos 25° = 02. The box has weight W and is held in place on the smooth inclined surface by rope AB. A rotatedcoordinate system is used as shown in the figure. Draw the free-body diagram of the box andselect the correct force equilibrium equation along the x direction.50A) -T sin 30° +W sin 50° = 0B) –T sin 30° +W cos 50° + N sin 50° = 0C) -T cos 30° + W sin 50° = 0D) -T cos 30° + W cos 50° + N sin 50° = 03. Determine the force in each strut and tell whether it is in tension or compression. Determinethe force in each strut and tell whether it is in tension or compression.E FacW=20 IbA) Fab = 11.47 Ib T, Fac = 25.0 lb C, Fad = 14.97 Ib CB) Fab = 1.76 Ib T, Fac = 5.00 lb T, Fad = 3.53 Ib CC) Fab = 11.47 Ib C, Fac = 25.0 lb T, Fad = 14.97 Ib CD) Fab = 1.76 Ib C, Fac = 5.00 lb T, Fad = 3.53 Ib C1

Since, the question number to be solved is not specified, the solution for question 1 is as follows:…

Answered: Multiple Choice Questions 1. Select the…

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

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

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Concept explainers

Management of Labor

Labor, worker or construction laborer are the skilled workers employed in the construction site who performs several works which demand physical strength. The laborer's task is very challenging and often leads to fatigue and exhaustion. Depending on the nature of work, a construction firm may employ many laborers. The laborers prepare sites for construction works, loading and unloading of materials, removing hazards, dismantling, overhauling of equipment, shifting materials, and assisting higher officials. The labor may also require to operate special equipment depending upon its expertise. In a nutshell, the laborer's work is essential and challenging sometimes, and their involvement help to finish the construction process on time.A labor manager undertakes the task of labor-management in a construction site. He helps in the organization of the project by wisely and systematically managing the laborers involved with the project. A labor manager hires people with the expertise, to deliver the right productivity. Proper labor management will ensure high productivity, right skills utilization, availability of workers on time, and timely project completion.

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

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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