IndiaBIX
IndiaBIX
Start typing & press "Enter" or "ESC" to close
  • Home
  • Jobs
  • Results
  • Current Affairs
  • GK
  • Online Test
  • HR Interview
  • BLOG

Force Vectors - General Questions (3)

  • Home
  • Mechanical Engineering
  • Engineering Mechanics Questions & Answers
  • Force Vectors - General Questions
Directions to Solve

Force Vectors - General Questions

17. 

pg2-25

Determine the projection of the position vector r along the aa axis.

A. raa = 35.3 m
B. raa = 6.28 m
C. raa = 5.42 m
D. raa = 5.61 m

Answer: Option C

Explanation:

No answer description available for this question. Let us discuss.

View Answer Discuss Workspace Report

18. 

pg2-14 (1)

Express the force F1 in Cartesian vector form.

A. F1 = (200 i - 200 j + 283 k) lb
B. F1 = (200 i + 200 j + 283 k) lb
C. F1 = (-200 i + 200 j + 565 k) lb
D. F1 = (-200 i + 200 j + 283 k) lb

Answer: Option D

Explanation:

No answer description available for this question. Let us discuss.

View Answer Discuss Workspace Report

19. 

pg2-15

The ball joint is subjected to the three forces shown. Find the magnitude of the resultant force.

A. R = 5.30 kN
B. R = 5.74 kN
C. R = 5.03 kN
D. R = 6.20 kN

Answer: Option D

Explanation:

No answer description available for this question. Let us discuss.

View Answer Discuss Workspace Report

20. 

pg2-23

The cable AO exerts a force on the top of the pole of F = {—120i — 90j — 80k} lb. If the cable has a length of 34 ft, determine the height z of the pole and the location (x,y) of its base.

A. x = 16 ft, y = 16 ft, z = 25 ft
B. x = 12 ft, y = 9 ft, z = 8 ft
C. x = 20 ft, y = 10 ft, z = 14 ft
D. x = 24 ft, y = 18 ft, z = 16 ft

Answer: Option D

Explanation:

No answer description available for this question. Let us discuss.

View Answer Discuss Workspace Report

21. 

pg2-22

The antenna tower is supported by three cables. The forces in these cables are as follows: FB = 520 N, FC = 680 N, and FD = 560 N. Write the resultant of these three forces as a vector.

A. R = (4i +16j-72k) N
B. R = (-120i +40 j-960k) N
C. R = (560i +720j+1440k) N
D. R = (80i +320j-1440k) N

Answer: Option D

Explanation:

No answer description available for this question. Let us discuss.

View Answer Discuss Workspace Report

22. 

pg2-14

Express the Force F2 in Cartesian vector form.

A. F2 = (155 i + 155 j + 300 k) lb
B. F2 = (212 i + 212 j - 519 k) lb
C. F2 = (155 i + 155 j - 300 k) lb
D. F2 = (367 i + 367 j - 300 k) lb

Answer: Option D

Explanation:

No answer description available for this question. Let us discuss.

View Answer Discuss Workspace Report

23. 

pg2-12

If F1 = F2 = 30lb, determine the angles thetaa and phii so that the resultant force is directed along the positive x axis and has a magnitude of FR = 20 lb.

A. thetaa = phii = 70.5°
B. thetaa = phii = 41.4°
C. thetaa = phii = 19.47°
D. thetaa = phii = 18.43°

Answer: Option A

Explanation:

No answer description available for this question. Let us discuss.

View Answer Discuss Workspace Report

24. 

pg2-17

Force F acts on peg A such that one of its components, lying in the x-y plane, has a magnitude of 50 lb. Express F as a Cartesian vector.

A. F = (43.3 i + 25.0 j + 25.0 k) lb
B. F = (43.3 i + 25.0 j + 28.9 k) lb
C. F = (43.3 i - 25.0 j + 25.0 k) lb
D. F = (43.3 i - 25.0 j + 28.9 k) lb

Answer: Option D

Explanation:

No answer description available for this question. Let us discuss.

View Answer Discuss Workspace Report

  • 1
  • 2
  • 3
  • 4

Questions & Answers

Aptitude Chemical Engineering Civil Engineering Computer Science & Engineering Current Affairs Data Interpretation Electrical & Electronics Engineering Electronics & Communication Engineering General Knowledge Logical Reasoning Mechanical Engineering Non Verbal Reasoning Verbal Ability Verbal Reasoning

Interviews

HR Interview

Jobs

Sarkari Jobs

Results

Rojgar ResultSarkari Result

Admission

Admission 2023

Admit Card

Admit Card 2023

Answer Key

Answer Key 2023
copyright
Privacy Policy
© 2026 IndiaBIX. All Rights Reserved.

Report