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Engineering Mechanics - Force Vectors

Exercise : Force Vectors - General Questions
  • Force Vectors - General Questions
6.

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

= = 70.5°
= = 41.4°
= = 19.47°
= = 18.43°
Answer: Option
Explanation:
No answer description is available. Let's discuss.
7.

Express the Force F2 in Cartesian vector form.

F2 = (155 i + 155 j + 300 k) lb
F2 = (212 i + 212 j - 519 k) lb
F2 = (155 i + 155 j - 300 k) lb
F2 = (367 i + 367 j - 300 k) lb
Answer: Option
Explanation:
No answer description is available. Let's discuss.
8.

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.

R = (4i +16j-72k) N
R = (-120i +40 j-960k) N
R = (560i +720j+1440k) N
R = (80i +320j-1440k) N
Answer: Option
Explanation:
No answer description is available. Let's discuss.
9.

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.

x = 16 ft, y = 16 ft, z = 25 ft
x = 12 ft, y = 9 ft, z = 8 ft
x = 20 ft, y = 10 ft, z = 14 ft
x = 24 ft, y = 18 ft, z = 16 ft
Answer: Option
Explanation:
No answer description is available. Let's discuss.
10.

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

R = 5.30 kN
R = 5.74 kN
R = 5.03 kN
R = 6.20 kN
Answer: Option
Explanation:
No answer description is available. Let's discuss.
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