400 lbF1 = 400 jF2 = -600 k37F3 = 500 sin 45 i -500 cos 45 k37r2=?600 lbr3=?45M- 11XF,, M,- r2x F, , M,- r3x F,M-M, + M,+ M,500 lb

Given data To determine the position vector and magnitude of the points where forces are acting?

Answered: 400 lb F1 = 400 i F2 = -600 k F3 = 500…

400 lb F1 = 400 i F2 = -600 k F3 = 500 sin 45 i-500 cos 45 k 7t r2=? 600 lb r3=? y 45 M,- riXF,, M,- r2* F,, M,- r3x F, M-M, + M, + M, 500 lb

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

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Question

400 lb
F1 = 400 j
F2 = -600 k
37
F3 = 500 sin 45 i -500 cos 45 k
3
7
r2=?
600 lb
r3=?
45
M- 11XF,, M,- r2x F, , M,- r3x F,
M-M, + M,+ M,
500 lb

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