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Scalars And Vectors Rectangular Coordinates: Solutions of Problems, Exercises of Electrical Engineering

Jaypee University of Engineering & TechnologyElectrical Engineering

Prof. Gulzarilal Shroff suggested this solution to problem related to Electrical Engineering course at Jaypee University of Engineering and Technology. It includes: Components, Projections, Scalar, Vector, Mass, Body, Weight, Newton, Second, Law

Typology: Exercises

2011/2012

Uploaded on 07/20/2012

sharman
sharman 🇮🇳

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Solutions Of Problems For Course ME-1002
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  • Solutions Of Problems For Course ME-
Solutions Of Problems For Course ME-1002 tie | Vy = Veosdy = 20C06) = —!2 1 Wag? > (ex-ae- te | we mg = 40(a31) = 322.8 $0 Y= lei - 12] w- 32" (Sogn) = Feel =mass of body = 50 k9 cee. of grovity at altitude we un sea (evel radtus of earth ittude ef bed . mass of earth, Gagner conston h _4f |lim g From Newton's second faw &@ grav, law mM _ mM 9 A\2 j mg= & feypje #99 Fee, 8 9° Soleen) 7 . 637) Biimedvarement a 32 welts Weight Wemg = 50(7.80665) agus) = 185.97" By cateaionon 2 VR%¢/* = 00 = 3.16 writs ziz]Fesomi4 Scalar} F, 8G)3) I ee F= Fnag ee E 5 = ART N = tooo BESS] walt og . aca) | 542% Stee a j Vector [ee -250i N \ ’ = [486i +594] N Components | Fy = 433 N ale Components, g z Sin 120". siniS* a of a8. Ff =6 / 7% Fa) ee OTEK sin 120" _ _sin45° z Fi @ EN s45° 14_ kN, Rocratiens 1.2 KN i PL= 2c 15° = 1.9392 kN & retayuen WP BTR =F, = 400 + 400 cas A= G00N 7 3 Ry ZR, = 400 + 490 eos 600 Y =0CEi- ti) Ry z= Fy = 400 sin GN = 346 N ~ s i-@i oN ook + S46 N = GIS N Bilt r Pe, = BS 3/4 = [Pal = 2 fies) = 40.N 30.8 in @ = 19 Sin 43.1" BAN eile Fars Law of sines + noo 200 2s R x si0°/ sin te Sin lo? ~ Sin 30° a ttl a on R= 200K ‘oN ie pen RE R, = G22n +b — ao Prajection P= Rees 30° = B00 cos 30"* 693 N F, mo ewe? mee 5 5.10 kN docsity.com