ME410_mid_soln1

# ME410_mid_soln1 - (FD magnitude and direction> restart>...

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> restart:with(LinearAlgebra); problem 1 Set a = 8 ft. (a) Moment due to the 210-lb force about A as a vector. (b) Moment about line AD. > ptF:=<-8.|0.0|6.>; ptF := [ K 8.00 0.00 6.00 ] > ptC:=<0.0|18.0|0.0>; ptC := [ 0.00 18.00 0.00 ] > vector_cf:=ptF-ptC; vector_cf := [ K 8.00 K 18.00 6.00 ] > mag2:=vector_cf.vector_cf;mag:=sqrt(mag2); mag2 := 424.00 mag := 20.59 > ramda:=vector_cf/mag;

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ramda := [ K .39 K .87 0.29 ] > F210:=ramda*210; F210 := [ K 81.59 K 183.57 61.19 ] > r:=<0.0|18.|0.>; r := [ 0.00 18.00 0.00 ] > vector_ma:=CrossProduct(r,F210); vector_ma := [ 1101.44 K 0.00 1468.58 ] > ad:=<4.5|0.0|-9>; ad := [ 4.50 0.00 K 9 ] > mag2:=ad.ad:mag:=sqrt(mag2); mag := 10.06 > ramda_ad:=ad/mag; ramda_ad := [ 0.45 0.00 K .89 ] > mad:=ramda_ad.vector_ma; mad := K 820.96 problem 2 P = 800 N (a) an equivalent force-couple system at C. (b) an equivalent system consisting of a vertical force at B (FB) and a second force at D

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Unformatted text preview: (FD, magnitude and direction). > restart: > p:=800:theta:=60*Pi/180.; q := 0.33 p > px:=evalf(p*cos(theta));py:=evalf(p*sin(theta)); px := 400.00 py := 692.82 > mc:=py*1.6-px*1.1; mc := 668.51 > md2:=-fby*2.4;md1:=-px*0.6;fb:=solve(md1=md2,fby); md2 := K 2.40 md1 := K 240.00 fb := 100.00 > fdy:=solve(fb+fdy=py,fdy); fdy := 592.82 > fdx:=px;alpha:=arctan(fdy/fdx); fdx := 400.00 a := 0.98 > a:=evalf(alpha*180/Pi); a := 55.99 > fd:=sqrt(fdx^2+fdy^2); fd := 715.15 problem 3 FBD Determine the tension in the cable and the pin reaction at B. To determine the tension T, set sum of moments about B = 0: > restart:T:='T':mb:=12*T-16*T+16*300;T:=solve(mb=0.,T); mb := K 4 T C 4800 T := 1200.00 > Bx:=solve(Bx-T=0.,Bx);By:=solve(By-300+T=0.,By); Bx := 1200.00 By := K 900.00...
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## This note was uploaded on 04/29/2008 for the course ME 410 taught by Professor Katsube during the Winter '08 term at Ohio State.

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ME410_mid_soln1 - (FD magnitude and direction> restart>...

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