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he answer in M1 is 4283.29lb, can you show me how to get this answer

he answer in M1 is 4283.29lb, can you show me how to get this answer

Elements Of Electromagnetics
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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the answer in M1 is 4283.29lb, can you show me how to get this answer?

093 M2 - D815 M3 =0 > 0.093 M2 +o.81S M3=0 From egn (ii) and Civ); M2= Z€3t 2036-77lb %3D M3 = -232.92b FBD of 400lb block Ja Tn 400lb fi M, ŹFy =0(1) M, -400 – M2 cos (is)-f28im (15)-0 M-400-2036-千子 cwis)-0-364×036.47 sinlis)-o > M, =4283.29 Ib S Fx -0(->) -P-M2Sin15)+f2ws(15)+ fi =0 O 364 x4283.29 + 0-364x 2036.77 coS CIS)-2036-7 Sun (15) P P = 1748.0sb foro P im moving :. The fhonizontal fo left to the P- 1748 -os lb
Transcribed Image Text:093 M2 - D815 M3 =0 > 0.093 M2 +o.81S M3=0 From egn (ii) and Civ); M2= Z€3t 2036-77lb %3D M3 = -232.92b FBD of 400lb block Ja Tn 400lb fi M, ŹFy =0(1) M, -400 – M2 cos (is)-f28im (15)-0 M-400-2036-千子 cwis)-0-364×036.47 sinlis)-o > M, =4283.29 Ib S Fx -0(->) -P-M2Sin15)+f2ws(15)+ fi =0 O 364 x4283.29 + 0-364x 2036.77 coS CIS)-2036-7 Sun (15) P P = 1748.0sb foro P im moving :. The fhonizontal fo left to the P- 1748 -os lb
(b) When 400 lb s movng deft : FBD for 2000 lb lock; 2000lb M3 fz M2 E Fy =0 (1) M2 cos (1s)+f2 sin (i5)- 2000+ M2 cos(70) + fz Sm (70) =0 M2 cos (1s) 40:364 Mz sino(is)+ M3 cos(70) +o364 M3 sin(70) = 2000 1.06 M2 + 0.684 M3 = 2000- E Fx =0 (>) M2simcis)-f2wslis)- M2 sinl70)+ f3 Cos (70) %3D M28 im(is)-0.364 Mq cos(is)-M3 sio() +0 364 Mg oS(70) =0.
Transcribed Image Text:(b) When 400 lb s movng deft : FBD for 2000 lb lock; 2000lb M3 fz M2 E Fy =0 (1) M2 cos (1s)+f2 sin (i5)- 2000+ M2 cos(70) + fz Sm (70) =0 M2 cos (1s) 40:364 Mz sino(is)+ M3 cos(70) +o364 M3 sin(70) = 2000 1.06 M2 + 0.684 M3 = 2000- E Fx =0 (>) M2simcis)-f2wslis)- M2 sinl70)+ f3 Cos (70) %3D M28 im(is)-0.364 Mq cos(is)-M3 sio() +0 364 Mg oS(70) =0.
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