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How to get that M1?? explain please

How to get that M1?? explain please

Structural Analysis
Structural Analysis
6th Edition
ISBN:9781337630931
Author:KASSIMALI, Aslam.
Publisher:KASSIMALI, Aslam.
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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How to get that M1?? explain please

(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.
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
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