Question

Im 11 m If the maximum force in each rod can not exceed F, determine the greatest mass of the crate that can be supported. Use F = 3,300 N, L1 = 1.2 m, L2 = 4.8 m. and 3 = 3.8 m for your calculations.

Answer 1

(-1, -2, 3-6) EBO (4.8, 2,3) Arco BO . (21.2 500 Ao Foz swimg -4.8 ê+ 27 + 3R ,~4.80 29+3R V. 9)? 0² 762 6.6033 • direction Foo - 10-29+3.818 -11 » 2h +3.8k VO2 + (2) 243-82 4.41 direction of apo - 21-1.29+1R = 2^-1.29+1 V2)*71.2)2+602 2.5371 let mag in Ao Bo, Co rod be FAO, Foo Fco F. (-0.824 0.330+0.58) JF) FBO (FB) (-0-226 2-0.4538+06168) FAO - FAO) (0.7880-0.4725 + 0.394R)

Ex=0 Fool(-0.82) + (FBo) (- 02260) - 1780) (-02268) +/F..) (0-188 1) = 0 Ery=0 (F) (0.331) + FBol to 39 lrgo) (-0-4531) + FP) (-0-4323) –o 22-0 Feol (0.5R) + 1 F ou Co.GG16R) FAN) C-394R) = renogging ♡ s 2 + FBd (-0.226) = -(FAO) 0.788 0.788 -1.015 Fe) + (-0.2869) F-Fad - © 0.33 0.472 -0.453 0.472 0.4 \Fed + (-0.96) Foolt (FAO) - © 4 +6 = 0 -0.315(Fco) + (- 1.2468 )|f80 = 0 Fcol = 1.2468 | FB Ecol= 3.961FBol 6 0.315

San from ® s© 0.7x3.967 Fol+ (-0.96)| Fool-(FA) Fol= 1.812 FBol - So lfcol is highest in magnitude Fol= 3300N IFBol = 3300 - 833.33 3.96 IFAO) = 1.812 | Fowl = 1509.99 N Plutting values in ♡ 3300 *0.5 + 83333 (06110) +1509.99x0-394) =w W=2796.26 m=2796.26-285.042 kg [285.042 leg