Question

If the maximum force in each rod can not exceed F, determine the greatest mass of the crate that can be supported. Use F = 2,300 N, L1 = 2.2 m, L2 = 3.8 m, and L3 = 2.8 m for your calculations.

3 m --2 m2 1 m 2 m 0 2 m 1 m L1- If the maximum force in each rod can not exceed F, determine the greatest mass of the crate that can be supported. Use F = 2,300 N, L1 = 2.2 m, L2 = 3.8 m, and L3 = 2.8 m for your calculations. Your answer: Greatest mass of the crate in kg (round the result to the nearest integer):

Answer 1

FOR = FOA DOA = FOR OR 10A) - = FOA 26-2-2}+k Ston D7C023372 FOA = FOA(21-2-2}+k) 19.84 Foe = FOB ROB = FOB OB = Fogf-1-23 +2.8K ) 1. Kes(-2; 2.82 ) = FOB/-(-2) +2-8k J12-84 Foca Foc doc a Foc od 1021 - Foc -381 +21+3k √ 63.852 +2²3² For = Foc/-3.80+27 +3k 127.44 WE-WK For equilibrium, Foat For Fodt n zo FOA (21-2.25 +") + Fool ---2] +28£.) + Foc (-3-86+2)+3k) - wkso 19.84 V12-84 127.804 į Equatog coefflients of cs sok lo zenu Garga) Foa + Gra) 0 + (parter) Foc 20

274 j: 0932.) Fox + lino) fos ) foc 30 K Centru a + (2) Fons ) foc - W 20 ) 64 (20) for Hem) foc = 9.6im solving 0.0 and ③. 12.84 27:44 GA = 18.472m FOB = -9.583m Foc - 19.922.. Sini Foc = (9.922m , Foc=2300N IL 2300 N = 19.922 m m =115.45 im = 116 kg /