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45° Figure 1 Question 2 HA=0.2 g=0.3 The linkage in Figure 2 connects Block A and Block B which each have a weight of 60 N. T

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Sd:- (tsFx=0 30 FAc Sin (us) – Fg sin(30) = 0 sin(30) sin(45) FAC= Fec FAC Cos (45) + Foc cos(30) - P=0 Fece COS(30) + cos (

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