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The force P is applied to the 42-kg block when it is at rest. Determine the magnitude and direction of the friction force exe

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TTT 190 mgsino att - to mgcoso mg For 20 mg sino - f .- Pcos 19° = 0 42x9.81xsin 12° - f = Pcos 19 ~ aty =o, Psinig of N = meano 4249.81% Sin 12 af = Pcos19 429181% sini2-ewxN - pcosig 42x9.81* sinni eixe 42x9.81 x cogi20 - Psin199 = pcos19 85.66

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