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Question Number 103739 by aurpeyz last updated on 17/Jul/20

Answered by Rio Michael last updated on 17/Jul/20

 at limiting equilibrium, F_f   + mg sin θ = P   but F_f  = Nμ  but N = mg cos θ = (163.5)cos 14.5 = 158.29 to the nearest   100^(th)  = 200 N  ⇒F_f  = 200 × 0.6 =120 N  ⇒ P = 120 + 163.5 sin 14.5 = 161 N

$$\:\mathrm{at}\:\mathrm{limiting}\:\mathrm{equilibrium},\:{F}_{{f}} \:\:+\:\mathrm{mg}\:\mathrm{sin}\:\theta\:=\:{P}\: \\ $$$$\mathrm{but}\:{F}_{{f}} \:=\:{N}\mu\:\:\mathrm{but}\:{N}\:=\:\mathrm{mg}\:\mathrm{cos}\:\theta\:=\:\left(\mathrm{163}.\mathrm{5}\right)\mathrm{cos}\:\mathrm{14}.\mathrm{5}\:=\:\mathrm{158}.\mathrm{29}\:\mathrm{to}\:\mathrm{the}\:\mathrm{nearest}\: \\ $$$$\mathrm{100}^{\mathrm{th}} \:=\:\mathrm{200}\:{N} \\ $$$$\Rightarrow{F}_{{f}} \:=\:\mathrm{200}\:×\:\mathrm{0}.\mathrm{6}\:=\mathrm{120}\:{N} \\ $$$$\Rightarrow\:{P}\:=\:\mathrm{120}\:+\:\mathrm{163}.\mathrm{5}\:\mathrm{sin}\:\mathrm{14}.\mathrm{5}\:=\:\mathrm{161}\:{N} \\ $$$$ \\ $$

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