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Question-193073




Question Number 193073 by Tawa11 last updated on 03/Jun/23
Answered by Subhi last updated on 03/Jun/23
  the system is in equalibrium  Σf = 0 at x,y axis  p.cos(30)+2.sin(45)−Q.sin(60)=0  ((√3)/2).p−((√3)/2)Q = −(√2)  P−Q = ((−2(√6))/3)  (i)  p.sin(30)+2.cos(45)+5−8+Q.cos(60)=0  (p/2)+(Q/2)=3−(√2)  P+Q = 6−2(√2)  (ii)  P =((6−2(√2)−((2(√6))/3))/2) =0 .769 N  Q = 2.4 N
$$ \\ $$$${the}\:{system}\:{is}\:{in}\:{equalibrium} \\ $$$$\Sigma{f}\:=\:\mathrm{0}\:{at}\:{x},{y}\:{axis} \\ $$$${p}.{cos}\left(\mathrm{30}\right)+\mathrm{2}.{sin}\left(\mathrm{45}\right)−{Q}.{sin}\left(\mathrm{60}\right)=\mathrm{0} \\ $$$$\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}.{p}−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{Q}\:=\:−\sqrt{\mathrm{2}} \\ $$$${P}−{Q}\:=\:\frac{−\mathrm{2}\sqrt{\mathrm{6}}}{\mathrm{3}}\:\:\left({i}\right) \\ $$$${p}.{sin}\left(\mathrm{30}\right)+\mathrm{2}.{cos}\left(\mathrm{45}\right)+\mathrm{5}−\mathrm{8}+{Q}.{cos}\left(\mathrm{60}\right)=\mathrm{0} \\ $$$$\frac{{p}}{\mathrm{2}}+\frac{{Q}}{\mathrm{2}}=\mathrm{3}−\sqrt{\mathrm{2}} \\ $$$${P}+{Q}\:=\:\mathrm{6}−\mathrm{2}\sqrt{\mathrm{2}}\:\:\left({ii}\right) \\ $$$${P}\:=\frac{\mathrm{6}−\mathrm{2}\sqrt{\mathrm{2}}−\frac{\mathrm{2}\sqrt{\mathrm{6}}}{\mathrm{3}}}{\mathrm{2}}\:=\mathrm{0}\:.\mathrm{769}\:{N} \\ $$$${Q}\:=\:\mathrm{2}.\mathrm{4}\:{N} \\ $$
Commented by Tawa11 last updated on 04/Jun/23
God bless you sir
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$

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