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Grade 11Mechanics

A light metal stick of square cross section area 5cm*5cm, length 4m, mass 2.5kg is hinged as shown in photo. Determine angle of inclination when water surface is 1m above hinge. What minimum depth of water above high will be required to bring the metal stick in vertical position?

Question image for A light metal stick of square cross section area 5
Profile image of Priyanka lalwani
9 Years agoGrade 11
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2 Answers

Profile image of Vikas TU
9 Years ago
the length of the rod is  4m.
The depth of water is  1m
Then tan Ѳ = Depth of water/ length of rod
  tan Ѳ = ¼
Ѳ = tan-1(1/4)
Then Inclination angle (Ѳ) = tan-1(1/4)
Profile image of Soumyobrata Das
9 Years ago
Volume of the liquid displaced =0.05*0.05*ycosecθ (assuming liquid surface is `y` m above pivot) Buyont force= 0.05*0.05*ycosec θ*9810Taking the moment of weight of stick about pivot equal to the moment of Buyont force about pivot, 24.525*2cosecθ= 0.05*0.05*ycosec θ*9810*(ycosecθ)/2Substituting y=1mcosecθ = 4θ=16.08°(angle of inclination of the stick) To bring the metal stick in vertical position, θ=90°Let`s y` is the new depth y`= 2m