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Bouyancy, Closed Vessel, Mass

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A) See attachment: Metal part, 2, is attached by a thin cord from a floating wood block, 1. The wood block has a specific gravity(S.G.)=0.3 and dimensions of 50x50x10mm. The metal part has a volume of 6600mm^3. Find mass, m2, of metal part and the tension, T, in the cord.

B) See attachment: The closed vessel has water of air pressure 10psi at the surface. One side of the vessel has a spout that is closed by a 6 inch diameter circular gate that is hinged along one side as illustrated. The horizontal axis of the hinge is located 10 feet below the water surface. Determine the minimum torque that must be applied at the hinge to hold the gate closed. Neglect the weight of the gate and friction at the hinge.

C) See attachment: A 1 meter diameter cylindrical mass, M, is connected to a 2 meter wide rectangular gate. The gate is to open when the water level, h, drops below 2.5 meters. Determine the required value for M. Neglect friction at the gate hinge and pulley.

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Solution Summary

The solution shows all calculations necessary to the questions posed about torque and tension on a buoyant closed vessel.

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For floating bodies
Upthrust or weight of the fluid displaced = weight of the floating body

=>weight of the H2O diplaced by (metal + wood block) U
= weight of the wood block + weight of the metal = Ww + Wm

Volume of metal Vm = 6600 mm^3 = 6.60*10^(-6) m^3
Volume of water displaced by metal Vm' = 6.60*10^(-6) m^3
Volume of wood block Vw = 50*50*10 = 25000 mm^3 = 2.50*10^(-5) m^3
Volume of water displaced by wood block Vw' = 50**50*(10-2.5) mm^3
=> Vw' = 18750 mm^3 = 1.875*10^(-5) m^3

weight of the H2O diplaced by ...

Solution provided by:
  • BEng, Allahabad University, India
  • MSc , Pune University, India
  • PhD (IP), Pune University, India
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