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A 1.1 ft diameter hollow sphere is made of steel (s.g.= 7.85)with a 0.015ft. wal

ID: 1761083 • Letter: A

Question

A 1.1 ft diameter hollow sphere is made of steel (s.g.= 7.85)with a 0.015ft. wall thickness a) how deep will the sphere sink in the water (i.e. find thedepth h)? b) how much weight must be added so that the sphere is justcompletely submerged (i.e. the total weight of the sphere plusadded weight equals the weight of water displaced)? A 1.1 ft diameter hollow sphere is made of steel (s.g.= 7.85)with a 0.015ft. wall thickness a) how deep will the sphere sink in the water (i.e. find thedepth h)? b) how much weight must be added so that the sphere is justcompletely submerged (i.e. the total weight of the sphere plusadded weight equals the weight of water displaced)?

Explanation / Answer

a)The buoyant force acting on the sphere is B = f * g * V f is the density of fluid(water),g is theacceleration due to gravity and V is the volume of the sphere we know that V = A * h or B = f * g * A * h The buoyant force acting on the sphere is equal to the weightof the sphere that is B = w = M * g or M * g = f * g * A * h or h = (M * g/f * g * A) =(M/f * A) ----------(1) M = s * Vs s = specific gravity * 997.6466kg/m3 specific gravity = 7.85 specific gravity = 7.85 or s = 7.85 * 997.6466 = 7831.5kg/m3 The volume of the sphere is Vs = (4/3)r3 where r = (d/2) d = 1.1 ft or r = (1.1/2) = 0.55 ft = 0.55 * 0.3048 m = 0.1676 m A = 4r2 From equation (1) we get h = (M/f * A) = (s *Vs/f * A) = (s *(4/3)r3/f *4r2) or h = (r/3) * (s/f) f = 1000 kg/m3 b)The weight that must be added so that the sphere is justcompletely submerged is w = weight of the sphere - bouyant force or w = o * g * V - f * g *V= (o - f) * g * V g = 9.8 m/s2 b)The weight that must be added so that the sphere is justcompletely submerged is w = weight of the sphere - bouyant force or w = o * g * V - f * g *V= (o - f) * g * V g = 9.8 m/s2