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[solution] » A block of wood of width d = 0.18m, length l = 0.2m, height h =

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A block of wood of width d = 0.18m, length l = 0.2m, height h =
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A block of wood of width d = 0.18m, length l = 0.2m, height h = 1m and weight m =27kg floats with its axis vertical in some liquid of density ? kg/m3. The equilibrium positioncorresponds to the centre of mass being 0.3m below the surface of the liquid. Initially, theblock is at its equilibrium position with a downward velocity of v0 = 0.5m/s. At the same timean external force G(t) = k cos(?t) Newton is applied to the block. Let x(t) denote the locationof the centre of mass of the block from its equilibrium position with x(t) > 0 corresponding tobelow the equilibrium position. The resistance and the kinetic energy of the liquid are bothneglected and the acceleration due to gravity is 9.8m/s2. Assume that the block never leavesor sinks in the liquid completely.[Hint: The upward force P from the liquid to the block is equal to the weight of the liquiddisplaced by the block. The block is homogenous and so the centre of mass is located at thegeometric centre.]


(a) Calculate the value of ?, the density of the liquid.

(b) Apply Newton?s law of motion on the vertical axis to prove that x(t) is described by the differential equation:     x¨ + 12.25x =k cos(?t)/27.

(c) Solve for x(t) for all ?.

(d) Let k = 5. Will ? = 2 or ? = 3.5 satisfy the assumption that the block never leaves or sinks in the liquid?[Hint: |x + y| ? |x| + |y|

 







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