Question

Archimedes Principle - Downward Push Additional Force As shown in the above figure, a solid block with a mass of 6.00 kg and a density of 790.00 kg/m^3 kg/m3 floats quietly in water (density of watser = 1000 kg/m3.) (a) Calculate the total volume (in m3) of the block. Keep 4 decimal places in all answers. (Don't use scientific notations.) Enter a number Submit (5 attempts remaining) Draw a force diagram (as practice, no submission) including all the forces on the solid object. (b) Find the volume (in m3) of the block that is under the water surface. Keep 4 decimal places in all answers. (Don't use scientific notations.) Enter a number Submit (5 attempts remaining)

Answer 1

(a) volume = mass / density

volume = 6 / 790

volume = 7.6e-3 m³

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(b) To find submerged volume

V_submerged / V_object = density of object / density of fluid

V_submerged / V_object = 790 / 1000

V_submerged / V_object = 0.79

therefore, 79 % of box is submerged under water

volume submerged = 0.79 * 7.6e-3

volume submerged = 6e-3 m³

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(c) Now, when object is fully submerged, then buoyant force is given as

F_b = density of fluid * volume displaced * gravity

F_b = 1000 * (6 / 790) * 9.8

F_b = 74.43 N

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(d) Using net force equation

F + mg = F_b

F = F_b - mg

F = 74.43 - 6 * 9.8

F = 15.6 N