Question

A wood cube 0.28m on each side has a density of 680kg/m3 and floats levelly in water.

What is the distance from the top of the wood to the water surface? (in m)

Express your answer using two significant figures

What mass has to be placed on top of the wood so that its top is just at the water level? (kg)

Express your answer using two significant figures.

Answer 1

You have :
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W sub Wood = ( V sub Wood ) ( rho sub Wood ) ( g / g sub C )

V sub Wood = ( 28 /100 )^3 = 0.021952 cu. m. = 21952 cu. cm.

W sub Wood = [ 0.021952] [ 680.0] [ 9.807 / 1.000 ]

W sub Wood = 146.392 N. <-------------------------

F sub B = ( V sub H2O displaced ) ( rho sub H2O ) ( g / g sub C )

V sub H2O displaced = [ F sub B ] / [ rho sub H2O ] [ g / g subC ]

F sub B = W sub Wood

V sub H2O displaced = [ 146.392 ] / ( 1000 ) ( 9.807 / 1.000 )

V sub H20 displaced = 0.0149272 cu. m. = 14927 cu. cm.

x submerged = [V sub H20 displaced ]/ [ A sub CS ]^2

x submerged = [ 14927 ] / [ 28 ^2 ] = 19.039 cm. submerged <-------------------------

x sub top to water = 28.00 - 19.039 = 8.96 cm. <---------------------------------------...
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F sub BMAX = [ V sub Wood ] [ rho sub H2O ] [ g / g sub C ]

F sub BMAX = [ 0.021952 ] [ 1000 ] [ 9.807 / 1.000 ]

F sub BMAX = 215.283 N.<------------------------

W sub Lead = F sub BMAX - W sub Wood

W sub Lead = 215.283 - 146.392 = 68.89 N <-----------------------

m sub Lead = [ W sub Lead ] [ g sub C / g ]

m sub Lead = [ 68.89 ] [ 1.000 / 9.807 ] = 7.0247 kg <-------------------------