Question

A block of wood has a mass of 3.54 kg and a density of 588 kg/m3. It is to be loaded with lead (1.13 × 104 kg/m3) so that it will float in water with 0.887 of its volume submerged. What mass of lead is needed if the lead is attached to (a) the top of the wood and (b) the bottom of the wood?

Answer 1

Total mass of wood and lead must equal mass of water displaced.

(a) volume of wood block = mass /density = 3.54 / 588 = 0.00602 kg/m³
mass of water displaced = density * volume = 1000 * 0.887 * volume of wood
= 1000 * 0.887 * 0.00602

= 5.3401 kg of water is displaced

therefore, mass of lead + mass of wood = 5.3401
or, mass of lead + 3.54 = 5.3401

or,   mass of lead = 1.8001 kg
.
(b) now... volume of water displaced = 0.887 times of volume of block + volume of lead (since it is submerged)
.
So... mass of water displaced = mass of block + mass of lead
.
or, density of water * V_w = 3.54 + M.................................................................(1)
.
Also... V_w = 0.887 x 0.00602 + V_L and V_L = mass of lead / density of lead ........................................(2)
.
From (1) : 1000 V_w = 3.54 + M
.
From (2) V_w = 0.005339 + V_L V_L = M / 11300
.
put (2) in (1)
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1000( 0.005339 + M /11300 ) = 3.54 + M

5.339 + M / 11.3 = 3.54 + M

M = 1.973 kg is the mass of the lead