1. A piece of wood has dimensions (LaWxT): (0.600m)x(0.250m)x(0.080m). Its densi

Question

1. A piece of wood has dimensions (LaWxT): (0.600m)x(0.250m)x(0.080m). Its density is 700[kg/m^3]. It is floating in water. a) How much of the wood's volume is over the waterline? b) We place a brass weight on top of the wood plank. If the waterline is now just even with the wood's top surface, what is the mass of the brass? Brass density is 8.60(10^3) (10^3) [kg/m^3] c) If we fasten the brass weight on the underside of the wood instead, how much of the wood's volume is now above (or below) the water surface?

Explanation / Answer

a)
Total volume of wood = 0.6*0.25*0.08 m^3 = 0.012 m^3
mass of wood = density * volume
                             = 700 * 0.012
                             = 8.4 Kg
This weight is balanced by buoyant force
Let volume outside be v m^3
then volume in water= 0.012 - v m^3
buoyant force = volume in water * density of water * g
                             = (0.012-v)*1000*9.8 {since density of water= 1000 Kg/m^3}
                             =9800*(0.012-v)

This is balancing the weight
9800*(0.012-v) = 8.4*9.8
117.6 - 9800 v = 82.32
v= 3.6*10^-3 m^3
Answer: 3.6*10^-3 m^3

b)
Let mass of brass be m Kg
then use:
buoyant force = total weight
volume in water * density of water * g = (8.4+m)*g
volume in water * density of water = (8.4+m)
0.012*1000 = 8.4+m
m= 3.6 Kg

c)
Let the above volume be v m^3
then volume of wood inside = 0.012 - v m^3
volume of brass inside = mass/density = 3.6/(8.6*10^3) = 4.19*10^-4 m^3
use:
buoyant force = total weight
volume in water * density of water * g = (8.4+m)*g
(4.19*10^-4 + 0.012-v)* 1000 = (8.4+3.6)
4.19*10^-4 + 0.012-v = 0.012
v = 4.19*10^-4 m^3
Answer: 4.19*10^-4 m^3

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