A piece of wood has dimensions (LxWxT): (0.600m)x(0.250m)x(0.080m). Its density

Question

A piece of wood has dimensions (LxWxT): (0.600m)x(0.250m)x(0.080m). Its density is 700[kg/m3 ]. 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(103 )[kg/m3 ].

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

from Density formula , we have Density = mass/Volume

Mass of Wood = density * volume = 700 * 0.012 = 8.4 Kg

here Weight is balnaced = Buoyant force

let V be the Volume then VOlume of Water = 0.012 - V

buoyant force = Rho V g

Where V is Volume in water

Rho is density of water = 1000 kg/m^3

g is accleration due to gravity

Fb = (0.012-v)*1000*9.8

Fb =9800*(0.012-V)

Now the Balancing Weight 9800*(0.012-V) = mg

9800*(0.012 -V) = 8.4*9.8

V = 3.6*10^-3 m^3

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b) now let mass of brass be m

then lets 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

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c) Let the above volume be V

then volume of wood inside = 0.012 - V

volume of brass inside = mass/density

Vbrass = 3.6/(8.6*10^3)

Vbrass = 4.19*10^-4 m^3

again 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

Volume of Wood = 4.19*10^-4 m^3

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