Answer to Question #274086 in Mechanics | Relativity for beya

Question

A block of cherry wood that is 20.0 cm long, 10.0 cm wide, and 2.00 cm thick has a density of 800. kg/m3. What is the volume of a piece of iron that, if glued to the bottom of the block, makes the block float in water with its top just at the surface of the water? The density of iron is 7860 kg/m3, and the density of water is 1000. kg/m3.

Answer 1

Normal force acting on the person is equal to difference between weight of the person and buoyant force acting on the person.

Buoyant force acting on the block of cherry wood and iron block which is attached at the bottom of the cherry (with top of the cherry is at the surface of the water) is

$F_b = ρ_{water}V_{immersed}g \\ = ρ_{water}(V_{cherry} + V_{iron})g$

Total weight of the blocks system is

W =(m_{cherry} +m_{iron})g

Relation between the mass of the block and volume of the block related to the density as

$ρ = \frac{m}{V}$

Volume of the cherry wood,

$V_{cherry}=Lwh \\ = (0.20 \;m)(0.10 \;m)(0.02\;m) \\ = 4.0 \times 10^{-4} \;m^3$

As the cherry block and iron block are floating in the water, weight of the blocks is equal to buoyant force.

$ρ_{water}(V_{cherry} + V_{iron})g = (m_{cherry} +m_{iron})g \\ ρ_{water}(V_{cherry} + V_{iron})g = (ρ_{wood}V_{cherry}+ρ_{iron}V_{iron}) g \\ ρ_{water} V_{cherry} + ρ_{water} V_{iron} = ρ_{wood}V_{cherry}+ρ_{iron}V_{iron}$

By solving above equation we get

$V_{iron} = \frac{ρ_{water} V_{cherry} -ρ_{wood}V_{cherry}}{ρ_{iron} -ρ_{water}}$

Density of water,

$ρ_{water} = 1.0 \times 10^3 \;kg/m^3$

Density of iron,

$ρ_{iron} = 7860 \; kg/m^3$

Density of wooden block,

$ρ_{wood} = 800\; kg/m^3$

Length of the block of cherry,

L=20.0 cm = 0.20 m

Width of the block of cherry,

w=10.0 cm = 0.10 m

Height of the block of cherry,

h=2.0 cm = 0.02 m

Substituting the values in equation we get

$V_{iron} = \frac{1.0 \times 10^3 \times 4.0 \times 10^{-4} -800 \times 4.0 \times 10^{-4}}{7860 -1.0 \times 10^3} \\ = 1.17 \times 10^{-5} \;m^3$