Answer:
(a) The weight of the wood is approximately 5,886 N
(b) The weight of the stone is approximately 1,962 N
Explanation:
The given parameters are;
The base area of the wood, A = 1 m × 1 m = 1 m²
The depth to which the wood is submerged below the water, d = 0.6 m
The depth further the wood sinks when a stone is placed on the wood, Δd = 0.2 m
The density of water, ρ = 1,000 kg/m³
(a) Archimedes' principle states that the up-thrust, or upward force acting on an object submerged in a fluid, is equal to the weight of the displaced fluid.
For a floating object such as the wood, the up-thrust is equal to the weight of the object
The weight of the water displaced, W = The volume of the displaced water, V × The density of the water, ρ × The acceleration due to gravity, g
W = V × ρ × g
'V' is the same as the volume of the submerged wood, therefore;
V = A × d = 1 m × 1 m × 0.6 m = 0.6 m³
g ≈ 9.81 m/s²
Therefore;
W = 0.6 m³ × 1,000 kg/m³ × 9.81 m/s² ≈ 5,886 Newtons
The weight of the water displaced, W ≈ 5,886 N = The up-thrust = The weight of the wood
∴ The weight of the wood, W ≈ 5,886 N
(b) Similarly, the weight of the stone, [tex]W_{stone}[/tex] is equal to the weight of the additional volume of water displaced
∴ [tex]W_{stone}[/tex] = A × Δd × ρ × g
By plugging in the values, we get
[tex]W_{stone}[/tex] ≈ 1 m² × 0.2 m × 1,000 kg/m³ × 9.81 m/s² = 1,962 N
The weight of the stone, [tex]W_{stone}[/tex] ≈ 1,962 Newtons
