A cuboid made of spruce wood has the edge lengths a = 20 cm, b = 25 cm and c = 30 cm. It is to be pressed completely under the water surface and then held in the water. a) Calculate the volume V of the cuboid!
b) With what force F must the cuboid be held in the water? (Density of spruce wood: p F = 500 kg/m³: Density of water: p w = 1000 kg/m³: g=9.81 N/ kg.)

a) The volume of the cuboid is 0.015 m³

b) The force that must be applied for the cuboid to be held in the water is 141.15 N

Explanation:

The given parameters of the cuboid made of spruce wood are;

The edge lengths dimensions of the cuboid are, a = 20 cm, b = 25 cm, c = 30 cm

The location of the cuboid = Pressed completely under water;

The density of spruce wood, ρF = 500 kg/m³

The density of water, ρw = 1,000 kg/m³

The acceleration due to gravity, g ≈ 9.81 N/kg

a) The volume of a cuboid = Length × Breadth × Height

Taking , a = The length of the cuboid, b = The breadth of the cuboid, and c = The height of the cuboid, we have;

The volume of the cuboid, V = a × b × c = 20 cm × 25 cm × 30 cm = 15,000 cm³

The volume of the cuboid, V = 15,000 cm³  = 0.015 m³

b) The force 'F' that must be applied for the cuboid to be held in the water is given by the up thrust on the cuboid by the water

∴ F = The up thrust on the cuboid by the water = The weight of the water displaced

The weight of the water displaced, Ww = mw × g = ρw × Vwater × g

Where;

mw = The mass of the water = ρw × Vwater

Vwater = The volume of the water displaced = The volume of the cuboid = V

∴ Vwater = V = 0.015 m³

∴ Ww =  1,000 kg/m³ × 0.015 m³ × 9.81 N/kg = 147.15 N

The force that must be applied, F = Ww = 147.15 N

The force that must be applied for the cuboid to be held in the water = 141.15 N.

the body's situation won't change the forces are equal opposit and on the same body

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