density and force

Question

A solid block weighs 1.80 N in air. When fully submerged in water, it displaces a volume of 2.0 x 10-5 m3. Density of water is 1000 kg/m3. Determine

a) the average density of the block

b) the normal force the floor of the vessel exerts on the submerged body

Answer 1

(a)

 

Get its mass from its weight, then get its density from its mass and the volume of the displaced water.

 

m = W/g = (1.80 N)/(9.8 m/s²) ≅ 0.184 kg

 

ρ = m/V = (0.184 kg)/(2.0 x 10^-5 m³) = 9200 kg/m³

 

(b)

 

That would be the difference between the body's weight and the buoyant force.  The buoyant force equals the weight of the displaced water.

 

F_b = ρ_H2OVg = (1000 kg/m³)(2.0 x 10^-5 m³)(9.8 m/s²) = 0.196 N

 

F_n = W - F_b = (1.80 - 0.196) N = 1.604 N