Question

A 10 cm x 10 cm x 10 cm block of steel (density of steel =7900 kg/m³) is suspended from a spring scale. The scaleis in newtons.

a.     What is the scale reading if theblock is in air?

b.     What is the scale reading after theblock has been lowered into a beaker of oil and is completelysubmerged?

Answer 1

V = 10 cm x 10 cm x 10 cm = .1m x .1m x .1m = .001m^3

density of steel = 7900kg/m³            

Part a

formula .... ρ=m/v.... desnity = ρ, mass=m,V=volume

ρV=m....solve for mass

7900 kg/m^{3  ×  .001m^3 = 7.9 kg}

m=7.9kg..therefore, the spring scale must read the masses'weight

F=mg= (7.9kg)(9.8m/s^2) = 77.42 N

Part b.

density of oil : 900kg/m^3 *(although the densities of oilvary)

Bouyant Force = Fb

w- weight of mass in air

w' - apparent weight (weight of mass in water)

w' = w-Fb

Fb = p*gV

*(of fliud)

Fb= (900kg/m^3)(9.8m/s^2)(.001m^3)

Fb = 8.82N

w= 77.42N (from part a)

w'= w-Fb

   = 77.42N - 8.82N

   = 68.6N