A 200-kg steel part is resting on a thin film of oil on a concrete floor. The oil is a Newtonian fluid and has a viscosity of μ = 6.52*10-2 N⋅s/m2. The film thickness is d = 0.5 mm. What horizontal force is required to push the part across the floor at a constant velocity of A = 0.65 m/s?

Question

Physics

Carmen Tran

25 November, 03:02

A 200-kg steel part is resting on a thin film of oil on a concrete floor. The oil is a Newtonian fluid and has a viscosity of μ = 6.52*10-2 N⋅s/m2. The film thickness is d = 0.5 mm. What horizontal force is required to push the part across the floor at a constant velocity of A = 0.65 m/s?

+1

Answers (1)

Know the Answer?

Noemi Watson · Answer 1

2.097*10^-5 N

Explanation:

Viscosity = mass/distance*time

time (t) = mass/viscosity*distance

Mass of steel part = 200 kg

Viscosity of oil = 0.065 N. s/m^2

Distance = 0.5 mm = 0.5/1000 = 5*10^-4 m

t = 200/0.065*5*10^-4 = 6.2*10^6 s

Force (F) = mass*velocity/time = 200*0.65/6.2*10^6 = 2.097*10^-5 N