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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?

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princessesther2011

Answer:

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

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