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Viscosity of oil on inclined plane

Short answer

Downslope force F = 300 sin30° = 150 N, shear stress τ = 150/0.64 = 234.375 Pa, gradient du/dy = 0.3/0.0015 = 200 s⁻¹. Dynamic viscosity μ = τ/(du/dy) = 1.172 N·s/m².

Calculate the dynamic viscosity of oil which is used for the lubrication between square plate of size 0.8 m × 0.8 m and an inclined plane with an angle of inclination 30°. The weight of the square plate is 300 N and it slides down the inclined plane with a uniform velocity of 0.3 m/s. The thickness of oil film is 1.5 mm.

The plate slides at uniform velocity, so the component of its weight along the plane is balanced by the viscous shear force of the oil film.

Given: A=0.8×0.8=0.64A = 0.8 \times 0.8 = 0.64 m², W=300W = 300 N, θ=30∘\theta = 30^\circ, u=0.3u = 0.3 m/s, t=1.5t = 1.5 mm =0.0015= 0.0015 m.

          plate (W = 300 N)
         _/_/
        /  /  ---> slides down at 0.3 m/s
       /  / oil film 1.5 mm
      /__/________
     /  30 deg

Shear force and shear stress

F=Wsin⁡30∘=300×0.5=150 Nτ=FA=1500.64=234.375 N/m2\begin{aligned} F &= W \sin 30^\circ = 300 \times 0.5 = 150\ \text{N} \\ \tau &= \frac{F}{A} = \frac{150}{0.64} = 234.375\ \text{N/m}^2 \end{aligned}

Velocity gradient (linear profile)

dudy=0.30.0015=200 s−1\frac{du}{dy} = \frac{0.3}{0.0015} = 200\ \text{s}^{-1}

Viscosity

μ=τdu/dy=234.375200=1.172 N⋅s/m2\mu = \frac{\tau}{du/dy} = \frac{234.375}{200} = 1.172\ \text{N·s/m}^2

Answer: dynamic viscosity of the oil μ=1.172\mu = 1.172 N·s/m² (= 11.72 poise).

✓ Answer checked by Claude Opus 5.5 (high effort)

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