Problem Statement
A square block weighing 1.15KN and 250mm on an edge slides down an incline on a film of oil 6µm thick. Assuming a linear velocity profile in the oil, calculate the terminal speed of the block. The viscosity of the oil is 0.007 NS/m2.
Solution
Given:
- Weight of block (W) = 1.15KN = 1150N
- Side of block (L) = 250mm = 0.25m
- Thickness (dy) = 6µm = 6 × 10-6m
- Viscosity of oil (µ) = 0.007 NS/m2
- Terminal velocity (u) = ?
Calculations:
Frictional force (F) = Shear stress(τ) at block surface × surface area of the block (A)
F = μ (du/dy) (L2) = 0.007 × u / (6 × 10-6) × (0.252) = 72.9u
Component of W in the direction of F is W Sin 20°
At the terminal condition, equilibrium occurs:
F = W sin 20°
72.9u = 1150 × Sin 20°
Result:
u = 5.4 m/s
Explanation
This problem involves calculating the terminal speed of a block sliding down an inclined plane on a thin film of oil. Here’s a breakdown of the solution:
- We first identify the forces acting on the block: the weight component parallel to the incline (W sin 20°) and the frictional force (F) due to the oil film.
- The frictional force is calculated using the concept of viscous shear stress in the oil film. We assume a linear velocity profile in the oil, which allows us to use the equation τ = μ (du/dy).
- We express the frictional force in terms of the unknown terminal velocity (u) by multiplying the shear stress by the block’s surface area.
- At terminal velocity, the forces are in equilibrium, so we equate the frictional force to the weight component.
- Solving this equation gives us the terminal velocity of 5.4 m/s.
This problem demonstrates the application of fluid mechanics principles (viscosity and shear stress) in a practical scenario involving an inclined plane, combining concepts from both fluid dynamics and classical mechanics.
