If a force of P=2 N causes the 30 -mm-diameter shaft to...

If a force of $P=2 \mathrm{~N}$ causes the 30 -mm-diameter shaft to slide along the lubricated bearing with a constant speed of $0.5 \mathrm{~m} / \mathrm{s}$, determine the viscosity of the lubricant and the constant speed of the shaft when $P=8 \mathrm{~N}$. Assume the lubricant is a Newtonian fluid and the velocity profile between the shaft and the bearing is linear. The gap between the bearing and the shaft is $1 \mathrm{~mm}$.

Key Concepts

Shear Stress and Velocity Gradient

Shear stress in a fluid is produced by the velocity gradient between adjacent layers of fluid. In scenarios where the velocity profile is linear, the shear rate is constant, leading directly to a constant shear stress for a Newtonian fluid. This relationship is fundamental in calculating forces in lubricated contacts.

Lubrication Theory

Lubrication theory deals with the behavior of thin fluid films between solid surfaces in relative motion. It simplifies the complex fluid dynamics by assuming a linear velocity profile and dominates in scenarios where the gap between surfaces is small, like in bearings. This theory is essential for calculating frictional forces and understanding how changes in applied force or geometry affect system behavior.

Newtonian Fluid

This concept refers to fluids whose viscosity remains constant irrespective of the shear rate applied. In Newtonian fluids, the shear stress is directly proportional to the velocity gradient, which simplifies the analysis of fluid flow between surfaces, such as in lubrication problems.

Viscosity

Viscosity is a measure of a fluid's resistance to flow and deformation under shear stress. It quantifies how easily the fluid layers slide past one another and is a critical property for predicting the behavior of fluids in lubrication and other engineering applications.

Transcript

00:01 Hello and welcome to this video solution of numerate.

00:04 Here it's given that if a force of 2 newton, right, p equal to 2 newton causes 30 millimeter diameter shaft to slide along the lubricated bearing with a constant speed of 5, 0 .5 meters per second.

00:17 So here we have got the diameter equal to 30 millimeter and speed v equal to 0 .5 meters per second.

00:27 You have to determine the shaft when the force 8 newton is acting on the fluid, right, and the gap between the bearing and the shaft is 1 millimeter given.

00:52 What i do is i take this as capital d and small d is the 1 millimeter, right, the gap.

01:06 Now from the expression of shear stress you will be having tau equal to mu du dy, right.

01:15 Now this shear stress is nothing but the force over the total surface area which is i d times the length, right.

01:36 If you see this diagram then it's given 50 millimeters inside which we have got the shaft, right.

01:43 This is 50 millimeters.

01:46 So this length will be, i'm writing l for the time being.

01:50 This is equal to mu.

01:52 The speed is 0 .5 and the surface inside it is, sorry, the layer of oil is 0 or maybe what i can do is small d i'm writing, right, small d.

02:07 Okay, this is, i will plug in the next step.

02:10 This is v by d, right.

02:13 So now we simply plug in the values.

02:15 P over pi times of 0 .03 diameter.

02:25 Everything i write in meters, right, 0 .05, 550 millimeters is this...

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