Question

a 90 N rectangular solid block slides a 30° inclined plane the plate is lubricated by a 3mm thick film of oil of specific gravity (relative density) 0.90 and viscosity 8.0 poise. If the contact area is 0.3m², estimate the terminal velocity of the block​

Answer 1

Explanation:

Terminal velocity means constant velocity, and acceleration is zero. An equation for the velocity, from viscosity, is given by:

τ = μ

umax

h

As τ is the shearing stress, the shearing force is:

Fτ = τA

The film contact area A is the area of the bottom of the block that is in contact with the fluid.

Another force is the normal force, and the last force is the weight of the block which may be given by:

Wb = mg

where m is mass and g is the gravitational acceleration. So a free body diagram is set up:

where N is the normal force. The shear force is of importance because the velocity is part of this quantity. The normal force has no relation to the shear force, so it can be forgotten. But, there is a component of the weight which counteracts the friction force. This component can be represented as:

mg sin θ

This is the force that counteracts the shear force, so they can be equated to each other:

Fτ = mg sin θ

But to get the velocity substitution is made for the shear force, where Fτ = τA and then further substitution is made for τ:

μ

umax

h

A = mg sin θ

finally solving for the velocity, we can get the equation we need to solve the problem:

umax =

hmg sin θ

μA

plugging in the values:

umax =

(0.00085 m)*(8 kg)*(9.8 m/s2)*(sin 20°)

(0.29 kg/ms)*(0.005 m2)

umax = 15.72 m/s

0.005 m2 is from dividing 50 cm2 by 100 twice to get area in m2

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