Question

1. A body weighing 1000N slides down at a uniform speed of 1m/s along a lubricated inclined
plane making 30° angle with the horizontal. The viscosity of lubricant is 0.1 kg/ms and the contact
area of the body is 0.25 m2. Determine the lubricant thickness by assuming a linear velocity
distribution. [5M]
1OOON
Im/s
t-thickness
of wlricant
30°​

Answer 1

Answer:

body weight sildes down at a uniform speed of along a lubricated inclined planeaking 30° angle with the horizontal.

Answer 2

The lubricant thickness by assuming a linear velocity in $0.05mm$.

Explanation:

speed$=1\frac{m}{s}$

contact area$=0.25m^{2}$

∅=$30^{0}$

viscosity$=0.1\frac{kg}{ms}$

τ$=\frac{fs}{A}$

$fs$=τ$A$

τ$=\frac{udu}{dy}$

$500=\frac{0.1*1*0.25}{y}$

$y=\frac{0.1*1*0.25}{500}$

$y=5*10^{-5}$

$y=5*10^{-5}*10^{3}$

$y=0.05mm$.