Question

A 3 ft x 2 ft block moves down a 15 degree inclined slope at a speed of V = 0.2 ft/s over a thin layer (h = 0.0125 ft) of oil with a dynamic viscosity of 8.2 x 10^-2 lbf-s/ft^2. What is the weight of the block? Draw FBD.

Answer 1

mg = 30.415 lbf

Step-by-step explanation:

from figure body of size 3ft*2ft is tend to move down side

weight is divided into two component

vertical component = mgcos15 and

horizontal component = mg sin15

considering horizontal component equal to shear force

mgsin15 = \tau A

mgsin15 =\mu \frac{dv}{dh} A

mg =\frac{ \mu v A}{h*sin15}

=\frac{8.2*10^{-2}*0.2*3*2}{0.0125*sin15}

mg = 30.415 lbf