A body weighing 100 lb with a flat surface area of 3 ft 2 slides down a lubricated inclined plane making a 35º angle with the horizontal. For viscosity of 0.002089 lb.s/ft 2 and a body speed of 3.5 ft/s, determine the lubricant film thickness.
Text Solution
Verified by ExpertsThe correct answer is:
CHECK THE SOLUTION.
Sol. F = weight of body along inclined plane = 100 sin 35º = 57.4 lb
τ = F/A =µ (dv/dx)
57.4/3 = (0.002089) (3.5/dx)
dx = 0.0003821 ft or 0.00459 in.
Prepare Smarter with CGP Edu
Get practice questions, solutions, and test series in one place.
Write a Review
Share your experience with this question and solution.
Commentary
Send your comment, doubt, correction, or feedback to admin.
Similar Questions
Explore conceptually related problems
A U-tube in which the cross-sectional area of the limb on the left is one quarter, the limb on the …
A wooden block, with a coin placed on its top, floats in water as shown in fig. the distance l and …
A body floats in a liquid contained in a beaker. The whole system as shown falls freely under gravi…
A liquid is kept in a cylindrical vessel which is being rotated about a vertical axis through the c…
Water is filled in a cylindrical container to a height of 3m. The ratio of the cross-sectional area…
A large open tank has two holes in the wall. One is a square hole of side L at a depth y from the t…