Published by:
CGP EDU Academic Team
Published on: September 12, 2026
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:
A
To determine the lubricant film thickness, we can use the formula related to viscous flow between two parallel plates. The film thickness can be calculated using the formula:
\[ h = \frac{F}{\mu A V} \]
Where:
- \( F \) is the weight of the body (100 lb)
- \( \mu \) is the viscosity (0.002089 lb.s/ft^2)
- \( A \) is the surface area (3 ft^2)
- \( V \) is the speed (3.5 ft/s)
Plugging in the values:
\[ h = \frac{100}{0.002089 \times 3 \times 3.5} \]
\[ h = \frac{100}{0.02185815} \approx 4571.63 \text{ ft} \]
The calculated lubricant film thickness indicates the required thickness for effective lubrication. Thus, the final lubricant film thickness is approximately \( 4571.63 \text{ ft} \).
Therefore, based on the calculations, the answer should be selected based on the closest answer available.
\[ h = \frac{F}{\mu A V} \]
Where:
- \( F \) is the weight of the body (100 lb)
- \( \mu \) is the viscosity (0.002089 lb.s/ft^2)
- \( A \) is the surface area (3 ft^2)
- \( V \) is the speed (3.5 ft/s)
Plugging in the values:
\[ h = \frac{100}{0.002089 \times 3 \times 3.5} \]
\[ h = \frac{100}{0.02185815} \approx 4571.63 \text{ ft} \]
The calculated lubricant film thickness indicates the required thickness for effective lubrication. Thus, the final lubricant film thickness is approximately \( 4571.63 \text{ ft} \).
Therefore, based on the calculations, the answer should be selected based on the closest answer available.
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