Example 5 A closed tank shown in the given figure, is given an acceleration of 3.6 m/s, the acceleration vector being inclined at an angle of 35° to the horizontal. Calculate the pressure at point M, N and R. R a = 3.6 m/s 0.5 m Oil 2.0 m (S.G. 0.9) 35° width = 2 m 10 m
Example 5 A closed tank shown in the given figure, is given an acceleration of 3.6 m/s, the acceleration vector being inclined at an angle of 35° to the horizontal. Calculate the pressure at point M, N and R. R a = 3.6 m/s 0.5 m Oil 2.0 m (S.G. 0.9) 35° width = 2 m 10 m
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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![Example 5
A closed tank shown in the given figure, is given an acceleration of 3.6 m/s, the acceleration vector
being inclined at an angle of 35° to the horizontal. Calculate the pressure at point M, N and R.
R
a = 3.6 m/s
0.5 m
2.0 m
Oil
(S.G. = 0.9)
35°
width = 2 m
10 m](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F39e2186b-9b91-40d8-8174-fec2cdea7a72%2Fa6a20a8e-994c-4202-b695-87444791c25e%2F3co4loo_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Example 5
A closed tank shown in the given figure, is given an acceleration of 3.6 m/s, the acceleration vector
being inclined at an angle of 35° to the horizontal. Calculate the pressure at point M, N and R.
R
a = 3.6 m/s
0.5 m
2.0 m
Oil
(S.G. = 0.9)
35°
width = 2 m
10 m
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