III. A triangular gate, 1.25 meter base x 0.90 meter height was vertically submerged at 2.5 meter below the water surface (as shown in Figure 1). Calculate the total pressure in the gate and location of the center of pressure. Given: Ig = bh/36 (for Triangle) Water surface 그 2.5m 0.9m 1.25m FIGURE 1
III. A triangular gate, 1.25 meter base x 0.90 meter height was vertically submerged at 2.5 meter below the water surface (as shown in Figure 1). Calculate the total pressure in the gate and location of the center of pressure. Given: Ig = bh/36 (for Triangle) Water surface 그 2.5m 0.9m 1.25m FIGURE 1
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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Question
![III.
A triangular gate, 1.25 meter base x 0.90 meter height was vertically submerged at 2.5
meter below the water surface (as shown in Figure 1). Calculate the total pressure in the
gate and location of the center of pressure.
Given: Ig = bh/36 (for Triangle)
Water surface
ㅏ
2.5m
1.25m
0.9m
FIGURE 1](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F5c6e3123-c8ef-4403-8508-bdf50f8af29b%2Fcd8ae7b1-b861-41eb-8299-49e153db3566%2Fpbki66xq_processed.jpeg&w=3840&q=75)
Transcribed Image Text:III.
A triangular gate, 1.25 meter base x 0.90 meter height was vertically submerged at 2.5
meter below the water surface (as shown in Figure 1). Calculate the total pressure in the
gate and location of the center of pressure.
Given: Ig = bh/36 (for Triangle)
Water surface
ㅏ
2.5m
1.25m
0.9m
FIGURE 1
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