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In-class Assign. Dr.waleed, Mob. 0100 4444 149 Page (1)
In-class Assignment
1- The two waterreservoirs shown in the figure are open to the atmosphere,
and the water has density 1000 kg/m3
. The manometer contains mercury
with a density of 13,600kg/m3
. What is the difference in elevation, h, if the
manometer reading is b = 25 cm? Take the gravitational acceleration, g,
equal to 9.8 m/s2
.
Solution
In-class Assign. Dr.waleed, Mob. 0100 4444 149 Page (2)
2- Gate AB shown in the figure is 1.2 m long and 0.8 m into the paper.
Neglecting atmospheric pressure, compute the force F on the gate and its
center-of-pressure position, distance X in the figure.
Solution
𝛾 𝑜𝑖𝑙 = 𝑆𝐺 𝑜𝑖𝑙 𝑥 𝛾 𝑤𝑎𝑡𝑒𝑟
𝛾 𝑜𝑖𝑙 = 0.82 𝑥 9.81 = 8.04 𝑘𝑁 𝑚3⁄
ℎ 𝐶𝐺 = 4 + (1 + 0.6) 𝑠𝑖𝑛40 = 5.028 𝑚
𝐹𝑅 = (𝑃𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝑎𝑡 𝑡ℎ𝑒 𝑐𝑒𝑛𝑡𝑒𝑟 𝑜𝑓 𝑡ℎ𝑒 𝑔𝑎𝑡𝑒)(𝐴𝑟𝑒𝑎 𝑜𝑓 𝑡ℎ𝑒 𝑔𝑎𝑡𝑒)
𝐹𝑅 = 𝛾ℎ 𝐶𝐺 𝐴
𝐹𝑅 = 8.04 𝑥 5.028 𝑥 (1.2 𝑥 0.8) = 38.83 𝑘𝑁
𝑦 𝐶𝑃 = 𝑦 𝐶𝐺 +
𝐼 𝑥𝑐
𝑦 𝐶𝐺 𝐴
In-class Assign. Dr.waleed, Mob. 0100 4444 149 Page (3)
𝑦 𝐶𝐺 =
ℎ 𝐶𝐺
𝑠𝑖𝑛40
𝐼 𝑥𝑐 =
(0.8)(1.2)3
12
= 0.1152 𝑚4
𝑦 𝐶𝑃 =
5.028
𝑠𝑖𝑛40
+
0.1152
(
5.028
𝑠𝑖𝑛40
)(0.96)
𝑦 𝐶𝑃 = 7.822 + 0.0153 = 7.837 𝑚
𝑋 = 0.6 + (𝑦 𝐶𝑃 − 𝑦 𝐶𝐺)
𝑋 = 0.6 + 0.0153 = 0.615 𝑚

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Inclass ass

  • 1. In-class Assign. Dr.waleed, Mob. 0100 4444 149 Page (1) In-class Assignment 1- The two waterreservoirs shown in the figure are open to the atmosphere, and the water has density 1000 kg/m3 . The manometer contains mercury with a density of 13,600kg/m3 . What is the difference in elevation, h, if the manometer reading is b = 25 cm? Take the gravitational acceleration, g, equal to 9.8 m/s2 . Solution
  • 2. In-class Assign. Dr.waleed, Mob. 0100 4444 149 Page (2) 2- Gate AB shown in the figure is 1.2 m long and 0.8 m into the paper. Neglecting atmospheric pressure, compute the force F on the gate and its center-of-pressure position, distance X in the figure. Solution 𝛾 𝑜𝑖𝑙 = 𝑆𝐺 𝑜𝑖𝑙 𝑥 𝛾 𝑤𝑎𝑡𝑒𝑟 𝛾 𝑜𝑖𝑙 = 0.82 𝑥 9.81 = 8.04 𝑘𝑁 𝑚3⁄ ℎ 𝐶𝐺 = 4 + (1 + 0.6) 𝑠𝑖𝑛40 = 5.028 𝑚 𝐹𝑅 = (𝑃𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝑎𝑡 𝑡ℎ𝑒 𝑐𝑒𝑛𝑡𝑒𝑟 𝑜𝑓 𝑡ℎ𝑒 𝑔𝑎𝑡𝑒)(𝐴𝑟𝑒𝑎 𝑜𝑓 𝑡ℎ𝑒 𝑔𝑎𝑡𝑒) 𝐹𝑅 = 𝛾ℎ 𝐶𝐺 𝐴 𝐹𝑅 = 8.04 𝑥 5.028 𝑥 (1.2 𝑥 0.8) = 38.83 𝑘𝑁 𝑦 𝐶𝑃 = 𝑦 𝐶𝐺 + 𝐼 𝑥𝑐 𝑦 𝐶𝐺 𝐴
  • 3. In-class Assign. Dr.waleed, Mob. 0100 4444 149 Page (3) 𝑦 𝐶𝐺 = ℎ 𝐶𝐺 𝑠𝑖𝑛40 𝐼 𝑥𝑐 = (0.8)(1.2)3 12 = 0.1152 𝑚4 𝑦 𝐶𝑃 = 5.028 𝑠𝑖𝑛40 + 0.1152 ( 5.028 𝑠𝑖𝑛40 )(0.96) 𝑦 𝐶𝑃 = 7.822 + 0.0153 = 7.837 𝑚 𝑋 = 0.6 + (𝑦 𝐶𝑃 − 𝑦 𝐶𝐺) 𝑋 = 0.6 + 0.0153 = 0.615 𝑚