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0 h So the volume of the trough is V = Lh²/tan 0 h = V tan L So p = pg V tan L Our results show that the pressure only depends on depth (h) and is not affected by the shape of the vessel, as we said in the lecture. However, the depth will depend on the shape! For example, if your pour a 200 ml of water into a narrow glass the depth would be greater than if you poured the same amount of water into a wide glass. Radioactive rods from a decommissioned nuclear power plants are stored in large troughs. The troughs are filled with water to absorb the emitted radiation. Two types of trough are used; a triangular shape and a rectangular shape, as shown in the figure below. a) Express the pressure at the bottom of the triangular and the rectangular troughs as a function of the depth of the fluid (h). b) Express the pressure at the bottom of the triangular and the rectangular troughs as a function of the volume of fluid in the trough (V). Part (a) The stagnation pressure is related to the depth according to p = pgh This is the same for both the square and the triangular troughs. Part (b) We have an expression for p in terms of h, so we will get h in terms of V and substitute it into the equation for p, so get p in terms of V. For the rectangular trough For the triangular one: The area of a triangle is V = wLh V h = WL p = pg \wL h² A = tan

0 h So the volume of the trough is V = Lh²/tan 0 h = V tan L So p = pg V tan L Our results show that the pressure only depends on depth (h) and is not affected by the shape of the vessel, as we said in the lecture. However, the depth will depend on the shape! For example, if your pour a 200 ml of water into a narrow glass the depth would be greater than if you poured the same amount of water into a wide glass. Radioactive rods from a decommissioned nuclear power plants are stored in large troughs. The troughs are filled with water to absorb the emitted radiation. Two types of trough are used; a triangular shape and a rectangular shape, as shown in the figure below. a) Express the pressure at the bottom of the triangular and the rectangular troughs as a function of the depth of the fluid (h). b) Express the pressure at the bottom of the triangular and the rectangular troughs as a function of the volume of fluid in the trough (V). Part (a) The stagnation pressure is related to the depth according to p = pgh This is the same for both the square and the triangular troughs. Part (b) We have an expression for p in terms of h, so we will get h in terms of V and substitute it into the equation for p, so get p in terms of V. For the rectangular trough For the triangular one: The area of a triangle is V = wLh V h = WL p = pg \wL h² A = tan

International Edition---engineering Mechanics: Statics, 4th Edition
International Edition---engineering Mechanics: Statics, 4th Edition
4th Edition
ISBN:9781305501607
Author:Andrew Pytel And Jaan Kiusalaas
Publisher:Andrew Pytel And Jaan Kiusalaas
Chapter8: Centroids And Distributed Loads
Section: Chapter Questions
Problem 8.90P: The hemispherical glass bowl is filled with water. Find the location y of the center of gravity of...
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i dont really understand the explanation of part b or how they used trigonometry to get that. can you please help.

0 h So the volume of the trough is V = Lh²/tan 0 h = V tan L So p = pg V tan L Our results show that the pressure only depends on depth (h) and is not affected by the shape of the vessel, as we said in the lecture. However, the depth will depend on the shape! For example, if your pour a 200 ml of water into a narrow glass the depth would be greater than if you poured the same amount of water into a wide glass.
Transcribed Image Text:0 h So the volume of the trough is V = Lh²/tan 0 h = V tan L So p = pg V tan L Our results show that the pressure only depends on depth (h) and is not affected by the shape of the vessel, as we said in the lecture. However, the depth will depend on the shape! For example, if your pour a 200 ml of water into a narrow glass the depth would be greater than if you poured the same amount of water into a wide glass.
Radioactive rods from a decommissioned nuclear power plants are stored in large troughs. The troughs are filled with water to absorb the emitted radiation. Two types of trough are used; a triangular shape and a rectangular shape, as shown in the figure below. a) Express the pressure at the bottom of the triangular and the rectangular troughs as a function of the depth of the fluid (h). b) Express the pressure at the bottom of the triangular and the rectangular troughs as a function of the volume of fluid in the trough (V). Part (a) The stagnation pressure is related to the depth according to p = pgh This is the same for both the square and the triangular troughs. Part (b) We have an expression for p in terms of h, so we will get h in terms of V and substitute it into the equation for p, so get p in terms of V. For the rectangular trough For the triangular one: The area of a triangle is V = wLh V h = WL p = pg \wL h² A = tan
Transcribed Image Text:Radioactive rods from a decommissioned nuclear power plants are stored in large troughs. The troughs are filled with water to absorb the emitted radiation. Two types of trough are used; a triangular shape and a rectangular shape, as shown in the figure below. a) Express the pressure at the bottom of the triangular and the rectangular troughs as a function of the depth of the fluid (h). b) Express the pressure at the bottom of the triangular and the rectangular troughs as a function of the volume of fluid in the trough (V). Part (a) The stagnation pressure is related to the depth according to p = pgh This is the same for both the square and the triangular troughs. Part (b) We have an expression for p in terms of h, so we will get h in terms of V and substitute it into the equation for p, so get p in terms of V. For the rectangular trough For the triangular one: The area of a triangle is V = wLh V h = WL p = pg \wL h² A = tan
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