For the cellar of a new house, a hole is dug in the ground with vertical sides g
ID: 2214130 • Letter: F
Question
For the cellar of a new house, a hole is dug in the ground with vertical sides going down 2.90 m . A concrete foundation wall is built all the way across the 10.70 m width of the excavation. This foundation wall is 0.181 m away from the front of the cellar hole. During a rainstorm, drainage from the street fills up the space in front of the concrete wall but not the cellar behind the wall. The water does not soak into the clay soil. Find the force that the water causes on the foundation wall. For comparison, the weight of the water is given by the formula below. 2.90 m 10.70 m 0.181 m 1000 kg/m3 9.80 m/s2 = 55.0 kN ___kNExplanation / Answer
For the cellar of a new house, a hole is dug in the ground, with vertical sides going down 2.40 m. A concrete foundation wall is built all the way across the 9.60-m width of the excavation. This foundation wall is 0.183 m away from the front of the cellar hole. During a rainstorm, drainage from the street fills up the space in front of the concrete wall, but not the cellar behind the wall. The water does not soak into the clay soil. Find the force the water causes on the foundation wall. For comparison, the weight of the water is given by 2.40 m X 9.60 m X 0.183 m X 1 000 kg/m3 X 9.80 m/s2 = 41.3 kN.I don't know where to start. The same question was asked by another member, but it has remained unanswered for 7 months. Diagrams would help. For the cellar of a new house, a hole is dug in the ground, with vertical sides going down 2.40 m. A concrete foundation wall is built all the way across the 9.60-m width of the excavation. This foundation wall is 0.183 m away from the front of the cellar hole. During a rainstorm, drainage from the street fills up the space in front of the concrete wall, but not the cellar behind the wall. The water does not soak into the clay soil. Find the force the water causes on the foundation wall. For comparison, the weight of the water is given by 2.40 m X 9.60 m X 0.183 m X 1 000 kg/m3 X 9.80 m/s2 = 41.3 kN.
I don't know where to start. The same question was asked by another member, but it has remained unanswered for 7 months. Diagrams would help. For the cellar of a new house, a hole is dug in the ground, with vertical sides going down 2.40 m. A concrete foundation wall is built all the way across the 9.60-m width of the excavation. This foundation wall is 0.183 m away from the front of the cellar hole. During a rainstorm, drainage from the street fills up the space in front of the concrete wall, but not the cellar behind the wall. The water does not soak into the clay soil. Find the force the water causes on the foundation wall. For comparison, the weight of the water is given by 2.40 m X 9.60 m X 0.183 m X 1 000 kg/m3 X 9.80 m/s2 = 41.3 kN.
I don't know where to start. The same question was asked by another member, but it has remained unanswered for 7 months. Diagrams would help. Best Answer Let me show you the normal way to show exponents here.
1 000 kg/m^3
where the ^ is shift+6.
Yes, a diagram would help. I will assume that only the one wall was built and somehow it seals against the sides of the hole. And the hole is rectangular.
The pressure is given by
p = rho*g*h
where rho = 1 000 kg/m^3, g = 9.80 m/s^2, h = 2.40 m.
The pressure builds linearly with depth, so the average pressure is the pressure at (1/2)*2.40 m. And that is 1/2 the pressure at the bottom. Calculate the average pressure, pave.
The area of that wall is
A = 2.40 m X 9.60 m.
So the force on the wall is
F = pave*A
The width of the gap does not matter. Best Answer Let me show you the normal way to show exponents here.
1 000 kg/m^3
where the ^ is shift+6.
Yes, a diagram would help. I will assume that only the one wall was built and somehow it seals against the sides of the hole. And the hole is rectangular.
The pressure is given by
p = rho*g*h
where rho = 1 000 kg/m^3, g = 9.80 m/s^2, h = 2.40 m.
The pressure builds linearly with depth, so the average pressure is the pressure at (1/2)*2.40 m. And that is 1/2 the pressure at the bottom. Calculate the average pressure, pave.
The area of that wall is
A = 2.40 m X 9.60 m.
So the force on the wall is
F = pave*A
The width of the gap does not matter. Let me show you the normal way to show exponents here.
1 000 kg/m^3
where the ^ is shift+6.
Yes, a diagram would help. I will assume that only the one wall was built and somehow it seals against the sides of the hole. And the hole is rectangular.
The pressure is given by
p = rho*g*h
where rho = 1 000 kg/m^3, g = 9.80 m/s^2, h = 2.40 m.
The pressure builds linearly with depth, so the average pressure is the pressure at (1/2)*2.40 m. And that is 1/2 the pressure at the bottom. Calculate the average pressure, pave.
The area of that wall is
A = 2.40 m X 9.60 m.
So the force on the wall is
F = pave*A
The width of the gap does not matter.