Compute the force the a coloumn of water exerts on an object.

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The discussion revolves around calculating the force exerted by a column of water on a sewer lid during a cloudburst in Sweden. The water rose 37.5 meters above the sewer lid, with a lid radius of 29 cm. Participants emphasize using the formula F = ρ * h * A * g, where ρ is the water density, h is the height of the water column, A is the area of the lid, and g is the gravitational acceleration. The conversation confirms that this formula accurately represents the force acting on the sewer lid, adhering to Pascal's law. Ultimately, the force exerted by the water on the lid can be calculated using this established relationship.
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When a cloudburst in the summer of 2011 in Sweden some sewer pipes were
completely filled with water. The sewer lids had been fastened, but in some
places the large amounts of water pushed anyway the sewer lids anyway.

At one point, the water in a sewer pipe, was pushed up 37.5 m above the level of the sewer lid, in a huge water column.
The radius of the sewer lid is 29 cm.

a) Determine the magnitude of the force of the water exerts on the sewer lid

I know that the answer somehow resolves around computing the pressure in the water column and then find the difference in pressure working respectively upwards and downwards on the lid, and the turn that pressure into force via the area of the lid.
 
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But if the water column persists, would there be some kind of force that will keep on working?

It has been given as a national exam a couple of years ago, so I do think that it must be possible to find a solution!
 
And btw, there is a little drawing included.
 

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Your picture makes a good point- the column of water will rise to exactly the height it reaches on the hill side. And the force pushing it up is simply the weight of the water itself. Multiply the height of the water by the cross sectional area and the density of water.
 
Thanks so much for your responses guys

But is that the force working ON the sewer lid?

Isoma
 
HallsofIvy said:
Multiply the height of the water by the cross sectional area and the density of water.

I think you forgot the gravitational acceleration, g..

IsomDan said:
Thanks so much for your responses guys

But is that the force working ON the sewer lid?

Isoma

Respecting Pascal's law, yes it is. :smile:
 
Okay ... So the whole force excerting on the lid is given by

Fwater-on-sewer = ρwater * hcolumn * Asewer * g

?
 
IsomDan said:
Okay ... So the force exerted by the water on the lid is given by

Fwater-on-sewer = ρwater * hcolumn * Asewer * g

?

Yes.
 
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