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

In summary: The water column exerts a force on the sewer lid that is equal to the weight of the water in the column.
  • #1
IsomDan
5
0
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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  • #2
deleting my post. Error
 
Last edited:
  • #3
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!
 
  • #4
And btw, there is a little drawing included.
 

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  • #5
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.
 
  • #6
Thanks so much for your responses guys

But is that the force working ON the sewer lid?

Isoma
 
  • #7
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:
 
  • #8
Okay ... So the whole force excerting on the lid is given by

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

?
 
  • #9
IsomDan said:
Okay ... So the force exerted by the water on the lid is given by

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

?

Yes.
 

1. How do you calculate the force of water on an object?

The force of water on an object can be calculated using the formula: F = ρghA, where F is the force, ρ is the density of water, g is the acceleration due to gravity, h is the height of the water column, and A is the cross-sectional area of the column.

2. Does the density of water affect the force it exerts on an object?

Yes, the density of water does affect the force it exerts on an object. The higher the density, the greater the force of the water column on the object.

3. What is the role of gravity in calculating the force of water on an object?

Gravity plays a crucial role in calculating the force of water on an object. This is because the force of gravity is what causes the water to exert pressure on the object, and the formula for calculating this force includes the acceleration due to gravity (g).

4. How does the height of the water column impact the force on an object?

The height of the water column has a direct impact on the force exerted on an object. The higher the water column, the greater the force it exerts on the object due to the increased pressure.

5. Is the cross-sectional area of the water column important in calculating the force on an object?

Yes, the cross-sectional area of the water column is an important factor in calculating the force on an object. A larger cross-sectional area means a larger surface area in contact with the object, resulting in a greater force being exerted on the object by the water column.

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