- #1

Rectifier

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*This problem was translated from Swedish, sorry for any grammatical errors present.*

The problem

The problem

The grey is liquid is water. The small container is a pipe made of glas and the big container is a barrel. Water is poured into the glas pipe. When water reaches 12m over the barrel top(lid) the barrel breaks.

Calculate

a) the mass of the water in the glass pipe (

*over the lid*)

b) the total force on the lid when the barrel breaks

**Relevant equations**

Liquid - water

h = 12m

R = 0.20m

r = 0.003m

(

*^^values from below the figure*in my book)

##p=dgh##

**The attempt**

**a)**

Density is described by ## d = \frac{m}{V} ##

Volume of a cylinder is ##V = \pi r^2 h##

density for water is ##d = 1000 kg/m^3##

Mass is thus:

## m = d \cdot V = d \cdot \pi r^2 h= 1000 kg/m^3 \cdot \pi (0.003)^2 12 = 0.339292... kg ##

**b)**

There is a hole in the lid so I calculate the area of the doughnut.

The area of the lid is

##A_L = \pi r_n^2 = \pi (R-r)^2 = \pi (0.20 - 0.003)^2 ##

The pressure is

##p = \frac{F}{A_L} \\ F=pA_L = \pi (0.20 - 0.003)^2p ##

I have only one more interesting formula in the book on this chapter - namely:

## p=dgh ##

where

d = density

g = gravity at the surface of the material

h = height

I know that

d for water is = 1000 kg/m^3

g = 9.82

h = 12m

then

##F=pA = A_Ldgh = 1000 \cdot 9.82 \cdot 12 \cdot \pi (0.20 - 0.003)^2 = 14367.3 N ##

It feels like I should add atmospheric pressure somwhere but I am not 100% sure.

*Please help*:,(

EDIT:

Apparently ## F_{total} = F_{water} + F_{air}##

and the foce from the air is

## F=p_{atm}A_L = 100kPa A_L = 100 \cdot 10^3 \pi (0.20 - 0.003)^2 = 12192.2...##

Thus ##F_{total} = F_{water} + F_{air} = 14367.3 N + 12192.2N =26559.5 N##

**Does it look right?**

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