Fluid mechanics mercury problem

In summary, the conversation discusses a problem in which Mercury is poured into a U-tube with different cross-sectional areas on each side. The question asks for the length of the water column on the right side and how high the Mercury rises in the left side. Using the equations for density and volume, the solution is found to be 20 cm for the water column and 0.49 cm for the height of the Mercury. This is determined by setting the masses on both sides equal to each other.
  • #1
clairez93
114
0

Homework Statement



Mercury is poured into a U-tube as in Figure P15.18a. [see the picture here: http://www.ux1.eiu.edu/~cfadd/1350/Hmwk/Ch15/Images/P15.18.gif] [Broken] The left arm of the tube has a cross-sectional area of A1 = 10.0 cm2, and the right arm has a cross-sectional area of A2 = 5.00 cm2. One hundred grams of water are then poured into the right arm as in Figure P15.18b.

(a) Determine the length of the water column in the right arm of the U-tube.

(b) Given that the density of mercury is 13.6 g/cm3, what distance, h, does the mercury rise in the left arm?


Homework Equations



[tex]\rho*gh_{i} = \rho*gh_{f}[/tex] (at same level)
[tex]\rho = \frac{m}{v}[/tex]


The Attempt at a Solution



I figured out a):
[tex]V = Ah[/tex]
[tex]\rho = \frac{m}{v}[/tex]
[tex]\rho = \frac{m}{Ah}[/tex]
[tex] 1000 = \frac{0.1}{(5*10^{-4}*h}[/tex]
[tex]h = 0.2 m = 20 cm[/tex]

I couldn't do part b), so I finally went to Google the problem and found a solution on this website: http://www.ux1.eiu.edu/~cfadd/1350/Hmwk/Ch15/Ch15.html

Their solution says this:

[tex]\rho*gl_{1} = \rho*gl_{2}[/tex]
This equation is taken at the point where the water on the right hand side meets the mercury as shown in the picture on the site. 1s is the distance from the top of the mercury on the left hand side to this point and l2 is the distance from the top of the water to that point.
[tex](13.6)(9.8)(l_{1}) = (1)(20)(9.8)[/tex]
[tex]l_{1} = 1.47 cm[/tex]

I understand this part. This is the part of the solution I do not understand:

[tex]A_{1}h = A_{2}(1.47 - h)[/tex]
[tex]10h = 5(1.47 - h)[/tex]
[tex]h = 0.49 cm[/tex]

Basically I don't understand why you would know to set the two volumes equal to each other? How do you know the volumes are the same?
 
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  • #2
The water on one side must have the same weight or mass as the Mercury on the other side (above the bottom of the water in both cases).
mass on left = mass on right
pV = 100 g
pAh = 100 g
solve for h. Looks like a little more than 0.7 cm.
 
  • #3


I would approach this problem by first understanding the concept of pressure and how it relates to the height and density of a fluid. In this case, the pressure at the bottom of the U-tube is the same on both sides, since the two arms are connected. This means that the pressure exerted by the mercury on the left arm is equal to the pressure exerted by the water on the right arm.

Next, I would use the equation for pressure, P = \rho*g*h, to set up an equation for the pressure on each side of the U-tube. Since the pressure is equal on both sides, we can set the two equations equal to each other:

\rho_{Hg}*g*h_{Hg} = \rho_{water}*g*h_{water}

We know the values of \rho_{Hg} (density of mercury), g (acceleration due to gravity), and h_{water} (height of water in the right arm). We are trying to solve for h_{Hg}, which is the height of the mercury in the left arm.

To solve for h_{Hg}, we need to rearrange the equation to isolate it on one side. This is where the concept of volume comes in. We know that the volume of a fluid is equal to its cross-sectional area multiplied by its height (V = A*h). In this case, we can use the cross-sectional areas of the two arms (A_{1} and A_{2}) to relate the heights of the two fluids:

A_{1}*h_{Hg} = A_{2}*h_{water}

We can then substitute this into our equation for pressure:

\rho_{Hg}*g*(A_{1}*h_{Hg}) = \rho_{water}*g*(A_{2}*h_{water})

Next, we can simplify by dividing both sides by g, and substituting in the known values for \rho_{Hg} and \rho_{water}:

(13.6)*A_{1}*h_{Hg} = (1)*A_{2}*h_{water}

Finally, we can solve for h_{Hg} by rearranging the equation:

h_{Hg} = \frac{A_{2}*h_{water}}{A_{1}} = \frac{(5.00 cm^2)*(1.47 cm)}{(10.0 cm^2)}
 

1. What is the Fluid Mechanics Mercury Problem?

The Fluid Mechanics Mercury Problem, also known as the Mercury Manometer Problem, is a classic physics problem that involves calculating the pressure exerted by a column of mercury in a U-shaped tube. It is used to demonstrate the principles of hydrostatics, specifically the relationship between pressure, density, and height in a fluid.

2. How do you solve the Fluid Mechanics Mercury Problem?

To solve the Fluid Mechanics Mercury Problem, you must first identify all the known variables, including the density of mercury, the height of the mercury column, and the atmospheric pressure. Then, you can use the equation P = ρgh to calculate the pressure at any point in the mercury column. This equation states that pressure (P) is equal to the density (ρ) of the fluid, times the acceleration due to gravity (g), times the height (h) of the fluid column.

3. What are the units of measurement used in the Fluid Mechanics Mercury Problem?

The units of measurement used in the Fluid Mechanics Mercury Problem depend on the system of units being used. Generally, the density of mercury is given in grams per cubic centimeter (g/cm3), the height is given in meters (m), and the pressure is given in pascals (Pa) or kilopascals (kPa). However, other units such as pounds per square inch (psi) or inches of mercury (inHg) may also be used.

4. How does the Fluid Mechanics Mercury Problem relate to real-world applications?

The Fluid Mechanics Mercury Problem has many real-world applications, particularly in the fields of engineering and meteorology. For example, it can be used to measure air pressure in barometers, or to calculate the pressure in a water tower or dam. It is also used in weather forecasting to measure atmospheric pressure and predict changes in weather patterns.

5. Are there any limitations to the Fluid Mechanics Mercury Problem?

The Fluid Mechanics Mercury Problem has some limitations, as it assumes an idealized scenario with a perfect fluid and a uniform gravitational field. In real-world situations, factors such as surface tension, viscosity, and variations in gravity may affect the accuracy of the calculations. Additionally, the use of mercury in manometers has been limited due to its toxicity, leading to the development of alternative measurement methods.

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