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$\fbox{\bf PROBLEM \thechapter.\arabic{probcount}}$

A steel railroad track is 1000 m long. When its temperature is raised from -30 o to + 30 o thermal expansion causes the length of the track to increase by .01%. What compressive force per unit area would be required to keep the track from expanding? (Hint: Calculate the force that would be required to compress it back to its original length. Young's modulus for steel is 2 x 1011 N/m 2.


$\displaystyle{F\over A}$ = Y$\displaystyle{\Delta L\over L}$   
  = 2 x 1011 N/m 2 x (1 x 10- 4)2   
  = 2 x 107 N/m 2.   
Note that this force is enough to break steel. This is the reason that railway tracks are spaced so as to allow for expansion and contraction due to temperature changes.

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$\fbox{\bf PROBLEM \thechapter.\arabic{probcount}}$

What is the approximate bulk modulus of the egg in our egg experiment?

We assume the egg to be a sphere of radius 5 cm, and that it is compressed by 2 mm in order to squeeze through the opening in the jar. If we assume the jar to be completely evacuated, the volume stress is simply the air pressure (1 Atm) and the volume strain is:

$\displaystyle{\Delta V\over V}$ = $\displaystyle{{4 \pi r_f^3\over 3} - {4\pi
r_i^3\over 3}
\over {4\pi r_i^3\over 3}}$   
  = 0.115.   
The bulk modulus is then given by:
B = - $\displaystyle{F/A\over \Delta V/V}$   
  = $\displaystyle{\frac{-1.01 \times 10^5}{-0.115}}$   
  = 8.76 x 105  Pa.   

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$\fbox{\bf PROBLEM \thechapter.\arabic{probcount}}$

A boy tries to use a garden hose to supply air for a swim at the bottom of a 50m deep pool. What goes wrong?

The pressure of air in his lungs and in the hose is at 1 atmosphere ( 1.01 x 105  Pa ), whereas the pressure of the water pushing in on his lungs and on the hose is:

P = P0 + $\displaystyle\rho$gh   
  = (1.01 x 105) + (1.00 x 103)(9.8)(50)   
  = 5.91 x 105  Pa.   
Therefore his lungs, and the hose through which he is breathing, may collapse.

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$\fbox{\bf PROBLEM \thechapter.\arabic{probcount}}$

A car weighing 1.2 x 104 N rests on four tires. If the gauge pressure in each tire is 200 kPa, what it the area of each tire in contact with the road?

Each tire must carry a weight of:

F = $\displaystyle{1.2 \times 10^4 \;{\:\rm N}\over 4}$ = 0.3 x 104  N.

The definition of pressure is Force per Area. Therefore:
P = $\displaystyle{\frac{F}{A}}$   
$\displaystyle\Rightarrow$ A = $\displaystyle{\frac{F}{P}}$   
  = $\displaystyle{\frac{0.3 \times 10^4}{200 \times 10^3}}$   
  = 0.015 m 2.   

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$\fbox{\bf PROBLEM \thechapter.\arabic{probcount}}$

The same car as in the problem above sits on a hydraulic press as shown below. If the area of the cylinder holding the car up is 4 times greater than the area of the cylinder on the other side of the press, what is the force that must be applied to the other side of the hydraulic press.

Figure 9.9: Problem 9.5
\epsfxsize=4 cm


P = $\displaystyle{\frac{F_1}{A_1}}$ = $\displaystyle{\frac{F_2}{A_2}}$   
$\displaystyle{\frac{1.2 \times 10^4}{4A}}$ = $\displaystyle{\frac{F_2}{A}}$   
$\displaystyle\Rightarrow$ F2 = 0.3 x 104  N .   

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$\fbox{\bf PROBLEM \thechapter.\arabic{probcount}}$

An aluminum object has a mass of 27.0 kg and a density of 2.70 x 103 kg/m 3. The object is attached to a string and immersed in a tank of water. Determine the a) volume of the object b) the tension in the string when it is completely immersed. Solution:

The density of aluminum is
$\displaystyle\rho_{\:\rm Al}^{}$ = 2.70 x 103 kg/m 3 = $\displaystyle{\frac{M_{\:\rm
$\displaystyle\Rightarrow$ V Al = $\displaystyle{\frac{M_{\:\rm Al}}{\rho_{\:\rm Al}}}$   
  = $\displaystyle{\frac{27.0}{2.70 \times 10^3}}$ = 0.01 m 3.   
Applying Newton's second law for the aluminum block and noting that it is completely submerged and in equilibrium, we have:
$\displaystyle\sum$F = 0 = T + B - mg   
$\displaystyle\Rightarrow$ T = mg - B   
  = mg - $\displaystyle\rho_{\;{\:\rm water}}^{}$gV  water   
  = (27.0)(9.8) - (1.00 x 103)(9.8)(0.01)   
  = 166.6  N   

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$\fbox{\bf PROBLEM \thechapter.\arabic{probcount}}$

A garden hose of diameter 2 cm is used to fill a 20 liter bucket. a) If it takes 1 minute to fill the bucket, what is the speed at which water enters the hose? b) A practical joker pinches the open end of the hose down to a diameter of 5 mm, and sprays his neighbour with it. What is the speed at which water comes out of the hose?


The cross-sectional area of the hose will be given by:

A1 = $\displaystyle\pi$r 2 = $\displaystyle\pi$(2  cm /2)2 = $\displaystyle\pi$ cm 2.

To find the velocity, v1 , we use
Flow Rate = A1v1 = 20.0  L/min = $\displaystyle{\frac{20.0 \times
{\:\rm cm}{}^3}{60.0 \;{\:\rm s}}}$   
$\displaystyle\Rightarrow$ v1 = $\displaystyle{\frac{20.0 \times 10^3 {\:\rm cm}{}^3 / 60.0
\;{\:\rm s}}{\pi {\:\rm cm}{}^2}}$   
  = 106.1  cm/s.   
The flow rate ( A1v1 ) of the water approaching the constriction must be equal to the flow rate leaving the hose ( A2v2 ). This gives:
v2 = $\displaystyle{\frac{A_1v_1}{A_2}}$   
  = $\displaystyle{\frac{(\pi)(106.1)}{(\pi)(0.5/2)^2}}$   
  = 1698  cm/s.   

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$\fbox{\bf PROBLEM \thechapter.\arabic{probcount}}$

If a 30 m/s wind blows across a 175 m 2 flat roof of a house a) what is the pressure difference between the inside of the house and the outside of the house just above the roof? (Assume that the air pressure inside the house is atmospheric pressure.) b) What is the force on the roof due to the pressure difference?


The difference in pressure will be given by:
$\displaystyle\Delta$P = P out - P in = $\displaystyle\left( \frac{1}{2}
\rho_{\:\rm air} v^2 +
P_O \right)-$PO   
  = $\displaystyle{\textstyle\frac{1}{2}}$$\displaystyle\rho_{\:\rm air}^{}$v 2   
  = $\displaystyle{\textstyle\frac{1}{2}}$(1.29)(302) = 580.5  Pa.   
This pressure difference produces a net force of
F = $\displaystyle\Delta$PA = (580.5)(175) = 1.02 x 105  N        
which is directed up from the inside of the house.

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$\fbox{\bf PROBLEM \thechapter.\arabic{probcount}}$

Water flows through the pipe shown below at a flow rate of 0.10 m 3 /s . The diameter at point 1 is 0.4 m. At point 2, which is 3.0 m higher than point 1, the diameter is 0.20 m. If the end at point 2 is open to the air, determine the gauge pressure at point 2.

Figure 9.10: Problem 9.9
\epsfxsize=2 in

First we use the fact that the flow rate must be the same at both ends of the pipe to determine the velocities v1 and v2 at points 1 and 2, respectively:

A1v1 = A2v2 = 0.1 m 3/ s   
$\displaystyle\Rightarrow$ v1 = $\displaystyle{0.1{\:\rm m}{}^3{\:\rm /s}\over \pi (0.2 \;{\:\rm m})^2}$   
  = 0.8  m/s   
  and $\displaystyle\qquad$v2 = $\displaystyle{0.1{\:\rm m}{}^3{\:\rm /s}\over \pi (0.1 \;{\:\rm m})^2}$   
  = 3.2  m/s (19)
Next we use Bernouilli's equation (9.17) to determine the pressure difference between the two ends of the pipe:
P1 - P2 = $\displaystyle{1\over 2}$$\displaystyle\rho$(v22 - v12) + $\displaystyle\rho$g(y2 - y1)   
  = $\displaystyle{1\over 2}$ $\displaystyle\cdot$ 103 kg/m 3 $\displaystyle\cdot$ ((3.2  m/s )2 - (0.8  m/s )2) + 103 kg/m 3 $\displaystyle\cdot$ 9.8 m/s 2 $\displaystyle\cdot$ 3.0  m   
  = 43,800 N/m 2 (20)

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Next: About this document ... Up: Solids and Fluids Previous: Fluids in Motion