Author Archives: Engineering Math

College Physics by Openstax Chapter 4 Problem 6

The same rocket sled drawn in Figure 4.30 is decelerated at a rate of 196 m/s2. What force is necessary to produce this deceleration? Assume that the rockets are off. The mass of the system is 2100 kg.

Solution:

Since the rockets are off, the only force acting on the sled is the friction f. This force is against the direction of motion. By using Newton’s Second Law of Motion, we have.

\begin{align*}
\Sigma F & = ma \\
-f & = ma \\
-f & = \left( 2100 \ \text{kg} \right)\left( -196 \ \text{m/s}^{2} \right) \\
-f & = -411600 \ \text{N} \\
f & = 411600 \ \text{N} \\ 
f & = 411.6 \ \text{kN} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

The force necessary to produce the given deceleration is 411.6 kN.

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College Physics by Openstax Chapter 4 Problem 5

In Figure 4.7, the net external force on the 24-kg mower is stated to be 51 N. If the force of friction opposing the motion is 24 N, what force F (in newtons) is the person exerting on the mower? Suppose the mower is moving at 1.5 m/s when the force F is removed. How far will the mower go before stopping?

Figure for College Physics by Openstax Chapter 4 Problem 5: A person is pushing a mower to the right.

Solution:

We can isolate the mower and expose the forces acting on it. This is the free-body diagram.

The free-body diagram of the mower: the force F exerted by the person and the friction force f exerted by the ground on the mower.

There are two forces acting on the mower in the horizontal directions:

  1. \textbf{F}:This is the force exerted by the person on the mower, and it is going to the right. This is the first unknown in the problem. We treat this as a positive force since it is directed to the right. The value of this force is F=51 \ \text{N}
  2. \textbf{f}: This is the friction force directed opposite the motion of the mower. We treat this as a negative force because it is directed to the left. The value of this force is f=24 \ \text{N}.

Part A. The third force in the figure is the net force, F_{net}. This is the vector sum of the forces F and f. That is

\begin{align*}
F_{net} & =F-f \\
51 \ \text{N} & = F-24 \ \text{N} \\
F & = 51 \ \text{N} +24 \ \text{N} \\
F & = 75 \ \text{N} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

The person should exert a force of 75 N to produce a net force of 51 N.

Part B. When the force F is removed, the friction is now the only force acting on the mower. The friction is acting opposite the direction of motion. The direction of motion is indicated by the blue arrow in the figure below, and the friction force is the red arrow.

The free-body diagram of the mower when force F is removed.

Using Newton’s Second Law, we can solve for the deceleration of the mower.

\begin{align*}
\Sigma F & = ma \\
-24 \ \text{N} & = \left( 24 \ \text{kg} \right) \ a \\
\frac{-24 \ \text{N}}{24 \ \text{kg}} & = \frac{\cancel{24 \ \text{kg}}\ \ a}{\cancel{24 \ \text{kg}}} \\
a & = \frac{-24 \ \text{N}}{24 \ \text{kg}} \\
a & = -1 \ \text{m/s}^{2} \\
\end{align*}

Using this deceleration computed above, we can solve for the distance traveled by the mower before coming to stop.

\begin{align*}
\left( v_{f} \right)^{2} & = \left( v_{o} \right)^{2}+2a\Delta x \\
\left( 0 \ \text{m/s} \right)^{2} & = \left( 1.5 \ \text{m/s} \right)^{2}+2\left( -1 \ \text{m/s}^{2} \right)\Delta x \\
 0 & = 2.25-2 \Delta x \\
2 \Delta x & = 2.25 \\
\frac{\cancel{2}\Delta x}{\cancel{2}} & =\frac{2.25}{2} \\
\Delta x & = 1.125 \ \text{m} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}
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College Physics by Openstax Chapter 4 Problem 4

Since astronauts in orbit are apparently weightless, a clever method of measuring their masses is needed to monitor their mass gains or losses to adjust diets. One way to do this is to exert a known force on an astronaut and measure the acceleration produced. Suppose a net external force of 50.0 N is exerted and the astronaut’s acceleration is measured to be 0.893 m/s2. (a) Calculate her mass. (b) By exerting a force on the astronaut, the vehicle in which they orbit experiences an equal and opposite force. Discuss how this would affect the measurement of the astronaut’s acceleration. Propose a method in which recoil of the vehicle is avoided.

Solution:

We are given the following: \sum F = 50.0 \ \text{N}, and a=0.893 \ \text{m/s}^{2}.

Part A. We can solve for the mass, m by using Newton’s second law of motion.

\begin{align*}
\sum F & = ma \\
50.0 \ \text{N} & = m \left( 0.893 \ \text{m/s}^{2} \right) \\
m & = \frac{50.0 \ \text{N}}{0.893 \ \text{m/s}^{2}} \\
m & = 56.0 \ \text{kg}\ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

Part B. The measured acceleration is equal to the sum of the accelerations of the astronauts and the ship. That is 

a_{measured}=a_{astronaut}+a_{ship}

If a force acting on the astronaut came from something other than the spaceship, the spaceship would not undergo a recoil. \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)

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College Physics by Openstax Chapter 4 Problem 3

A cleaner pushes a 4.50-kg laundry cart in such a way that the net external force on it is 60.0 N. Calculate its acceleration.

Solution:

From Newton’s Second Law of Motion, \sum F =ma. Substituting the given values, we have

\begin{align*}
\sum F & = ma \\
60.0 \ \text{N} & = \left( 4.50 \ \text{kg} \right) \ a \\
a & = \frac{60.0 \ \text{N}}{4.50 \ \text{kg}} \\
a & = 13.3 \ \text{m/s}^{2} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}
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College Physics by Openstax Chapter 4 Problem 2

If the sprinter from the previous problem accelerates at that rate for 20 m, and then maintains that velocity for the remainder of the 100-m dash, what will be his time for the race?

Solution:

Solving for the time it takes to reach the first 20 meters.

\begin{align*}
\Delta x & =v_{0}t+\frac{1}{2}at^{2} \\
20 \ \text{m} & = \left( 0 \ \text{m/s} \right)t + \frac{1}{2}\left( 4.20 \ \text{m/s}^{2} \right)t^{2} \\
20 & = 2.1 t^{2} \\
\frac{20}{2.1}& = \frac{\cancel{2.1} \ t^{2}}{\cancel{2.1}} \\
t^{2} & = 9.5238 \\
\sqrt{t^{2}} & = \sqrt{9.5238} \\
t_{1} & = 3.09 \ \text{s}
\end{align*}

We can compute the velocity of the sprinter at the end of the first 20 meters.

\begin{align*}
v^2 & = v_{0}^2 + 2ax \\
v^2 & = \left( 0 \ \text{m/s} \right)^2 + 2\left( 4.20 \ \text{m/s}^2 \right) \left( 20 \ \text{m} \right) \\
v & = \sqrt{2\left( 4.20 \right)\left( 20 \right)} \ \text{m/s} \\
v & = 12.96 \ \text{m/s}
\end{align*}

For the remaining 80 meters, the sprinter has a constant velocity of 12.96 m/s. The sprinter’s time to run the last 80 meters can be computed as follows.

\begin{align*}
\Delta x & = vt \\
80 \ \text{m} & = \left( 12.96 \  \text{m/s} \right)\  t_{2} \\
\frac{80 \ \text{m}}{12.96 \ \text{m/s}} & =\frac{\cancel{12.96 \ \text{m/s} } \ \ t_{2}}{\cancel{12.96 \ \text{m/s}}} \\
t_{2} & = \frac{80}{12.96} \ \text{s} \\
t_{2} & = 6.17 \ \text{s} \\
\end{align*}

The sprinter’s total time, t_{T}, to finish the 100-m race is the sum of the two times. 

\begin{align*}
t_{T} & = t_{1} + t_{2} \\
t_{T} & = 3.09 \ \text{s} + 6.17 \ \text{s} \\
t_{T} & = 9.26 \ \text{s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}
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Hydrology and Floodplain Analysis by Bedient et.al. Chapter 1 Problem 7

Clear Lake has a surface area of 708,000 m2 (70.8 ha.). For a given month, the lake has an inflow of 1.5 m3/s and an outflow of 1.25 m3/s. A +1.0-m storage change or increase in lake level was recorded. If a precipitation gage recorded a total of 24 cm for this month, determine the evaporation loss (in cm) for the lake. Assume that seepage loss is negligible.

Solution:

We are given the following values:

\begin{align*}
\text{Area}, \ A&=708,000 \ \text{m}^2 \\
\text{Inflow}, \ I&=1.5 \ \text{m}^3/\text{s} \\
\text{Outflow}, \ O & = 1.25 \ \text{m}^3/\text{s} \\
\text{change in storage}, \ \Delta S & = 1.0 \ \text{m} \\
\text{Precipitation}, \ P&=24 \ \text{cm} \\
\text{time}, \ t &= 1 \ \text{month} = 30 \ \text{days}
\end{align*}

The required value is the \text{Evaporation}, \ E.

We shall use the formula

\Delta S=I+P-O-E

Solving for E in terms of the other variables, we have

E=I+P-O-\Delta S

Before we can substitute all the given values, we need to convert everything to the same unit of cm.

\begin{align*}
\text{Inflow}&=\frac{1.5\:\frac{\text{m}^3}{\text{s}}\cdot \frac{100\:\text{cm}}{1\:\text{m}}\cdot \frac{3600\:\text{s}}{1\:\text{hr}}\cdot \frac{24\:\text{hr}}{1\:\text{day}}\cdot \frac{30\:\text{days}}{1\:\text{month}}\cdot 1\:\text{month}}{708,000\:\text{m}^2} \\
\text{Inflow}&=549.1525 \ \text{cm}
\end{align*}
\begin{align*}
\text{Outflow}&=\frac{1.25\:\frac{\text{m}^3}{\text{s}}\cdot \frac{100\:\text{cm}}{1\:\text{m}}\cdot \frac{3600\:\text{s}}{1\:\text{hr}}\cdot \frac{24\:\text{hr}}{1\:\text{day}}\cdot \frac{30\:\text{days}}{1\:\text{month}}\cdot 1\:\text{month}}{708,000\:\text{m}^2} \\
\text{Outflow}&=457.6271 \ \text{cm}
\end{align*}
\Delta S=1.0 \ \text{m} \times \frac{100 \ \text{cm}}{1.0 \ \text{m}}=100 \ \text{cm}

Now, we can substitute the given values in the formula

\begin{align*}
E & =I+P-O-\Delta S \\
E& =549.1525 \ \text{cm}+24 \text{cm}-457.6271 \ \text{cm}-100 \ \text{cm} \\
E& =15.5254 \ \text{cm} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}
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Hydrology and Floodplain Analysis by Bedient et.al. Chapter 1 Problem 6

A lake with a surface area of 1050 acres was monitored over a period of time. During a one-month period, the inflow was 33 cfs, the outflow was 27 cfs, and a 1.5-in. seepage loss was measured. During the same month, the total precipitation was 4.5 in. Evaporation loss was estimated as 6.0 in. Estimate the storage change for this lake during the month.

Solution:

We are given the following values:

\begin{align*}
\text{Area}, \ A&=1050 \ \text{acres} \\
\text{Time}, \ t&=1 \ \text{month} \\
\text{Inflow}, \ I&=33 \ \text{cfs} \\
\text{Outflow}, \ O&=27 \ \text{cfs} \\
\text{Ground seepage}, \ G&=1.5 \ \text{in} \\
\text{Precipitation}, \ P&=4.5 \ \text{in} \\
\text{Evaporation}, \ E&=6.0 \ \text{in}
\end{align*}

The formula that we are going to use is:

\sum \text{Inflows}-\sum \text{Outflows}=\text{Change in Storage}, \Delta S \\
\\ 
\sum I-\sum Q=\Delta S

In this case, the inflows are \text{Inflow} \ I and \text{Precipitation}, \ P, while the others are outflows. Our formula now becomes

\color{Blue} \sum \text{Inflow}-\color{Red} \sum \text{Outflow}=\color{Green} \Delta S
\\
\color{Blue}(I+P)- \color{Red}(O+G+E)=\color{Green}\Delta S

Before substituting, we need to convert all the given to inches. More specifically the outflow O and inflow I.

The inflow and outflow, in cfs, will be divided by the given are to come up with units of inches.

The inflow is

\begin{align*}
\text{Inflow}&=\frac{33\:\frac{\text{ft}^3}{\text{s}}\cdot \frac{1\:\text{acre}}{43560\:\text{ft}^2}\cdot \frac{12\:\text{in}}{1\:\text{ft}}\cdot \frac{3600\:\text{s}}{1\:\text{hr}}\cdot \frac{24\:\text{hr}}{1\:\text{day}}\cdot \frac{30\:\text{days}}{1\:\text{month}}\cdot 1\:\text{month}}{1050\:\text{acres}} \\
\text{Inflow}& =22.4416 \ \text{in}
\end{align*}

The outflow is

\begin{align*}
\text{Outflow}&=\frac{27\:\frac{\text{ft}^3}{\text{s}}\cdot \frac{1\:\text{acre}}{43560\:\text{ft}^2}\cdot \frac{12\:\text{in}}{1\:\text{ft}}\cdot \frac{3600\:\text{s}}{1\:\text{hr}}\cdot \frac{24\:\text{hr}}{1\:\text{day}}\cdot \frac{30\:\text{days}}{1\:\text{month}}\cdot 1\:\text{month}}{1050\:\text{acres}} \\
\text{Outflow}& =18.3613\ \text{in}
\end{align*}

Now that everything is in inches, we can now substitute the values in the formula

\begin{align*}
\Delta S&=\left( I+P \right)-\left( O+G+E \right) \\
\Delta S&=\left( 22.4416 \ \text{in}+4.5 \ \text{in} \right)-\left( 18.3613 \ \text{in}+1.5 \ \text{in}+6 \ \text{in} \right) \\
\Delta S&=1.0803 \ \text{in} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

We can also state the change in storage in terms of volume by multiplying the given area

\begin{align*}
\Delta S \ \text{in volume} & =1.0803 \ \text{in}\times 1050 \ \text{acres}\times \frac{1 \ \text{ft}}{12 \ \text{in}}\\
\Delta S \ \text{in volume} & = 94.5263 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}
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Hydrology and Floodplain Analysis by Bedient et.al. Chapter 1 Problem 5

List seven major factors that determine a watershed’s response to a given rainfall.

Solution:

The seven major factors that determine a watershed’s response to a given rainfall are:

  1. Drainage Area
  2. Channel Slope
  3. Soil Types
  4. Land Use
  5. Land Cover
  6. Main Channel and tributary characteristics-channel morphology
  7. The shape, slope, and character of the floodplain
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Hydrology and Floodplain Analysis by Bedient et.al. Chapter 1 Problem 4

Explain how air masses are classified. Where are these types of air masses located?

Solution:

They are classified in two ways: the source from which they are generated, land (continental) or water (maritime), and the latitude of generation (polar or tropical).

These air masses are present in the United States. The Continental polar emanates from Canada and passes over the northern United States. The maritime polar air mass also comes southward from the Atlantic Coast of Canada and affects the New England states. Another maritime polar comes from the Pacific and hits the extreme northwestern states. The maritime tropical air masses come from the Pacific, the Gulf of Mexico, and the Atlantic (these affect the entire Southern United States). Continental tropical air masses form only during summer. They originate in Texas and affect the states bordering the north.

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Hydrology and Floodplain Analysis by Bedient et.al. Chapter 1 Problem 3

Explain the difference between humidity and relative humidity.

Solution:

Humidity is a measure of the amount of water vapor in the atmosphere and can be expressed in several ways. Specific humidity is a mass of water vapor in a unit mass of moist air while relative humidity is a ratio of the air’s actual water vapor content compared to the amount of water vapor at saturation for that temperature.

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