Author Archives: Engineering Math

Strength of Materials Problem 101 – Stress in each section of a composite bar

A composite bar consists of an aluminum section rigidly fastened between a bronze section and a steel section as shown in Fig. 1-8a. Axial loads are applied at the positions indicated. Determine the stress in each section.

Strength of Materials by Andrew Pytel and Ferdinand Singer Problem 101
Figure 1.8a

Solution:

We must first determine the axial load in each section to calculate the stresses. The free-body diagrams have been drawn by isolating the portion of the bar lying to the left of imaginary cutting planes. Identical results would be obtained if portions lying to the right of the cutting planes had been considered.

Solve for the internal axial load of the bronze

Free body diagram for the internal axial load of the bronze section for Problem 101 of Strength of Materials by Ferdinand Singer and Andrew Pytel
The free-body diagram of the bronze section
\begin{align*}
\sum_{}^{}F_x & = 0  \to  \\
-4000\ \text{lb}+P_{br} & = 0 \\
P_{br} & = 4000 \ \text{lb} \ \text{(tension)}
\end{align*}

Solve for the internal axial load of the aluminum

Free-body diagram of the aluminum section for problem 101 of Strength of materials by Andrew Pytel and Ferdinand Singer
The free-body diagram of the aluminum section
\begin{align*}
\sum_{}^{}F_x & = 0 \\
-4000 \ \text{lb} + 9000 \ \text{lb} - P_{al} & = 0 \\
P_{al} & = 5000 \ \text{lb} \ \text{(Compression)}
\end{align*}

Solve for the internal axial load of the aluminum

The free-body diagram of the steel section
\begin{align*}
\sum_{}^{}F_x & = 0 \\
-4000\ \text{lb} + 9000 \ \text{lb} + 2000\ \text{lb} - P_{st} & =0 \\
P_{st} & = 7000 \ \text{lb} \ \text{(Compression)}
\end{align*}

We can now solve the stresses in each section.

For the bronze

\begin{align*}
\sigma_{br} & = \frac{P_{br}}{A_{br}} \\
& = \frac{4000\ \text{lb}}{1.2 \ \text{in}^2} \\
& = 3330 \ \text{psi}\ \text{(Tension)} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

For the aluminum

\begin{align*}
\sigma_{al} & = \frac{P_{br}}{A_{al}} \\
& = \frac{5000\ \text{lb}}{1.8 \ \text{in}^2} \\
& = 2780 \ \text{psi}\ \text{(Compression)} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

For the steel

\begin{align*}
\sigma_{st} & = \frac{P_{st}}{A_{st}} \\
& = \frac{7000\ \text{lb}}{1.6 \ \text{in}^2} \\
& = 4380\ \text{psi}\ \text{(Compression)} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}
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College Physics by Openstax Chapter 4 Problem 8

What is the deceleration of the rocket sled if it comes to rest in 1.1 s from a speed of 1000 km/h? (Such deceleration caused one test subject to black out and have temporary blindness.)

Solution:

We are given the following: v_{0}=1000 \ \text{km/h}, v_{f}=0 \ \text{km/h}, \Delta t = 1.1 \ \text{s}.

The acceleration is computed as the change in velocity divided by the change in time.

\begin{align*}
a & = \frac{\Delta v}{\Delta t} \\
a & = \frac{v_{f}-v_{o}}{\Delta t} \\
a & = \frac{\left( 0\ \text{km/h}-1000 \ \text{km/h} \right)\left( \frac{1000 \ \text{m}}{1\ \text{km}} \right) \left( \frac{1\ \text{h}}{3600\ \text{s}} \right)}{1.1\ \text{s}} \\
a & = -252.5\ \text{m/s}^2 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)\\
\end{align*}
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Probability & Statistics for Engineers & Scientists Ninth Edition by Walpole Exercise 1.2

According to the journal Chemical Engineering, an important property of a fiber is its water absorbency. A random sample of 20 pieces of cotton fiber was taken and the absorbency on each piece was measured. The following are the absorbency values:

18.7121.4120.7221.8119.2922.4320.17
23.7119.4420.5018.9220.3323.0022.85
19.2521.7722.1119.7718.0421.12

(a) Calculate the sample mean and median for the above sample values.
(b) Compute the 10% trimmed mean.
(c) Do a dot plot of the absorbency data.
(d) Using only the values of the mean, median, and trimmed mean, do you have evidence of outliers in the data?

Solution:

Part A. The sample mean is computed as follows:

\begin{align*}
\bar x & = \frac{\Sigma x_{i}}{n} \\
\bar x & = \frac{18.71+21.41+20.72+\cdots +21.12}{20} \\
\bar x & = 20.77 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

We can solve for the sample median by arranging the data in increasing order first.

18.04, \
18.71, \
18.92, \
19.25, \
19.29, \
19.44, \
19.77, \
20.17, \
20.33, \
20.50 \\
20.72, \
21.12, \
21.41, \
21.77, \
21.81, \
22.11, \
22.43, \
22.85, \
23.00, \
23.71

Since there are 20 measurements (even), the middle measurements are the (20/2) 10th and the (20/2 + 1) 11th measurement. The 10th measurement is 20.50 and the 11th measurement is 20.72. The median is the average of these two measurements. 

\begin{align*}
\tilde x & = \frac{20.50+20.72}{2} \\
\tilde x & = 20.61 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

Part B. The 10% trimmed mean is calculated by removing 10% of the lowest data and 10% of the highest data. That is, removing the 2 lowest and 2 highest data. We are left with the following:

18.92, \
19.25, \
19.29, \
19.44, \
19.77, \
20.17, \
20.33, \
20.50 \\
20.72, \
21.12, \
21.41, \
21.77, \
21.81, \
22.11, \
22.43, \
22.85

The 10% trimmed mean, \bar x _{tr10} is

\begin{align*}
\bar x _{tr10} & = \frac{\Sigma x_{i}}{n} \\
\bar x _{tr10} & = \frac{18.92+19.25+\cdots+22.85}{16} \\
\bar x _{tr10} & = 20.74 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

Part C. The dot plot is shown

dot plot for Probability & Statistics for Engineers & Scientists Ninth Edition by Walpole Exercise 1.2

Part D. Since the values of the mean, median, and trimmed mean are not actually far from each other, we can conclude that there are no outliers in the given measurements.

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College Physics by Openstax Chapter 4 Problem 7

(a) If the rocket sled shown in Figure 4.31 starts with only one rocket burning, what is the magnitude of its acceleration? Assume that the mass of the system is 2100 kg, the thrust T is 2.4 \times 10^{4} N, and the force of friction opposing the motion is known to be 650 N. (b) Why is the acceleration not one-fourth of what it is with all rockets burning?

Solution:

Considering the direction of motion as the positive direction, we are given the following: T=2.4 \times 10^4 \ \text{N}, f=-650 \ \text{N}, and mass, m=2100 \ \text{kg}.

Part A. The magnitude of the acceleration can be computed using Newton’s Second Law of Motion.

\begin{align*}
\Sigma F & =ma \\
2.4\times 10^4 \ \text{N}-650 \ \text{N} & = 2100 \ \text{kg}\times a \\
23350 & = 2100 a \\
\frac{23350}{2100} & = \frac{\cancel{2100} a}{\cancel{2100}} \\
a & = \frac{23350}{2100} \\
a & = 11 \ \text{m/s}^2 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

Part B. The acceleration is not one-fourth of what it was with all rockets burning because the frictional force is still as large as it was with all rockets burning. \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)

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Probability & Statistics for Engineers & Scientists Ninth Edition by Walpole Exercise 1.1

The following measurements were recorded for the drying time, in hours, of a certain brand of latex paint.

3.42.54.82.93.6
2.83.35.63.72.8
4.44.05.23.04.8

Assume that the measurements are a simple random sample.
(a) What is the sample size for the above sample?
(b) Calculate the sample mean for these data.
(c) Calculate the sample median.
(d) Plot the data by way of a dot plot.
(e) Compute the 20% trimmed mean for the above data set.
(f) Is the sample mean for these data more or less descriptive as a center of location than the trimmed mean?

Solution:

Part A. Sample size, n is the total number of measurements.

n=15 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)

Part B. The sample mean, \bar x is computed as follows:

\begin{align*}
\bar x & = \sum_{i=1}^{n}\frac{x_{i}}{n} \\
& = \frac{3.4+2.5+4.8+2.9+3.6+2.8+3.3+5.6+3.7+2.8+4.4+4.0+5.2+3.0+4.8}{15} \\
& = \frac{56.8}{15} \\
& = 3.79 \ \text{hours} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

Part C. The sample median, \tilde x is the number at the middle of the arranged measurements in increasing magnitude. There are 15 measurements, n=15. If we arranged the data in increasing magnitude, the median is the measurement in the middle.

2.5, \ 2.8, \ 2.8, \ 2.9, \ 3.0, \ 3.3, \ 3.4, \ \underset{\color{Blue} \text{middle number}}{3.6}, \ 3.7, \ 4.0, \ 4.4, \ 4.8, \ 4.8, \ 5.2, \ 5.6

The middle number is 3.6. That is

\tilde x= 3.6 \ \text{hours} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)

Part D. This dot plot has been created using the Statistical Software Rguroo

a dot plot for the data: 3.4, 2.5, 4.8, 2.9, 3.6, 2.8, 3.3, 5.6, 3.7, 2.8, 4.4, 4.0, 5.2, 3.0, 4.8. this dot plot was made possible through Rgurro at https://www.rguroo.com/

Part E. The 20% trimmed mean means the average of the measurements left after removing 20% highest and 20% lowest data. This means we remove the 3 highest and 3 lowest numbers. Therefore, the data becomes

\ 2.9, \ 3.0, \ 3.3, \ 3.4, \ 3.6,  \ 3.7, \ 4.0, \ 4.4, \ 4.8,

The 20% trimmed mean, \bar x _{tr20} is

\begin{align*}
\bar x _{tr20} & = \frac{2.9+3.0+ \cdots+4.8}{9} \\
\bar x _{tr20} & = \frac{33.1}{9} \\
\bar x _{tr20} & = 3.678 \ \text{hours} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

Part F. The sample mean for these data is \bar x = 3.79 \ \text{hours} while the 20% trimmed mean is \bar x _{tr20} = 3.678 \ \text{hours}. Seems like the two means are not really far from each other, but because of the elimination of the extreme values, we can treat the trimmed mean as a better descriptive mean.

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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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