Author Archives: Engineering Math

College Physics by Openstax Chapter 2 Problem 33

An unwary football player collides with a padded goalpost while running at a velocity of 7.50 m/s and comes to a full stop after compressing the padding and his body 0.350 m.

a) What is his deceleration?

b) How long does the collision last?

Solution:

We are given the following: v_0=7.50\:\text{m/s}; v_f=0.00\:\text{m/s}; and\Delta x=0.350\:\text{m}.

Part A

We are going to use the formula

\left(v_f\right)^2=\left(v_0\right)^2+2a\Delta x

Solving for the acceleration a in terms of the other variables:

a=\frac{\left(v_f\right)^2-\left(v_0\right)^2}{2\Delta x}

Substituting the given values:

\begin{align*}
a & =\frac{\left(v_f\right)^2-\left(v_0\right)^2}{2\Delta x} \\
a & =\frac{\left(0\:\text{m/s}\right)^2-\left(7.50\:\text{m/s}\right)^2}{2\left(0.350\:\text{m}\right)} \\
a & =-80.4\:\text{m/s}^2 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

Part B

We are going to use the formula

\Delta x=v_{ave}t

Since v_{ave}=\frac{v_f+v_0}{2}, we can write the formula as

\Delta x=\frac{v_f+v_0}{2}\cdot t

Solving for time t in terms of the other variables:

\:t=\frac{2\Delta x}{v_f+v_0}

Substituting the given values:

\begin{align*}
t&=\frac{2\Delta x}{v_f+v_0} \\
t&=\frac{2\left(0.350\:\text{m}\right)}{0\:\text{m/s}+7.50\:\text{m/s}} \\
t& =9.33\times 10^{-2}\:\text{s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}
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College Physics by Openstax Chapter 2 Problem 32

A woodpecker’s brain is specially protected from large decelerations by tendon-like attachments inside the skull. While pecking on a tree, the woodpecker’s head comes to a stop from an initial velocity of 0.600 m/s in a distance of only 2.00 mm.

a) Find the acceleration in m/s2 and in multiples of g (g=9.80 m/s2).

b) Calculate the stopping time.

c) The tendons cradling the brain stretch, making its stopping distance 4.50 mm (greater than the head and, hence, less deceleration of the brain). What is the brain’s deceleration, expressed in multiples of g?

Solution:

We are given the following values: v_0=0.600 \ \text{m/s}; v_f=0.000\:\text{m/s}; and \Delta x=0.002\:\text{m}.

Part A

The acceleration is computed based on the formula, 

\left(v_f\right)^2=\left(v_0\right)^2+2a\Delta x

Solving for acceleration a in terms of the other variables, we have

a=\frac{\left(v_f\right)^2-\left(v_0\right)^2}{2\Delta x}

Substituting the given values, 

\begin{align*}
a & =\frac{\left(v_f\right)^2-\left(v_0\right)^2}{2\Delta x} \\
a & =\frac{\left(0.000\:\text{m/s}\right)^2-\left(0.600\:\text{m/s}\right)^2}{2\left(0.002\:\text{m}\right)} \\
a & =-90.0\:\text{m/s}^2 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

In terms of g taking absolute values of the acceleration , we have

\begin{align*}
a & = 90.0 \ \text{m/s}^2 \cdot \left( \frac{g}{9.80 \ \text{m/s}^2} \right) \\
a & = \frac{90.0}{9.80}g \\
a & = 9.18g \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

Part B

We shall use the formula

\Delta x=v_{ave}t

where v_{ave} is the average velocity computed as 

v_{ave}=\frac{v_0+v_f}{2}

Solving for time t in terms of the other variables, we have

t=\frac{2\Delta x}{v_0+v_f}

Substituting the given values, we have

\begin{align*}
t & =\frac{2\Delta x}{v_0+v_f} \\
t & =\frac{2\left(0.002\:\text{m}\right)}{0.600\:\text{m/s}+0.000\:\text{m/s}} \\
t & =6.67\times 10^{-3}\:\text{s}  \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

Part C

Employing the same formula we used in Part A, we have

\begin{align*}
a & =\frac{\left(v_f\right)^2-\left(v_0\right)^2}{2\Delta x} \\
a & =\frac{\left(0.000\:\text{m/s}\right)^2-\left(0.600\:\text{m/s}\right)^2}{2\left(0.0045\:\text{m}\right)} \\
a & =-40.0\:\text{m/s}^2 
\end{align*}

In terms of g, taking absolute values of the acceleration

\begin{align*}
a & = 40.0 \ \text{m/s}^2 \cdot \left( \frac{g}{9.80 \ \text{m/s}^2} \right) \\
a & =\frac{40.0}{9.80}g \\
a & = 4.08g \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}
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Mechanics of Materials: An Integrated Learning Approach 3rd Edition by Timothy A. Philpot Problem P1.2

A 2024-T4 aluminum tube with an outside diameter of 2.50 in. will be used to support a 27-kip load. If the axial normal stress in the member must be limited to 18 ksi, determine the wall thickness required for the tube.

Solution:

We are given the following values:

\begin{align*}
\text{Outside Diameter}, D & = 2.50\ \text{in} \\
\text{Axial Load}, P & = 27\ \text{kips} \\
\text{Maximum Axial Stress}, \sigma & = 18\ \text{ksi}\\
\text{Inside Diameter}, d & = D-2t
\end{align*}

We are solving for the unknown wall thickness of the tube, t.

From the definition of normal stress, solve for the minimum area required to support a 27-kip load without exceeding a stress of 18 ksi

\begin{align*}
\sigma & =\frac{P}{A} \\ 
A_{\text{min}} & =\frac{P}{\sigma } \\ 
A_{\text{min}} & = \frac{27\:\text{kips}}{18\:\text{ksi}} \\ 
A_{\text{min}} & = 1.500\:\text{in}^2 
\end{align*}

The cross-sectional area of the aluminum tube is given by

A=\frac{\pi }{4}\left(D^2-d^2\right)

Set this expression equal to the minimum area and solve for the maximum inside diameter, d

\begin{align*}
A & =\frac{\pi }{4}\left(D^2-d^2\right) \\
A & = \frac{\pi }{4}\left[\left(2.50\ \text{in}\right)^2-d^2\right] \\
1.500\ \text{in}^2 & = \frac{\pi }{4}\left[\left(2.50\ \text{in}\right)^2-d^2\right] \\
4 \left( 1.500\ \text{in}^2 \right) & = \pi \left[\left(2.50\ \text{in}\right)^2-d^2\right] \\
\frac{4 \left( 1.500\ \text{in}^2 \right)}{\pi} & = \left(2.50\ \text{in}\right)^2-d^2 \\
d^{2} & = \left( 2.50\ \text{in}^{2} \right) - \frac{4 \left( 1.500\ \text{in}^2 \right)}{\pi} \\
d & = \sqrt{\left( 2.50\ \text{in}^{2} \right) - \frac{4 \left( 1.500\ \text{in}^2 \right)}{\pi}} \\
d & = 2.08330\ \text{in}
\end{align*}

The outside diameter D, the inside diameter d, and the wall thickness t are related by

D = d+2t

Therefore, the minimum wall thickness required for the aluminum tube is

\begin{align*}
t & =\frac{D-d}{2} \\
t & = \frac{2.50\:\text{in}-2.08330\:\text{in}}{2} \\
t & = 0.208\ \text{in} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

Mechanics of Materials: An Integrated Learning Approach 3rd Edition by Timothy Philpot Problem P1.1

A stainless steel tube with an outside diameter of 60 mm and a wall thickness of 5 mm is used as a compression member. If the axial normal stress in the member must be limited to 200 MPa, determine the maximum load P that the member can support.

Solution:

We are given the following values:

\begin{align*}
\text{Outside Diameter}, D &= 60\ \text{mm} \\
\text{Wall Thickness}, t & = 5\ \text{mm} \\
\text{Inside Diameter}, d & = D-2t = 60\ \text{mm}-2\left( 5\ \text{mm} \right) = 50\ \text{mm}\\
\text{Maximum Axial Stress}, \sigma & =200\ \text{MPa} = 200\ \frac{\text{N}}{\text{mm}^2}
\end{align*}

The cross-sectional area of the stainless-steel tube is

\begin{align*}
A & = \frac{\pi}{4}\left( D^2 - d^2 \right) \\
A & = \frac{\pi}{4}\left[ \left( 60\ \text{mm} \right)^2 - \left( 50\ \text{mm} \right)^2 \right] \\
A & = 863.938\ \text{mm}^2
\end{align*}

The normal stress in the tube can be expressed as

\sigma =\frac{P}{A}

The maximum normal stress in the tube must be limited to 200 MPa. Using 200 MPa as the allowable normal stress, rearrange this expression to solve for the maximum load P.

\begin{align*}
P_{max} & = \sigma _{max} A \\
P_{max} & = \left( 200\ \text{MPa} \right)\left( 863.938\ \text{mm}^2 \right)\\
P_{max} & = \left( 200\ \frac{\text{N}}{\text{mm}^2} \right)\left( 863.938\ \text{mm}^2 \right)\\
P_{max} & = 172788\ \text{N} \\
P_{max} & = 172.8\ \text{kN}\ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

Differential and Integral Calculus by Feliciano and Uy, Exercise 1.2, Problem 3

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PROBLEM:

Evaluate \displaystyle \lim\limits_{x\to \frac{\pi }{4}}\left(\tan\:x+\sin\:x\right).

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SOLUTION:

\begin{align*}

\lim\limits_{x\to \frac{\pi }{4}}\left(\tan\:x+\sin\:x\right) & =\lim\limits_{x\to \frac{\pi }{4}}\left(\tan\:x\right)+\lim\limits_{x\to \frac{\pi }{4}}\left(\sin\:x\right)\\

& =\tan\:\frac{\pi }{4}+\sin\:\frac{\pi }{4}\\

& =1+\frac{\sqrt{2}}{2}\\

& =\frac{2+\sqrt{2}}{2} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)\\

\end{align*}
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Differential and Integral Calculus by Feliciano and Uy, Exercise 1.2, Problem 2

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PROBLEM:

Evaluate \displaystyle \lim\limits_{x\to 3}\left(\frac{4x+2}{x+4}\right).

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SOLUTION:

\begin{align*}

\lim_{x\to 3}\left(\frac{4x+2}{x+4}\right)& =\frac{\lim\limits_{x\to 3}\left(4x+2\right)}{\lim\limits_{x\to 3}\left(x+4\right)}\\

& =\frac{\lim\limits_{x\to 3}\left(4x\right)+\lim\limits_{x\to 3}\left(2\right)}{\lim\limits_{x\to 3}\left(x\right)+\lim\limits_{x\to 3}\left(4\right)}\\

& =\frac{4\cdot \lim\limits_{x\to 3}\left(x\right)+2}{3+4}\\

& =\frac{4\cdot 3+2}{3+4}\\

& =\frac{12+2}{7}\\

& =\frac{14}{7}\\

& =2 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)\\

\end{align*}
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Differential and Integral Calculus by Feliciano and Uy, Exercise 1.2, Problem 1

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PROBLEM:

Evaluate \displaystyle \lim _{x\to 2}\left(x^2-4x+3\right).

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SOLUTION:

\begin{align*}

\lim_{x\to 2}\left(x^2-4x+3\right)& = \lim_{x\to 2}\left(x^2\right)-\lim_{x\to 2}\left(4x\right)+\lim_{x\to 2}\left(3\right)\\

& =\left[\lim_{x\to 2}\left(x\right)\right]^2-4\lim_{x\to 2}\left(x\right)+3\\

& =\left(2\right)^2-4\left(2\right)+3\\

& =-1 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)\\

\end{align*}
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Differential and Integral Calculus by Feliciano and Uy, Exercise 1.1, Problem 10

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PROBLEM:

If  \displaystyle f\left(x\right)=\frac{4}{x+3} and \displaystyle \:g\left(x\right)=x^2-3 , find \displaystyle f\left[g\left(x\right)\right] and \displaystyle g\left[f\left(x\right)\right].

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SOLUTION:

Part A

\begin{align*}

f\left[g\left(x\right)\right] & =\frac{4}{\left(x^2-3\right)+3}\\

& =\frac{4}{x^2} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)\\

\end{align*}

Part B

\begin{align*}

g\left[f\left(x\right)\right] & =\left(\frac{4}{x+3}\right)^2-3\\

& =\frac{16}{\left(x+3\right)^2}-3\\

& =\frac{16-3\left(x+3\right)^2}{\left(x+3\right)^2}\\

& =\frac{16-3\left(x^2+6x+9\right)}{\left(x+3\right)^2}\\

& =\frac{16-3x^2-18x-27}{\left(x+3\right)^2}\\

& =\frac{-3x^2-18x-11}{\left(x+3\right)^2} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)\\

\end{align*}
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Differential and Integral Calculus by Feliciano and Uy, Exercise 1.1, Problem 9

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PROBLEM:

If \displaystyle f\left(x\right)=3x^2-4x+1, find \displaystyle \frac{f\left(h+3\right)-f\left(3\right)}{h},\:h\ne 0.

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SOLUTION:

\begin{align*}
\frac{f\left(h+3\right)-f\left(3\right)}{h} & =\frac{\left[3\left(h+3\right)^2-4\left(h+3\right)+1\right]-\left[3\left(3\right)^2-4\left(3\right)+1\right]}{h} \\

& =\frac{3\left(h^2+6h+9\right)-4h-12+1-16}{h}\\

& =\frac{3h^2+18h+27-4h-12+1-16}{h}\\

& =\frac{3h^2+14h}{h}\\

& =\frac{h\left(3h+14\right)}{h}\\
& =3h+14 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)\\
\end{align*}
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Differential and Integral Calculus by Feliciano and Uy, Exercise 1.1, Problem 8

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PROBLEM:

If \displaystyle f\left(x\right)=x^2+1, find \displaystyle \frac{f\left(x+h\right)-f\left(x\right)}{h},\:h\ne 0.

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SOLUTION:

\begin{align*}
\displaystyle \frac{f\left(x+h\right)-f\left(x\right)}{h} & =\frac{\left[\left(x+h\right)^2+1\right]-\left(x^2+1\right)\:}{h}\\ \\
& =\frac{x^2+2xh+h^2+1-x^2-1}{h}\\ \\
& =\frac{2xh+h^2}{h}\\ \\
& =\frac{h\left(2x+h\right)}{h}\\ \\
& =2x+h \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \\
\end{align*}
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