Author Archives: Engineering Math

Problem 6-1: Odometer reading based on the number of wheel revolutions

Semi-trailer trucks have an odometer on one hub of a trailer wheel. The hub is weighted so that it does not rotate, but it contains gears to count the number of wheel revolutions—it then calculates the distance traveled. If the wheel has a 1.15 m diameter and goes through 200,000 rotations, how many kilometers should the odometer read?

Solution:

The formula for the total distance traveled is

Δs=Δθ×r\Delta s=\Delta \theta \times r

Therefore, the total distance traveled is

Δs=(200000 rotations ×2π radian1 rotation)(1.15 m2)Δs=722566.3103 mΔs=722.6 km  (Answer)\begin{align*} \Delta s & =\left(200000\:\text{rotations}\:\times \frac{2\pi \:\text{radian}}{1\:\text{rotation}}\right)\left(\frac{1.15\:\text{m}}{2}\right) \\ \Delta s & =722566.3103\:\text{m} \\ \Delta s & =722.6\:\text{km} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}
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College Physics by Openstax Chapter 5 Problem 1

A physics major is cooking breakfast when he notices that the frictional force between his steel spatula and his Teflon frying pan is only 0.200 N. Knowing the coefficient of kinetic friction between the two materials, he quickly calculates the normal force. What is it?

Solution:

The formula for friction is

f=μk Nf=\mu _{k\:}N

When we solve for the normal force, NN, in terms of the other variables, we have

N=fμkN=\frac{f}{\mu _k}

The coefficient of kinetic friction is 0.04. Therefore, the normal force is

N=fμkN=0.200 newton0.04N=5.00 newton  (Answer)\begin{align*} N & =\frac{f}{\mu _k} \\ N & =\frac{0.200\:\text{newton}}{0.04} \\ N & =5.00\:\text{newton} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}
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College Physics by Openstax Chapter 4 Problem 1

A 63.0-kg sprinter starts a race with an acceleration of 4.20 m/s2. What is the net external force on him?

Solution:

So, we are given mass, m=63.0 kg m = 63.0 \ \text{kg} , and acceleration, a=4.20 m/s2a = 4.20 \ \text{m/s}^2.

The net force has a formula 

F=ma\text{F}=\text{m}a

Substituting the given values, we have

F=(63.0 kg)(4.20 m/s2)F=265 kg⋅m/s2F=265 N  (Answer)\begin{align*} F & = \left( 63.0 \ \text{kg} \right)\left( 4.20 \ \text{m/s}^2 \right) \\ F & = 265 \ \text{kg}\cdot \text{m/s}^2 \\ F & = 265 \ \text{N} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)} \end{align*}
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College Physics by Openstax Chapter 3 Problem 1

Find the following for path A in Figure 3.52:
(a) The total distance traveled, and
(b) The magnitude and direction of the displacement from start to finish.

Figure 3.54 The various lines represent paths taken by different people walking in a city. All blocks are 120 m on a side
Figure 3.52 The various lines represent paths taken by different people walking in a city. All blocks are 120 m on a side

Solution:

Part A

The total distance traveled is 

d=(3×120 m)+(1×120 m)d=480 m  (Answer)\begin{align*} \text{d} & =\left(3\times 120\ \text{m}\right)+\left(1\times 120\:\text{m}\right) \\ \text{d} & =480\:\text{m} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)}\\ \end{align*}

Part B

The magnitude of the displacement is 

s =(sx)2 +(sy)2s =(1×120 m)2+(3×120 m)2s =379 m  (Answer)\begin{align*} \text{s }& =\sqrt{\left( s_x \right)^{2\:}+\left( s_y \right)^2} \\ \text{s }& = \sqrt{\left(1\times 120\:\text{m}\right)^2+\left(3\times 120\:\text{m}\right)^2} \\ \text{s }& = 379\ \text{m} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)} \end{align*}

The direction is

θ=tan⁡−1(sxsy)θ=tan⁡−1(1×120 m3×120 m)θ=71.6∘, E of N  (Answer)\begin{align*} \theta & = \tan^{-1}\left(\frac{s_x}{s_y}\right) \\ \theta & = \tan^{-1}\left(\frac{1\times 120\:\text{m}}{3\times 120 \ \text{m}}\right) \\ \theta & =71.6^{\circ} ,\:\text{E of N} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)} \end{align*}
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