Category Archives: Sciences

Includes Engineering Sciences such as Statics, Dynamics, Strength of Materials, Physics, Fluid Mechanics, Chemistry

Grantham Week 1 Discussion

If an equation is dimensionally correct, does this mean that the equation must be true? If an equation is not dimensionally correct, does this mean that the equation cannot be true? Explain.

Answer 1:

If an equation is dimensionally correct, it does not mean that the equation must be true. On the other hand, when the equation is dimensionally correct, the equation cannot be true.

Dimensional analysis is a technique used to check whether a relationship is correct. So, it can only tell you if a relationship is correct or not, but it can not tell you if it is completely right because of the numerics that may be involved in the calculations.

Answer 2:

An equation being dimensionally correct doesn’t mean that the equation is true. For instance, when calculating the area of a circle you can replace pi with another number and it would still be wrong while being dimensionally correct. However, if an equation is not dimensionally correct, the equation cannot be true. This is because if it is not dimensionally correct it would equate into something looking like 4 grapes = 4 bananas which could not happen.

Answer 3:

In order for an equation to be valid, the dimensions on the left side must match the dimensions on the right side, in which case it is dimensionally correct. An equation can be dimensionally correct but still can be wrong. However, if an equation is dimensionally incorrect, it must be wrong.

Answer 4:

No, a true equation must be dimensionally correct but some dimensionally correct equations are not true. Yes, unless the results of the equation produces the correct units, the equation cannot be correct.

Answer 5:

For an equation to be valid, the dimensions on the left side must match the dimensions on the right side (just like our oranges example.) It is then dimensionally correct.However an equation can be dimensionally correct but still wrong.
For example if I say the area of a circle = 2 x radius^2:
– this is dimensionally correct (both sides have dimensions L^2)
– but it is wrong, as ‘2’ should be ‘pi’.
On the other hand, if an equation is dimensionally incorrect, it must be wrong.

College Physics by Openstax Chapter 2 Problem 49

You throw a ball straight up with an initial velocity of 15.0 m/s. It passes a tree branch on the way up at a height of 7.00 m. How much additional time will pass before the ball passes the tree branch on the way back down?

Solution:

The known values are a=-9.80\:\text{m/s}^2; v_o=15.0\:\text{m/s}; y=7.00\:\text{m}

The applicable formula is.

y=v_ot+\frac{1}{2}at^2

Using this formula, we can solve it in terms of time, t.

t=\frac{-v_0\pm \sqrt{v_0^2+2ay}}{a}

Substituting the known values, we have

\begin{align*}
t & =\frac{-v_0\pm \sqrt{v_0^2+2ay}}{a} \\
t & =\frac{-15.0\:\text{m/s}\pm \sqrt{\left(15.0\:\text{m/s}\right)^2+2\left(-9.80\:\text{m/s}^2\right)\left(7.00\:\text{m}\right)}}{-9.80\:\text{m/s}^2} \\
t&=\frac{-15.0\:\text{m/s}\pm 9.37\:\text{m/s}}{-9.80\:\text{m/s}^2}
\end{align*}

We have two values for time, t. These two values represent the times when the ball passes the tree branch.

 t_1=\frac{-15.0\:m/s+9.37\:m/s}{-9.80\:m/s^2}=0.57\:sec \\
t_2=\frac{-15.0\:m/s-9.37\:m/s}{-9.80\:m/s^2}=2.49\:sec

Therefore, the total time between passing the branch is the difference between 2.49 seconds and 0.57 seconds.

t_2-t_1=2.49  \ \text{s} - 0.57 \ \text{s}=1.92 \ \text{s}  \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
Advertisements
Advertisements

Hibbeler Statics 14E P2.2 — Resultant of a System of Two Forces

Determine the magnitude of the resultant force \textbf{F}_{\text{R}} = \textbf{F}_1 + \textbf{F}_2 and its direction, measured counterclockwise from the positive x axis. 

Engineering Mechanics: Statics figure for Problem 2-3

Engineering Mechanics: Statics 13th Edition by RC Hibbeler, Problem 2-1
Engineering Mechanics: Statics 14th Edition by RC Hibbeler, Problem 2-3

SOLUTION:

The parallelogram law of the force system is shown.

Consider the triangle AOB.

Using cosine law to solve for the resultant force \textbf{F}_{\text{R}}

\begin{align*}
\textbf{F}_\text{R} & =\sqrt{\left(250\right)^2+\left(375\right)^2-2\left(250\right)\left(375\right) \cos\:75^{\circ} }\\
& =393.2 \ \text{lb}\\
& =393\:\text{lb}\\
\end{align*}

The value of angle θ can be solved using sine law. 

\begin{align*}
\frac{393.2}{\sin\:\left(75^{\circ} \right)} & = \frac{250}{\sin\:\theta } \\
\sin \theta & = \frac{250 \ \sin75 \degree}{393.2}\\
\theta & =\sin^{-1} \left(\frac{250 \ \sin75 \degree}{393.2}\right)\\
\theta & = 37.89^{\circ}\\
\end{align*}

Solve for the unknown angle \phi .

\phi =360^{\circ} -45^{\circ} +37.89^{\circ} =353^{\circ}

The resultant force has a magnitude of 393 lb and is located 353º measured counterclockwise from the positive x-axis.

Advertisements
Advertisements

Computing the mass and weight of a man on earth and on the moon

If a man weighs 155 lb on earth, specify (a) his mass in slugs, (b) his mass in kilograms, and (c) his weight in newtons. If the man is on the moon, where the acceleration due to gravity is gm=5.30 ft/s², determine (d) his weight in pounds, and (e) his mass in kilograms.

Engineering Mechanics: Statics 13th Edition by RC Hibbeler, Problem 1-21
Engineering Mechanics: Statics 14th Edition by RC Hibbeler, Problem 1-20

Solution:

Part A

From the formula, \text{W}=\text{mg}, we can solve for the mass by dividing the weight by the acceleration due to gravity. That is

\begin{align*}
\text{m} & = \frac{\text{W}}{\textbf{g}}\\
& = \frac{155\ \text{lb}}{32.2 \ \text{ft/s}^2}\\
& = 4.81 \ \text{slug}\\
\end{align*}

Part B

Convert the slug to kilograms, knowing that 1 slug = 14.59 kg. 

\begin{align*}
\begin{align*}
\text{m} & = \left( \frac{155}{32.2} \text{slug}\right)\left( \frac{14.59 \ \text{kg}}{1 \ \text{kg}} \right)\\
& = 70.2 \ \text{kg}\\
\end{align*}
\end{align*}

Part C

Convert the 155 lb to newtons using 1 lb = 4.448 N. 

\begin{align*}
\textbf{W} & = 155 \ \text{lb}\times \frac{4.448 \ \text{N}}{1 \ \text{lb}}\\
& = 689 \ \text{N}\\
\end{align*}

Part D

Using the same formulas, but now \textbf{g}=5.30 \ \text{ft/s}^2.

\textbf{W}=155\left(\frac{5.30}{32.2}\right)=25.5\:\text{lb}

Part E

\textbf{m}=155\left(\frac{14.59\:\text{kg}}{32.2}\right)=70.2\:\text{kg}
Advertisements
Advertisements

The force of gravity acting between two particles

Two particles have a mass of 8 kg and 12 kg, respectively. If they are 800 mm apart, determine the force of gravity acting between them. Compare this result with the weight of each particle.

Engineering Mechanics: Statics 13th Edition by RC Hibbeler, Problem 1-20
Engineering Mechanics: Statics 14th Edition by RC Hibbeler, Problem 1-21

Solution:

The force of gravity acting between them:

\begin{align*}
\textbf{F} & =\textbf{G}\cdot \frac{\text{m}_1\text{m}_2}{\text{r}^2}\\
& =66.73\left(10^{-12}\right) \text{m}^3/ \left( \text{kg} \cdot \text{s}^2 \right)   \left[\frac{8 \  \text{kg} \left(12\ \text{kg}\right)}{\left(0.8\ \text{m} \right)^2}\right]\\
&=10\left(10^{-9}\right)\ \text{N}\\
& =10.0 \ \text{nN}\\
\end{align*}

The weight of the 8 kg particle

\textbf{W}_1=8\left(9.81\right)=78.5\:\text{N}

Weight of the 12 kg particle

\textbf{W}_2=12\left(9.81\right)=118\:\text{N}
Advertisements
Advertisements

Expressing the Density of Water in SI Units

Water has a density of 1.94 slug/ft³. What is the density expressed in SI units? Express the answer to three significant figures.

Engineering Mechanics: Statics 13th Edition by RC Hibbeler, Problem 1-19
Engineering Mechanics: Statics 14th Edition by RC Hibbeler, Problem 1-17

Solution:

\begin{align*}
\rho _w & =\left(\frac{1.94\:\text{slug}}{1\:\text{ft}^3}\right)\left(\frac{14.59\:\text{kg}}{1\:\text{slug}}\right)\left(\frac{1\:\text{ft}^3}{0.3048^3\:\text{m}^3}\right) \\
& =\left(\frac{1.94\:\text{slug}}{1\:\text{ft}^3}\right)\left(\frac{14.59\:\text{kg}}{1\:\text{slug}}\right)\left(\frac{1\:\text{ft}^3}{0.3048^3\:\text{m}^3}\right) \\
& =999.6\:\frac{\text{kg}}{\text{m}^3}\\
& =1.00\:\text{Mg/m}^3\\
\end{align*}
Advertisements
Advertisements

Showing How an Equation is Dimensionally Homogeneous

Using the SI system of units, show that Eq. 1–2 is a dimensionally homogeneous equation which gives F in newtons. Determine to three significant figures the gravitational force acting between two spheres that are touching each other. The mass of each sphere is 200 kg and the radius is 300 mm.

Engineering Mechanics: Statics 13th Edition by RC Hibbeler, Problem 1-18
Engineering Mechanics: Statics 14th Edition by RC Hibbeler, Problem 1-15

Solution:

To prove that F is in Newtons, we have

\begin{align*}
\text{F} & =\text{G}\cdot \frac{\text{m}_1\text{m}_2}{\text{r}^2}\\
& =\left(\frac{\text{m}^3}{\text{kg}\cdot \text{s}^2}\right)\left(\frac{\text{kg}\cdot \text{kg}}{\text{m}^2}\right)\\
& =\frac{\text{kg}\cdot \text{m}}{\text{s}^2}\\
& =\text{N}
\end{align*}

Now, if we substitute the given values into the equation

\begin{align*}
\text{F} & = 66.73\left(10^{-12}\right)\left[\frac{200\left(200\right)}{0.6^2}\right]\\
& = 7.41\left(10^{-6}\right) \text{N}\\
& =7.41\ \mu  \text{N}\\
\end{align*}
Advertisements
Advertisements

Problem 1-17| General Principles| Engineering Mechanics: Statics| RC Hibbeler

If an object has a mass of 40 slugs, determine its mass in kilograms.

Continue reading →

Problem 1-16| General Principles| Engineering Mechanics: Statics| RC Hibbeler

What is the weight in newtons of an object that has a mass of: (a) 10 kg, (b) 0.5 g, and (c) 4.50 Mg? Express the result to three significant figures. Use an appropriate prefix.

Continue reading →

Problem 1-15| General Principles| Engineering Mechanics: Statics| RC Hibbeler

Determine the mass of an object that has a weight of (a) 20 mN, (b) 150 kN, and (c) 60 MN. Express the answer to three significant figures.

Continue reading →