applications

New Release for Kindle: Math Shorts – Integrals

Yesterday I released the second part of my “Math Shorts” series. This time it’s all about integrals. Integrals are among the most useful and fascinating mathematical concepts ever conceived. The ebook is a practical introduction for all those who don’t want to miss out. In it you’ll find down-to-earth explanations, detailed examples and interesting applications. Check out the sample (see link to product page) a taste of the action.

Important note: to enjoy the book, you need solid prior knowledge in algebra and calculus. This means in particular being able to solve all kinds of equations, finding and interpreting derivatives as well as understanding the notation associated with these topics.

Click the cover to open the product page:

cover

Here’s the TOC:

Section 1: The Big Picture
-Anti-Derivatives
-Integrals
-Applications

Section 2: Basic Anti-Derivatives and Integrals
-Power Functions
-Sums of Functions
-Examples of Definite Integrals
-Exponential Functions
-Trigonometric Functions
-Putting it all Together

Section 3: Applications
-Area – Basics
-Area – Numerical Example
-Area – Parabolic Gate
-Area – To Infinity and Beyond
-Volume – Basics
-Volume – Numerical Example
-Volume – Gabriel’s Horn
-Average Value
-Kinematics

Section 4: Advanced Integration Techniques
-Substitution – Basics
-Substitution – Indefinite Integrals
-Substitution – Definite Integrals
-Integration by Parts – Basics
-Integration by Parts – Indefinite Integrals
-Integration by Parts – Definite Integrals

Section 5: Appendix
-Formulas To Know By Heart
-Greek Alphabet
-Copyright and Disclaimer
-Request to the Reader

Enjoy!

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What is Torque? – A Short and Simple Explanation

Often times when doing physics we simply say “a force is acting on a body” without specifying which point of the body it is acting on. This is basically point-mass physics. We ignore the fact that the object has a complex three-dimensional shape and assume it to be a single point having a certain mass. Sometimes this is sufficient, other times we need to go beyond that. And this is where the concept of torque comes in.

Let’s define what is meant by torque. Assume a force F (in N) is acting on a body at a distance r (in m) from the axis of rotation. This distance is called the lever arm. Take a look at the image below for an example of such a set up.

Great Formulas II_html_m7f65e4ec

(Taken from sdsu-physics.org)

Relevant for the rotation of the body is only the force component perpendicular to the lever arm, which we will denote by F’. If given the angle Φ between the force and the lever arm (as shown in the image), we can easily compute the relevant force component by:

F’ = F · sin(Φ)

For example, if the total force is F = 50 N and it acts at an angle of Φ = 45° to the lever arm, only the the component F’ = 50 N · sin(45°) ≈ 35 N will work to rotate the body. So you can see that sometimes it makes sense to break a force down into its components. But this shouldn’t be cause for any worries, with the above formula it can be done quickly and painlessly.

With this out of the way, we can define what torque is in one simple sentence: Torque T (in Nm) is the product of the lever arm r and the force F’ acting perpendicular to it. In form of an equation the definition looks like this:

T = r · F’

In quantitative terms we can interpret torque as a measure of rotational push. If there’s a force acting at a large distance from the axis of rotation, the rotational push will be strong. However, if one and the same force is acting very close to said axis, we will see hardly any rotation. So when it comes to rotation, force is just one part of the picture. We also need to take into consideration where the force is applied.

Let’s compute a few values before going to the extremely useful law of the lever.

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We’ll have a look at the wrench from the image. Suppose the wrench is r = 0.2 m long. What’s the resulting torque when applying a force of F = 80 N at an angle of Φ = 70° relative to the lever arm?

To answer the question, we first need to find the component of the force perpendicular to the lever arm.

F’ = 80 N · sin(70°) ≈ 75.18 N

Now onto the torque:

T = 0.2 m · 75.18 N ≈ 15.04 Nm

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If this amount of torque is not sufficient to turn the nut, how could we increase that? Well, we could increase the force F and at the same time make sure that it is applied at a 90° angle to the wrench. Let’s assume that as a measure of last resort, you apply the force by standing on the wrench. Then the force perpendicular to the lever arm is just your gravitational pull:

F’ = F = m · g

Assuming a mass of m = 75 kg, we get:

F’ = 75 kg · 9.81 m/s² = 735.75 N

With this not very elegant, but certainly effective technique, we are able to increase the torque to:

T = 0.2 m · 735.75 N = 147.15 Nm

That should do the trick. If it doesn’t, there’s still one option left and that is using a longer wrench. With a longer wrench you can apply the force at a greater distance to the axis of rotation. And with r increased, the torque T is increased by the same factor.

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This was an excerpt from my Kindle ebook More Great Formulas Explained.

Check out my BEST OF for more interesting physics articles.