Vectors in Physics

These are my notes and thoughts on vectors in physics.

Quantities like mass and temperature are described with a number called a scalar. Other quantities like displacement, velocity, or force have a direction associated with them and are called vectors. When looking at a vector on paper, a vector is an arrow with the length of the arrow representing the number and direction of the vector. 

A vector can be described with a number and an angle, like \( A=19 \angle 25 \deg \). When doing basic math operations, you should use the component form to write vectors. So, if a line were placed with the tail of the vector at the origin of the coordinate system, x and y portions can be calculated like this:

\( y = 19 \sin 25^{\circ} = 8.02 \)

\( x = 19 \cos 25^{\circ} =  17.21 \)

This is how you calculate vectors. 

If you have the x and y vectors, you can calculate the length of the corresponding line by:

\( \sqrt{x^{2} + y{2}} \)

\( \sqrt{8.02^{2} + 17.21^{2}} \)

\( \sqrt{64.32 + 296.18} \)

\( \sqrt{360.5} = 19  \)

To find the original angle we use:

\( \theta = \tan^{-1} \frac{y}{x} \)

So:

\( \theta = \tan^{-1} \frac{8.02}{17.21} \)

\( \theta = 25^{\circ} \)

Example 1

Write out the components of the vector \( c = 47 \angle 193^{\circ} \)

The x component is:

\( x = 47 \cos 193^{\circ} = -45.8 \)

The y component is:

\( y = 47 \sin 193^{\circ} = -10.6 \)

The use of unit vectors simplifies the mathematical operations on vectors. In two dimensions, unit vectors are vectors of unit value and in the +x and +y directions. 

So the first vector would be written as \( 18.4i + 13.8j \) and the second vector is 

\( -45.8i - 10.6j \).

You can add vectors by adding the individual components. If you have:

\( (18.4i + 13.8j) + (-45.8i - 10.6j) \)

You just add the individual components together. The combined vector becomes:

\( -27.4i + 3.2j \)

Subtracting vectors can also be done. It is a bit quirky, however. You will use the opposite sign of the second vector.

\( (18.4i + 13.8j) - (-45.8i - 10.6j) \)

This gives the result of:

\( 64.2i + 24.4j \)

We should also know how to work backward when given the components. When we do this properly it will give us the length of the vector and its angle.

Let us use the last vector:

\( 64.2i + 24.4j \)

\( V = \sqrt{64.2^{2} + 24.4^{2}} = 68.7 \)

\( \theta = \tan^{-1}\frac{24.4}{64.2} = 20.8^{\circ} \)

When put in vector form:

\( V = 68.7 \angle 20.8^{\circ} \)

Example

Add the vectors: 

\(A=13 \angle 50^{\circ}\)

\(B=15 \angle -60^{\circ} \)

\(C=17 \angle 20^{\circ} \)

The first step is to put each vector in its component form.

\( A = \sin 50 * 13 = 10 = y \)

\( A = \cos 50 * 13 = 8.4 = x \)

\( A = 8.4i + 10.0j \)

\( B = \sin(-60) * 15 = -13 = y \)

\( B = \cos(-60) * 15 = 7.5 = x \)

\( B = 7.5i -13.0j \)

\( C = \sin 120 * 17 = -8.5 = y \)

\( C = \cos 120 * 17 = -14.7 = x \)

\( C = -14.7i -8.5 j \)

Sum all the i’s and all the j’s. You get:

\( 1.2i -11.5j \)

That is your x and y dimensions

Now use the Pythagorean theorem with those values

\( \sqrt{11.5^{2} + 1.2^{2}} = 11.6 \)

That gives you the length of the new vector

Now, let’s find the new angle

We use the inverse tangent to do this

\( \tan^{-1} \frac{-11.5}{1.2} = -84^{\circ} \)

So the new vector is:

\( 11.6 \angle -84^{\circ} \)

Use as many diagrams as possible when doing problems like this. Diagrams help keep your thought processes straight and help prevent silly errors. 

Example

Find the resultant of the two forces \(800(N) \angle 47^{\circ}\) and \(600(N) \angle 140^{\circ}\)

I will write out all the calculation steps first, before I actually compute them

\(800 \cos 47^{\circ} = 545.6 \)

\(800 \sin 47^{\circ} = 585.1 \)

\(600 \cos 140^{\circ} = -460.0 \)

\(600 \sin 140^{\circ} = 385.7 \)

Vector 1 is: \(545.6i + 585.1j\)

Vector 2 is: \(-460.0i + 385.7j\)

Add the vectors together

We get:

\(85.6i + 970.8j\)

Use Pythagorean theorem

\(\sqrt{85.6^{2} + 970.8^{2}}\)

The length of our new vector is:

\(= 974.6 \)

The new angle of our vector is:

\(\tan^{-1}\frac{970.8}{85.6} = 85^{\circ} \)

So:

\(974.6(N) \angle 85^{\circ}\)

Scalars

The dot product of two vectors produces a scalar. Scalars, as you know, do not have a direction associated with them. There are two ways to compute a dot product.

\(A*B = AB \cos \theta\)

And 

\(A*B = a_xb_x + a_yb_y\)

Example

Form the dot product of \(A = 23 \angle 37^{\circ}\) and \(B = 14 \angle -35^{\circ}\)

Use the first definition

\(A*B = A*B \cos \theta = 23*14 \cos 72^{\circ} = 100\)

Now use the second definition

This requires the components of both vectors

We learned how to do this earlier, it is the same process

We get:

\(18.4i + 13.8j\) and \(11.5i + (-8.0)j\)

\(A*B=100\)

Cross Product

The cross product of A*B also produces a vector. 

\(A*B \sin \theta\)

Example

Form the cross product of \(A = 23 \angle 37^{\circ}\) and \(B = 47 \angle 193^{\circ}\)

This is A*B* sine of the difference in the angles.

The difference in angles is 156 degrees.

\(A*B = 23*47*\sin 156^{\circ} = 440\)

We can also use determinants to get the same result

Calculate the components of each vector:

Cross multiply. This is why it is called the cross product

\(A*B = 18.4(-10.6) - 13.8(-45.8) = 440\)

If it is a simple problem then use the first definition any time you can. Otherwise, use the second definition.