Vectors and angles

While this section has been an introduction to vectors and slightly theoretical primer you should move on to the next chapter Vectors and Angles to see how useful vectors are when they are combined with angles.

When you are solving a physics problem, especially a realistic one, you will be given angles. In the last section we have learned that vectors look like this: 574m/s [E66°N] where 66° is the angle going from the horizontal East direction towards the vertical North direction.

Most of the time you will have to simplify complicated physics situation into several simple triangles, representing the addition of several magnitudes and the sum.

Vectors and angles -

In a setup such as this you will be able to use angles that are given to you not only to find a magnitude but also its direction. To find missing vector quantities you need to be proficient with trigonometry. Lets review some trigonometric formulae, where Я is an angle in a triangle:

Right triangles

· sin Я = opposite / hypotenuse

· cos Я = adjacent / hypotenuse

· tan Я = opposite / adjacent

Other triangles

· aІ = bІ Ч cІ – 2ac cos( A );

· a/sin A = b/sin B = c/sin C

Lets plunge into an example invloving vectors and angles

The Thompsons drive to their friends’ lodge but they do not take a straight path. They travel at a velocity of 70km/h [W20°N] for and hour and then they stop and travel at a velocity of 80km/h [N20°E] for half an hour. Using the data given find the displacement of the lodge in relation to the Thompsons’ house.Since we are to find the displacement of the lodge ( how far is it, and what angle it is from the house when travelling there on a straight path ) we convert the velocities and times given into displacements.

Vectors and angles - 1 = 70km/h [W20°N] Ч 1h

Vectors and angles - 1 = 70km [W20°N]

Vectors and angles - 2 = 80km/h [N20°E] Ч 1/2 h

Vectors and angles - 2 = 40km [N20°E]

Vectors and angles - = Vectors and angles - 1 + Vectors and angles - 2

Our resultant displacement is the sum of the two other displacements. Lets sketch the situation so we can better understand the angle at which the lodge is to the Thompsons’ house.

Vectors and angles -

The black vectors are the two displacements through which the Thompsons went to get to the lodge, the blue vector is the shorter, direct path to the lodge (the resultant vector Vectors and angles - ). When looking at the angles in relation to the directions (the “cross”) we can deduct that the angle between the two displacements is 90°.

Vectors and angles -

This is convenient for us, as we can simply use the Pythagorean theorem to get the magnitude of Vectors and angles - , the direct distance to the lodge.

| Vectors and angles - |І = | Vectors and angles - 1|І + | Vectors and angles - 2|І

| Vectors and angles - |І = 70І + 40І

| Vectors and angles - |І = 6500

| Vectors and angles - | = 80.622 km

The direct distance to the lodge is 81 km (significant digits). Since we have used the absolute value, | |, of the resultant displacement we don’t have an angle yet (we need the angle to know exactly in which direction to drive). Since we know the magnitude of all three sides of our “triangle” we can now use a trigonometric ratio to find any angle. The angle we are interested in is the angle between the first displacement and the resultant displacement, when we add 20° to it we will be able to say that the direction is [West 20° + ?° N].

Let Я be the angle between Vectors and angles - 1 and Vectors and angles -
Cos Я = Vectors and angles - 1 / Vectors and angles -

Cos Я = 70 / 81

Cos Я = 0.8641

Я = 30°

The angle of the displacement is 20° + 30° = 50°
We can now state the displacement of the lodge from the Thompson’s house as: 81km [W50°N].

You can also state the direction as going from north to west, [N40°W].

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