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UExcel Physics: Study Guide & Test Prep17 chapters | 188 lessons

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Lesson Transcript

Instructor:
*David Wood*

David has taught Honors Physics, AP Physics, IB Physics and general science courses. He has a Masters in Education, and a Bachelors in Physics.

After watching this lesson, you will understand how we analyze physics involving two-dimensional vectors and be able to split a vector at an angle into a vertical and horizontal component. A short quiz will follow.

In physics, we love to keep things simple. But not everything goes in nice, neat straight lines. When you shoot a cannonball at a pirate ship, that pirate ship moves in at least two dimensions -- the *x*-direction and the *y*-direction. But so many of our physics equations only work in one dimension. So what do we do?

We break vectors into *x* and *y* components. We look at the motion in the *x*-direction and the motion in the *y*-direction separately.

A **vector** is a quantity that has both magnitude (numerical size) and direction. So for example, 20 meters per second is not a vector and is called speed, but 20 meters per second south is a vector and is called velocity. If that vector is pointed diagonally, to analyze the motion, we need to break it into vertical (*y*) and horizontal (*x*) components.

If your cannonball is shot at a shallow angle, close to the ground, you're going to have a bigger *x*-component and a smaller *y*-component. If your cannonball is shot almost upwards, you're going to have a bigger *y*-component and a smaller *x*-component. And we can put numbers to these things and treat the motion in the two dimensions (or three dimensions for that matter) separately. This is incredibly helpful for solving problems.

Okay, so how do we do it? We use a bit of geometry. The word that you may have heard before is: SOHCAHTOA. This helps us find the sides of a right-angled triangle when we know one of the sides and we know an angle. The SOH part reminds us that sine of the angle is equal to the opposite side of the triangle divided by the hypotenuse side of the triangle. The CAH part tells us that the cosine of the angle is equal to the adjacent side of the triangle divided by the hypotenuse. And the TOA part says that the tangent of the angle is equal to the opposite side of the triangle divided by the adjacent.

As you can see from this graphic, the opposite side is the side directly opposite from the angle that you're given. The hypotenuse is the longest side, the diagonal, which is always directly across from the right angle. And the adjacent side is the side that is adjacent to, or next to, both the right angle and the angle you're given.

Okay, but this is physics, not math. So why do we need to know that?

Well, if you're shooting something up at an angle, or have any kind of physics vector, like a force vector, electric field vector, or whatever it is. If you draw it to scale, you can create a right-angled triangle from that vector. One side of the triangle should point directly in the *x*-direction -- perfectly horizontal. The other side should point directly in the *y*-direction -- vertical. And if you know the angle it's being shot at, you can use this SOHCAHTOA geometry to find the *x* and *y* components of the velocity, or force, or electric field, or whatever it is.

So for example, with this cannonball being fired at an angle, we're firing it at a velocity *v*, that's the hypotenuse, and that has an *x*-component, *vx*, and a *y*-component, *vy*. We can write a couple of equations using SOHCAHTOA. Or really just SOHCAH, because we only use sine and cosine for breaking into components.

We can write an equation using sine. Sine of the angle is equal to the opposite side, which is *vy*, divided by the hypotenuse, which is *v*. And we can also write an equation using cosine. Cosine of the angle is equal to the adjacent side, which is *vx*, divided by the hypotenuse, which is *v*. Rearrange these equations, and we get expressions for the *x* and *y* components of the velocity, *v*.

So that's how we find our two components. But maybe this would be easier if we looked at an example problem.

Let's say you're pushing a table across the floor, but you're pushing down and at an angle. That means you're pushing partly down and partly to the side -- partly in the *x*-direction and partly in the *y*-direction. You're pushing at an angle of 30 degrees to the horizontal and with a force of 200 newtons. And you're asked to figure out how much you're pushing down and how much you're pushing sideways -- or in other words, you're asked to break that force into components.

We can turn this pushing force into a right-angled triangle, just like before, and find the other two sides of that triangle -- the *x*-component, which we'll call *Fx* (force-*x*), and the *y*-component, which we'll fall *Fy*. You could do SOHCAHTOA again just to check, but this one actually turns out exactly the same as the cannonball example. *Fx* is equal to *F* multiplied by cosine of the angle, and *Fy* is equal to *F* multiplied by sine of the angle. It very often turns out to be this way around, but do be careful -- if you knew the other angle in the triangle instead, or if you're working on a problem on a slope, the sine and cosine might reverse. So it's always good to check your geometry.

Anyway, we have our equations. Now let's write down what we know. We know that the angle, *theta*, is 30 degrees, and we know that the force *F* is 200 newtons. So we can plug those numbers into the two equations and solve.

The *x*-component, *Fx*, equals 200 cosine 30, which gives a force of 173 newtons. The *y*-component, *Fy*, equals 200 sine 30, which gives a force of 100 newtons. So there we have it. If you push down on a table at an angle of 30 degrees to the horizontal, with a force of 200 newtons, you're pushing sideways with a 173 newton force and down with a 100 newton force. Those are our two components.

Many things in real life happen in two (or even three) dimensions. The path of a cannonball is both up and sideways, for example. We can analyze this motion by breaking the velocity vector into *x* and *y* components. We can do the same thing with forces applied diagonally, or electric field vectors, or really anything that's a vector quantity and is acting diagonally. We look at the motion in the *x*-direction and the motion in the *y*-direction separately.

If your cannonball is shot at a shallow angle, close to the ground, you're going to have a bigger *x*-component and a smaller *y*-component. If your cannonball is shot almost upwards, you're going to have a bigger *y*-component and a smaller *x*-component. We can figure out the sizes of those components using the geometry of a right angle. If we draw a vector to represent the velocity or force, or whatever it is we're splitting into components, we can then make a right-angled triangle and use SOHCAHTOA to figure out the sizes of the other two sides of the triangle. This will give us our *x* and *y* components, which can be incredibly useful in understanding physical phenomena.

Make it your mission to achieve these goals when the lesson concludes:

- Remember what a vector is
- Determine the components of a vector

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UExcel Physics: Study Guide & Test Prep17 chapters | 188 lessons

- What Is a Vector? - Definition & Types 5:10
- Vector Addition (Geometric Approach): Explanation & Examples 4:32
- Resultants of Vectors: Definition & Calculation 6:35
- Scalar Multiplication of Vectors: Definition & Calculations 6:27
- Vector Subtraction (Geometric): Formula & Examples 5:48
- Standard Basis Vectors: Definition & Examples 5:48
- How to Do Vector Operations Using Components 6:29
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- Vector Resolution: Definition & Practice Problems 5:36
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