# Newton‘s 2nd Law: Example, 2 Dimensions Part 1

by Jared Rovny

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00:01 Now that we've done our one dimensional example, let's move to two dimensions and see how we solve a Newton's second law problem in two dimensions.

00:07 Here's a comparable example to what we just did with a slight difference.

00:13 Now, we are going to accelerate the same 1.000 kilograms spaceship but we're going to apply a two Newton's force, slightly more than before and again it will be starting from rest and floating in space, no gravity or anything to worry about.

00:24 But now we say that a leak exerts a force of one Newton at an angle of 30 degrees from the direction of motion where the ship was normally going in.

00:32 So we could ask a few questions, we could start with what is the space ship's velocity after one hour of this and then we can ask what is its speed as well which we will do next.

00:40 First, it's always important get an idea of what this problem looks like.

00:45 So we have our two Newton force, which is doing what the original one Newton force was doing in a simple one d problem except now we also have one Newton force acting up in an angle of 30 degrees relative to the original direction of motion.

00:58 We might expect is that the final velocity of our spaceship would be somewhere between the two that it would be going somewhat directly to the right and somewhat upwards leaving it in some new trajectory.

01:10 We're going to try to find some things about this trajectory.

01:13 The way to solve any Newton's second law of problem in two dimensions is first to choose a coordinate system and we've seen already many times that if I have vector pointing in some strange direction, the first thing I do is I find the components of the vector in a vertical direction and in a horizontal direction.

01:30 To do that, the first thing we have to do is pick what does vertical and what does horizontal mean.

01:35 In some problems we will pick a coordinate system that is slightly at an angle if we're dealing this problem having to do with slopes and in this problem it's a little more simple.

01:42 We'll have a horizontal x direction, and a vertical y direction as well.

01:46 The second thing you'll do now that you have your forces all figured out in two different directions, is to draw what we call a free body diagram which is just the way to account for all the different forces acting on your object.

01:58 Once you have a free body diagram showing all your forces acting in many different directions, and you know where your coordinate system is, you can do what we say breaking up all the vectors into the perpendicular components along the axis you've chosen.

02:11 So we'll do that using trigonometry in the same way is that we've seen already so far.

02:15 Then you'll write Newton's second law, for each axis, or each direction independently, so for example, we're writing Newton's second law for the x direction, and then we'll write it again for the y direction and just like we saw with the projectile motion problem, we'll treat the x and y directions independently of each other.

02:31 Then finally, once you have Newton's second law written down for each axis, you can solve for the acceleration. Once you have the acceleration, you're ready to put the equations of motion into using that acceleration that you found.

02:44 So let's do that with this problem, we are gonna do our first step and introduce our axis so this might be the easiest step you do.

02:52 You just have a positive axis which I'm calling to the right and in a positive y axis which I'm calling a vertically upwards.

02:58 What we have in this problem as I've already shown is these forces so this would be something like a free body diagram where we have force acting purely horizontally and then we have this new force acting at an angle which we're a calling theta, an angle which was 30 degrees in our problem acting in both the x and y directions.

03:14 Again what we want to do when we have an angle, a force pointing at an angle, is to break that force up to its two components so we have a force acting at an angle that we don't like so we find the horizontal component, and the vertical component.

03:28 Again using the same trigonometry, that we introduced when talking about vectors in the equation of motion, so we know that the vertical component of our force will be the force times the sine of theta.

03:39 We know the horizontal component of the force will be the force times the cosine of theta.

03:44 Once we've done that, instead of having one force, at an angle, we'll now have two separately considered forces, one component in a vertical direction and one component in a horizontal direction.

03:55 One last thing to say about this which can be confusing sometimes, is that we did say that vectors have a magnitude and a direction, but one thing we never said is that they have a position.

04:06 They do not have a location in space that they have to stay at, so this vector that was on far right here, we can move it to the left to sort of consider it in our axis, and that doesn't change the physical meaning because the physical meaning is of course, a force acting on an object.

04:19 The only reason we draw in the triangle the way we did, was to find the vertical and the horizontal components, the length of this different sides of the triangle, once we have those we don't need to consider the arrow showing this force to be acting on the right or anything because we know where the force is acting, the force is acting on our body, what we are considering our mass, on the spaceship in this case.

04:39 So now we have our final free body diagram, with everything broken up into its different horizontal and vertical components.

04:46 You see on the horizontal axis, we have the original force, which is not being broken up into components because it was already purely in the horizontal direction, but now we have a new force which is partly vertical which is why we have F times the sine of theta and we also have a somewhat horizontal aspect which is F times the cosine of theta.

05:04 So now we can treat this as two different one dimensional problems one in the x direction just horizontally with two forces, and one just in the y direction with a vertical force acting on it.

05:14 We'll do as I have said, we'll just write down Newton's second law.

05:18 And we'll do it for each one independently.

05:20 If we wrote down Newton's second law as a vector that's what these bold letters mean, we could consider the vectors separately so we have force equals mass times acceleration, where force and acceleration are vectors.

05:33 They have different components in different directions.

05:35 Notice that mass is not a vector, mass is just some number telling you how heavy your object is.

05:40 Mass doesn't have different directions.

05:42 You don't have a different mass in one direction than another.

05:44 If we write these forces all the way out so we can write the x and the y components, we can see that Fx, Fy, the two components of your force, equal mass times the two components of your acceleration.

05:57 And so when you read it like this, what we can see is that we can write the x component of your equation and the y component of your equation separately from each other.

06:05 Now, it look something like this.

06:07 In the x direction we have the force equals mass times acceleration, so we're simply writing Newton's second law only in the x direction, the x direction forces equals the mass of the object, times the acceleration, in the x direction.

06:20 And we can write up all the forces that are acting in the x direction, on the left hand side of the equation, which is your Fa and Fb cosine of theta, and that's going to be equal to the mass times the acceleration in just the x direction.

06:33 In our other one dimensional direction, our y direction we can do the exact same thing, and right on Newton's second law just in the y direction, so that the force in the y direction is equal to the mass times the acceleration in the y direction.

06:43 Writing that out with our y direction force being Fb times the sine of theta.

06:48 We have Fb times the sine of theta equals mass times acceleration in the y direction only. Now this is something we can solve quite quickly.

06:56 All we have to do to find the accelerations from these two equations is to divide each one by the mass, because on the right hand side of each equation, we already have the mass times acceleration, so we just need to divide it by the mass.

07:07 And we can see below we've already solved for our accelerations in the x and the y directions.

### About the Lecture

The lecture Newton‘s 2nd Law: Example, 2 Dimensions Part 1 by Jared Rovny is from the course Force.

### Included Quiz Questions

1. aₓ = 0.005 m/s² , aᵧ = 0.009 m/s²
2. aₓ = 0.05 m/s² , aᵧ = 0.09 m/s²
3. aₓ = 0.002 m/s² , aᵧ = 0.006 m/s²
4. aₓ = 0.001 m/s² , aᵧ = 0.002 m/s²
5. aₓ = 0.005 m/s² , aᵧ = 0.01 m/s²

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By Janine T. on 30. October 2020 for Newton‘s 2nd Law: Example, 2 Dimensions Part 1

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