·In each of the cases that follow, the magnitude of a vector is given along with the countercloc

step 1 This is chapter 1, problem 40. In this problem, for each part of the problem, we're given the ma

·In each of the cases that follow, the magnitude of a vector...

Key Concepts

Angle Measurement in Standard Position

Understanding angle measurement in standard position is crucial, as the angle is typically measured counterclockwise from the positive x-axis. This standard reference ensures consistency when applying trigonometric functions to resolve the vector into its components, and it is particularly important when dealing with vectors in different quadrants.

Scale Drawing of Vectors

Scale drawings of vectors provide a visual check of computed results by approximating the vector’s direction and magnitude on graph paper. This graphical representation helps verify that the calculated components and overall vector orientation are reasonable, enhancing the understanding of vector behavior in a physical context.

Polar to Cartesian Conversion

Polar to Cartesian conversion is the method of converting a vector expressed in magnitude and angle (polar form) to its x and y components (Cartesian form). This transformation is important not only in physics but also in mathematics and engineering to facilitate calculations involving linear operations.

Trigonometric Decomposition

Trigonometric decomposition involves using sine and cosine functions to determine the components of a vector based on its magnitude and the angle it makes with a reference axis. In this process, the cosine of the angle is associated with the x-component and the sine with the y-component, enabling the conversion from polar (magnitude and angle) to rectangular (x and y) coordinates.

Vector Components

The process of breaking a vector into its x (horizontal) and y (vertical) components is critical for analyzing its effect along the coordinate axes. This decomposition allows for the separation of a vector's influence in independent directions, making it easier to apply mathematical operations and solve problems related to motion and force in two dimensions.

Vectors

Vectors are quantities characterized by both magnitude and direction. They are fundamental in physics and engineering, representing concepts such as force, velocity, and displacement. Understanding vectors involves knowing how these two properties interact, which is essential for further manipulation and application in various problems.

Transcript

00:01 This is chapter 1, problem 40.

00:04 In this problem, for each part of the problem, we're given the magnitude and direction of a vector, and we are asked to find the x and y components of the vector.

00:15 And then we're also asked to sketch the vector and see if our answers seem reasonable, which, by the way, seeing if your answer seem reasonable is something that you should always do in any problem.

00:27 So for vectors, when we're given magnitude and direction, you can find the x and y components using the equations in the top left of the screen here.

00:35 The x component, r sub x, is equal to r, the vector magnitude times cosine of theta, where theta is the angle measured counterclockwise from the x -axis in the direction of the red arrow.

00:47 And the y component is magnitude times sine of theta.

00:53 So in each case, luckily, the angle that we're given is measured counterclockwise from the x -axis.

00:58 So we can just use the angle as is, except that some of the angles are given in radiance.

01:03 So we may want to convert them to degrees.

01:07 So let's get started.

01:10 Vector a is 50 newtons at 60 degrees.

01:15 So measuring 60 degrees in the direction of the red arrow, that's going to be about here.

01:23 If this vector length is 50 newtons, then my x component is going to be this long on the x -axis, so i expect it to be about half of 50 newtons.

01:37 And my y component going up here should be larger than the x component, and close to 50, close, larger than the x component, less than 50 newtons still.

01:51 So let's see what we get.

01:53 Our x here is going to be our magnitude 50 newtons times the cosine of 60 degrees.

02:16 And you do the math and you get 25 newtons.

02:19 That looks reasonable.

02:26 Y component, we're going to use same magnitude, same angle, but now we've got a sign in there.

02:41 And if you calculate it out, you should get 43 .3 newtons.

02:47 And both these components are positive, which makes sense because this is in the first quadrant.

02:55 So that's good for part a.

02:57 Let's move on.

02:58 Part b, we've got a vector of 75 meters per second.

03:02 So in our magnitude is in totally different units.

03:05 I'm still going to draw it on these same axes, but we shouldn't actually compare the magnitude of vector b with the magnitude of vector a.

03:14 They're unrelated.

03:16 What is related is the magnitude of vector b and then the lengths of vector b's x and y components.

03:23 That's something we can look at in our diagram.

03:26 So 75 meters per second at 5 pi over 6 radians.

03:30 To draw that, let's first convert 5 pi over 6 radians to degrees.

03:37 And the way we do that is you take the value in radians, multiply by 180 degrees over pi, and we get 150 degrees.

04:02 So it measured counterclockwise from the x -axis.

04:06 That's going to take us over here, 30 degrees away from the negative x -axis.

04:16 And our magnitude is this 75 meters per second.

04:25 So to actually confine the components, we again use these equations.

04:28 You've got 75 meters per second times cosine of 150 degrees and then times sign of 150 degrees.

04:51 And if you're doing this on a calculator and things aren't coming out right.

04:54 If your calculator has degree mode and radian mode, this would be a good time to check that your calculator is in the mode you expect.

05:01 So it should be in degree mode here, unless you chose not to convert from radiance to degrees, and then it should be in radian mode.

05:08 In any case, plug this in, and you end up with negative 65 meters per second.

05:23 And that seems reasonable.

05:24 We expect the x component to be negative, because it's pointing back to the left, and we expect it to be pretty large, almost as large as the vector magnitude, but still less than it...

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