$19.5 \textrm{ m, } 4.65^\circ \textrm{ S of W}$

### Solution video

# OpenStax College Physics, Chapter 3, Problem 5 (Problems & Exercises)

### Calculator Screenshots

*x*component of the resulting vector will be the sum of the

*x*components of the other vectors,

*Ax*and

*Bx*. So to calculate them, notice first of all that there is a bit of a tricky thing here that we use sine on the one hand and then we use cosine on the other. So you can't just follow a recipe where you memorize that the

*x*component is always cosine and the

*y*component is always sine. That's not necessarily true. It often is, but in this particular question, though the way they've drawn it and the angles that they've given you, it turns out that the

*x*component of the

*A vector*is using sine because it's the opposite leg of this triangle here. So this is the triangle that makes up

*vector A*and its opposite leg is opposite the angle given and so to find the opposite leg of the triangle we use the trigonometric function sine, multiplied by the hypotenuse. Then we're finding the adjacent here which will be the

*y*component of

*A*, we're going to end up using cosine. Okay. So we go 12 meters times sine twenty and we put a negative in front of it because we can see that it is directed to the left. We'll assume that to the right is positive, and so we have a minus in front of that. Then likewise, minus 20 meters magnitude of

*vector B*, multiplied by cosine of 40 because the

*x*component of

*vector B*is the adjacent leg of this triangle, and this gives us negative 19.43 meters as the

*x*component of the resultant. Now, for the y component we have 12 meter magnitude of

*vector A*, times by the cosine of 20 to get the adjacent leg of this triangle, that's positive and so we did not put a minus in front of -- we did not put a minus in front of this term. But we do put a minus in front of the

*y*component of

*vector B*because it is directed downwards. Here is the

*y*component of

*vector B*directed down here. So we have a minus 20 meters times sine 40, and that gives negative 1.579 meters and that's the resultant

*y*component. So if we were to draw the resultant

*y*component it would be like this. What I mean to say is I draw the components of the resulting vector it would look like this, this being the

*x*component of the resultant, negative 19.43, and then the

*y*component directed down, negative 1.579, and this is the resultant here. Okay. So now that we know the legs of this triangle, we can now calculate the length of the hypotenuse which is the magnitude of the resultant, and then we can also find the angle in here. So the size of the resultant is the square root of the sum of the squares of the components, so that's the square root of negative 19.43 meters squared, plus negative 1.579 meters squared, giving us 19.5 meters is the length of the resultant. Then for the angle, we take the inverse tangent of its

*y*component divided by its

*x*component and there is little vertical bars here to say absolute value or to say ignore the negative sign. We're just going to take the magnitude of these components. Then the direction we'll figure out by looking at the picture. We don't need positives or negatives to tell us direction. So, we have 4.65 degrees because that's a little bit in here, this is 4.65 degrees towards the south compared to west. We're south of west. So our resultant is 19.5 meters, 4.65 degrees south of west.

## Comments

why do you do Rx= Ax+Bx and same with y? and not just do R= square root of A^2+B^2?

Hello, that's an excellent question, since there's a very good and important answer: The vectors $\vec{A}$ and $\vec{B}$ are **not perpendicular**. The Pythagoras Theorem you mentioned works only when the lines are perpendicular. When resolving both vectors into *x* and *y* components, then adding corresponding components together, we are finding the components of the resultant. These total components of the resultant **are perpendicular**, which is why the Pythagoras Theorem works in that case.

I keep putting this exactly the same in my TI-84 and im not getting the same answer...

Hi kaileyeddy, some suggestions would be to check that your calculator is in "degree mode" (this showed up on a Google search for how to do that: https://www.youtube.com/watch?v=tAqk7g9oXFY) and also double check brackets.

Good luck!

How do you know when to use SIN vs. COS in the Ax, Bx, Ay, By calculations?

Disregard: I watched further and saw you explained it.....whoops!!

I am confused as to why we you use sin on the x axis here. Is there a place in the book where I can look at more examples of when this is used?

Hi fabricio, thanks for the question. I can understand that it's less common to see sine used for finding an "x" component. This is because problems are traditionally set up with the angle given with respect to the horizontal, in which case the *vertical* component is the opposite leg of the triangle. Recall that sine is used to find the **opposite** leg of the triangle. In this problem, vector A has its angle given with respect to the vertical, which makes the opposite leg *horizontal*. Sine is used to find opposite legs, so sine is used to find the horizontal component for vector A.

Hope this helps,

Shaun