Calculating Vector Components: Finding B_y with A-B Parallel to -x Axis

[jamesbrewer]

Homework Statement

The diagram below shows two vectors, A and B, and their angles relative to the coordinate axes as indicated.

[URL]http://loncapa.physics.mun.ca/res/mun/PHYSICS/msuphysicslib/Graphics/Gtype07/prob01a_vectors2.gif[/URL]

DATA: α=43.7°; β=53.4°; |A|=4.30 cm. The vector A−B is parallel to the −x axis. Calculate the y-component of vector B.

Homework Equations

A_x = |A| cos \theta, A_y = |A| sin \theta

The Attempt at a Solution

I easily found that A_x = 4.30 cm cos(43.7) = 3.10 cm and A_y = 4.30 cm sin(43.7) = 2.97 cm.

The problem I'm having is figuring out how to calculate B_x with only the information I have. If I had |B| then I could easily solve for B_x. The problem states that \vec{A-B} || -x-axis, but what role does this play in solving the problem?

 

[jncarter]

Try drawing a picture. It can be very difficult to sort this sort of thing without a good reference.

The key to this problem is the statement that \vec{A} - \vec{B} is // to the -x axis. What does it mean if a vector is parallel to an axis? Also, how do you subtract two vectors? If you can answer these two questions, you should be able to figure out B_y.

 

[jamesbrewer]

A vector being parallel to an axis can tell you something about it's direction. In this case, the vector is parallel to the -x axis. Is the fact that it's parallel to the negative x-axis (as opposed to the positive x axis) important here? If so then that means that \theta_{A-B} = 180 deg.

When you subtract a vector \vec{A} from another vector \vec{B}, you are essentially adding the negative of \vec{B} to \vec{A}.

So, from this I gather that \theta_{A-B} = 180 deg, but I'm still having trouble seeing how that helps me find B_y. To find B_y, I need to find the magnitude of B, don't I?

 

[jamesbrewer]

A vector being parallel to an axis can tell you something about it's direction. In this case, the vector is parallel to the -x axis. Is the fact that it's parallel to the negative x-axis (as opposed to the positive x axis) important here? If so then that means that \theta_{A-B} = 180 deg.

When you subtract a vector \vec{A} from another vector \vec{B}, you are essentially adding the negative of \vec{B} to \vec{A}.

So, from this I gather that \theta_{A-B} = 180 deg, but I'm still having trouble seeing how that helps me find B_y. To find B_y, I need to find the magnitude of B, don't I?

 

[jncarter]

A vector being parallel to an axis can tell you something about it's direction. In this case, the vector is parallel to the -x axis. Is the fact that it's parallel to the negative x-axis (as opposed to the positive x axis) important here? If so then that means that \theta_{A-B} = 180 deg.

Technically, the positive x-axis and neg x-axis would both be parallel. The fact that the problem specifies the negative is suggestive, but I would ask the professor if I were you. There's a more important point here though, and it has to do with how you subtract vectors.

When you subtract a vector \vec{A} from another vector \vec{B}, you are essentially adding the negative of \vec{B} to \vec{A}.

Remember that you have to perform algebraic operations on the components of vectors. \vec{A} - \vec{B} is not the same as \vec{A-B}. So, if \vec{A} - \vec{B} is parallel to an axis, what does that tell you about the components? If your having a problem with this, I suggest drawing a picture or going over vector calculus (search the web, look to your textbooks or ask your professor).