parallel/perpendicular vectors...

[amberlianne]

I've already just skipped this problem and moved on in the homework, but I'd like to learn it for future reference:

A block weighing 11 lb rests on an inclined plane, as indicated in the figure. Determine the components of the weight perpendicular to and parallel to the inclined plane. Round your answers to two decimal places.

I tried solving for the sides of the triangle created by the 11lb downward vector, and got a hypotenuse of 17.87 and x-axis of 13.96, but those are nowhere near the answer choices given.

I'm assuming I don't know what the question wants when it asks for parallel and perpendicular vectors. I thought that parallel vectors were always the same, therefore at least the hypotenuse one should be right, but no such luck. None of the examples in the text address this type of problem (the problem sets are online and vary from the material in the text to differing degrees).

Would someone be so kind as to explain to me what this problem is asking me to do?

Thanks.[/img]

 

[amberlianne]

Sorry for the double-post; the edit function isn't working for me right now --

I don't want an answer to the problem per se, I just want clarification of the problem so I can figure it out myself. Thanks!

 

[soroban]

Hello, amberlianne!

You may be drawing the wrong right triangle . . .

A block weighing 11 lb rests on an inclined plane, as indicated in the figure.
Determine the components of the weight perpendicular to and parallel to
the inclined plane. .Round your answers to two decimal places.

                                *
                              * |
                            *   |
                        A *     |
                        *       |
                      * |*      |
                    *   | *     |
                  *     |  *    |
                *       |38°*   |
              *       11|    *  |
            *           |     *C|
          *             |   *   |
        *               | *     |
      *                 *       |
    *                   B       |
  * 38°                         |
* - - - - - - - - - - - - - - - *

The block is at \(\displaystyle A.\)
Its weight is line segment \(\displaystyle AB\,=\,11\)

The component of the weight perpendicular to the plane is: \(\displaystyle AC\,=\,11\cdot\cos38^o\)

The component of the weight parallel to the plane is: \(\displaystyle CB\,=\,11\cdot\sin38^o\)

 

[amberlianne]

Wow. I was totally on the wrong track. Thank you.

How can I apply this to future problems, e.g. how do I know where/how to draw the new triangle?