Three-Dimensional Vectors

Hi, Reema.

This is a tough one to visualize! You need to break it down into
several diagrams - projections of the vectors into different planes.
For instance, imagine looking directly down on the stump from a great
height. What you see won't involve the up/down dimension (which I'd
call z) at all; you'll only see the x and y components of the vectors
(which are parallel to the ropes). We have projected the vectors into
the x-y plane.

I would first draw the projection in the x-y plane, and label each
vector with the magnitude and the angle from the horizontal. I can't
really draw the lengths of these vectors in this projection yet; the
magnitudes of the vectors are not the lengths of their projections in
the x-y plane. But at least I can get the angles (60 degrees
clockwise, etc.), and I can write down all the information. You'll
want to pick a direction for the first vector, say, along the positive
x-axis. This establishes the coordinate system.

Now you'll need a separate projection for each force. You will be
considering the vertical plane that contains the vector. For instance,
for the second vector (155 N, 56 degrees to the horizontal), you'll
draw a right triangle:

*
/|
/ |
155 N/ |z
/ |
/ |
/56 |
*------+
r

From this diagram you can find the z component of the force vector,
and what I labeled as the r component of the vector. What is this
component? It is the length of the vector in the x-y projection that
I drew first. I can go back to that projection and label the length of
the vector. From this (and the angle that I indicated in that diagram)
I can determine the x and y components of the vector.

Once you have the x, y, and z components of each vector, you can add
them component-wise. You're asked for the magnitude and the z
component of the resultant force.

If you had been asked for the direction (angle clockwise or
anticlockwise from the first person, and angle from the horizontal),
you'd want to work backward through the process by which we obtained
the components. In the x-y projection, you can find the angle and the
"r component" (that's cylindrical coordinate notation, by the way).
Then transfer the r component to a new vertical-plane diagram for this
vector, and use the r and z components to find the angle from the
horizontal.

I hope this clarifies a hard-to-visualize problem. The usual trick in
working with three dimensions, as I have shown, is to find a way to
work in two dimensions at a time, choosing a plane that contains
information you can use. Let me know if you'd like to discuss it
further.