Appendix B in the back of the 170 text has a good description of how to graphically add vectors. Carefully drawing the vectors is important since any errors will add together and create an inaccurate resultant vector.

Graphically adding vectors:

• Establish the scale for the magnitude of the vector, such as 1 cm = 10 lbs. Sometimes the problem specifies the scale to use. If you have a choice, then try to pick a scale so that the vectors are as long as possible while still fitting in the available space. The longer the drawn vectors, the greater the accuracy.
• Establish the convention for stating direction, such as North or South, or in degrees. One convention for showing direction in degrees has due right (the positive direction of the x-axis) as 0°, with increasing angles going counter-clockwise. Straight up (the positive direction of the y-axis) would be 90°, left would be 180°, and down would be 270°. 360° would be back at the beginning.
• Draw the first vector with the proper length and direction.
• Draw the next vector with its starting point ("tail") at the ending point ("head") of the first vector, again with the proper length and direction.
• Continue with any additional vectors, placing the tail of the next vector at the head of the previous vector.
• After all of the vectors are drawn, then connect the tail of the first vector to the head of the last vector. This is the resultant vector. Convert its length to magnitude according to the scale and find its direction.

To subtract a vector from another vector, take the negative of the vector being subtracted and add it to the first vector. The negative of a vector has the same magnitude but the opposite direction. For example, the negative of 5 lb to the north would be 5 lb to the south, or 3 N at 30° would the the negative of 3 N at 210°.

Another way to add vectors is to break the vectors down into their x and y components and then add up the x components and the y components. Find the magnitude with Pythagorean's Theorem (c² = a² + b²). Find the angle of the resultant vector by taking the inverse tangent (arctangent) of the total y component divided by the total x component. This process involves trigonometry and requires a calculator in all but the easiest of problems (and how many are there of those?). This method can be used as a check after graphically adding the vectors to see how close the graphical result is to the actual values of magnitude and direction of the resultant vector. Anyone interested in this method can can continue on below. It is not, however, required for the class.

Let's start off with a fairly easy example. Vector A is 3 lb at 0° (0° is to the right). Vector B is 4 lb at 90° (straight up). The x component of vector A is 3 lb while the x component of vector B is 0 lb. The y component of vector A is 0 lb while the y component of vector B is 4 lb. The resultant vector therefore has an x component of 3 lb and a y component of 4 lb. The magnitude of the resultant vector (using the Pythagorean theorem) is 5 lb. Divide 4 lb by 3 lb and then take the arctangent to find the angle of the resultant vector, which is 53.1°.

Not all vectors are at such nice angles and more trigonometry is required. The x component of any vector is the cosine of the angle of the vector. The y component of the vector is the sine of the angle of the vector. Add all the x components together and all the y components together to find the resultant vector. Then do the same as the simple example and use the Pythagorean theorem to find the magnitude and use the arctangent of the y component divided by the x component to find the angle.

Be careful when finding the angle of the resultant vector. In the simple example above, the resultant vector had positive x and y components and the calculator returned the correct angle. But if the resultant vector had negative x and y components the calculator would still give 53.1°, which would be wrong in this case. The actual angle would be 53.1° + 180°, or 233.1°. If the resultant vector had one negative component and the other positive, then the arctangent would give -53.1°. The usual angle convention for vectors is 0° is to the right, and the angle increases going counterclockwise until it reaches 360° (which is back to where we started - pointing to the right). Add 360° to the -53.1° and we get 306.9°. This would be the correct angle only for a resultant vector with a positive x component and a negative y component. If, however, the situation was reversed and the x component was negative and the y component was positive, then the angle would have to be 180° out from 306.9°, or 126.9°.

Apparently it is easier sometimes to just draw the darn vectors and add them graphically.