1.6.  Vector Addition and Subtraction

ADDITION

Often it is necessary to add one vector to another, and the process of addition must take into account both the magnitude and the direction of the vectors. The simplest situation occurs when the vectors point along the same direction—that is, when they are colinear, as in Figure 1.9. Here, a car first moves along a straight line, with a displacement vector A of 275 m, due east. Then, the car moves again in the same direction, with a displacement vector B of 125 m, due east. These two vectors add to give the total displacement vector R, which would apply if the car had moved from start to finish in one step. The symbol R is used because the total vector is often called the resultant vector. With the tail of the second arrow located at the head of the first arrow, the two lengths simply add to give the length of the total displacement. This kind of vector addition is identical to the familiar addition of two scalar numbers (2+3=5) and can be carried out here only because the vectors point along the same direction. In such cases we add the individual magnitudes to get the magnitude of the total, knowing in advance what the direction must be. Formally, the addition is written as follows:

Figure 1.9    Two colinear displacement vectors A and B add to give the resultant displacement vector R.

Perpendicular vectors are frequently encountered, and Figure 1.10 indicates how they can be added. This figure applies to a car that first travels with a displacement vector A of 275 m, due east, and then with a displacement vector B of 125 m, due north. The two vectors add to give a resultant displacement vector R. Once again, the vectors to be added are arranged in a tail-to-head fashion, and the resultant vector points from the tail of the first to the head of the last vector added. The resultant displacement is given by the vector equation

Figure 1.10  The addition of two perpendicular displacement vectors A and B gives the resultant vector R.

The addition in this equation cannot be carried out by writing R=275 m+125 m, because the vectors have different directions. Instead, we take advantage of the fact that the triangle in Figure 1.10 is a right triangle and use the Pythagorean theorem (Equation 1.7). According to this theorem, the magnitude of R is

The angle q  in Figure 1.10 gives the direction of the resultant vector. Since the lengths of all three sides of the right triangle are now known, either sin q , cos q , or tan q  can be used to determine q . Noting that tan q =B/A and using the inverse trigonometric function, we find that:

Thus, the resultant displacement of the car has a magnitude of 302 m and points north of east at an angle of 24.4°. This displacement would bring the car from the start to the finish in Figure 1.10 in a single straight-line step.

When two vectors to be added are not perpendicular, the tail-to-head arrangement does not lead to a right triangle, and the Pythagorean theorem cannot be used. Figure 1.11a illustrates such a case for a car that moves with a displacement A of 275 m, due east, and then with a displacement B of 125 m in a direction 55.0° north of west. As usual, the resultant displacement vector R is directed from the tail of the first to the head of the last vector added. The vector addition is still given according to

Figure 1.11  (a) The two displacement vectors A and B are neither colinear nor perpendicular but even so they add to give the resultant vector R. (b) In one method for adding them together, a graphical technique is used.

However, the magnitude of R is not , because the vectors A and B are not perpendicular and the Pythagorean theorem does not apply. Some other means must be used to find the magnitude and direction of the resultant vector.

One approach uses a graphical technique. In this method, a diagram is constructed in which the arrows are drawn tail to head. The lengths of the vector arrows are drawn to scale, and the angles are drawn accurately (with a protractor, perhaps). Then, the length of the arrow representing the resultant vector is measured with a ruler. This length is converted to the magnitude of the resultant vector by using the scale factor with which the drawing is constructed. In Figure 1.11b, for example, a scale of one centimeter of arrow length for each 10.0 m of displacement is used, and it can be seen that the length of the arrow representing R is 22.8 cm. Since each centimeter corresponds to 10.0 m of displacement, the magnitude of R is 228 m. The angle q , which gives the direction of R, can be measured with a protractor to be q =26.7° north of east.

Two vectors, A and B, are added by means of vector addition to give a resultant vector R: R=A+B. The magnitudes of A and B are 3 and 8 m, but they can have any orientation. What is (a) the maximum possible value and (b) the minimum possible value for the magnitude of R? These questions deal with adding two vectors by means of the tail-to-head method. They illustrate how the two vectors must be oriented relative to each other to produce a resultant vector that has the greatest possible magnitude and one that has the least possible magnitude. For similar questions (including calculational counterparts), consult Self-Assessment Test 1.1. The test is described at the end of this section.

SUBTRACTION

The subtraction of one vector from another is carried out in a way that depends on the following fact. When a vector is multiplied by –1, the magnitude of the vector remains the same, but the direction of the vector is reversed. Conceptual Example 6 illustrates the meaning of this statement.

Consider the two vectors described as follows:    1. A woman climbs 1.2 m up a ladder, so that her displacement vector D is 1.2 m, upward along the ladder, as in Figure 1.12a. Figure 1.12  (a) The displacement vector for a woman climbing 1.2m up a ladder is D. (b) The displacement vector for a woman climbing 1.2m down a ladder is –D.    2. A man is pushing with 450 N of force on his stalled car, trying to move it eastward. The force vector F that he applies to the car is 450 N, due east, as in Figure 1.13a. Figure 1.13  (a) The force vector for a man pushing on a car with 450N of force in a direction due east is F. (b) The force vector for a man pushing on a car with 450N of force in a direction due west is –F. What are the physical meanings of the vectors –D and –F? Reasoning and Solution A displacement vector of –D is (–1)D and has the same magnitude as the vector D, but is opposite in direction. Thus, –D would represent the displacement of a woman climbing 1.2 m down the ladder, as in Figure 1.12b. Similarly, a force vector of –F has the same magnitude as the vector F but has the opposite direction. As a result, –F would represent a force of 450 N applied to the car in a direction of due west, as in Figure 1.13b. Related Homework: Conceptual Question 15 , Problem 62

In practice, vector subtraction is carried out exactly like vector addition, except that one of the vectors added is multiplied by a scalar factor of –1. To see why, look at the two vectors A and B in Figure 1.14a. These vectors add together to give a third vector C, according to C=A+B. Therefore, we can calculate vector A as A=C–B, which is an example of vector subtraction. However, we can also write this result as A=C+(–B) and treat it as vector addition. Figure 1.14b shows how to calculate vector A by adding the vectors C and –B. Notice that vectors C and –B are arranged tail to head and that any suitable method of vector addition can be employed to determine A.

Figure 1.14  (a) Vector addition according to C = A + B. (b) Vector subtraction according to A = C – B = C + (–B).

Test your understanding of the concepts discussed in Sections 1.5 and 1.6: · Adding and Subtracting Vectors by the Tail-to-Head Method  · Using the Pythagorean Theorem and Trigonometry to Find the Magnitude and Direction of the Resultant Vector

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