Adding Vectors Using Analytical Methods

Add the vector $A$ to the vector $B$ shown in Figure 3.33, using perpendicular components along the x- and y-axes. The x- and y-axes are along the east–west and north–south directions, respectively. Vector $A$ represents the first leg of a walk in which a person walks $53 . 0 m$ in a direction $20.0 º$ north of east. Vector $B$ represents the second leg, a displacement of $34 . 0 m$ in a direction $63.0 º$ north of east.

Two vectors A and B are shown. The tail of the vector A is at origin. Both the vectors are in the first quadrant. Vector A is of magnitude fifty three units and is inclined at an angle of twenty degrees to the horizontal. From the head of the vector A another vector B of magnitude 34 units is drawn and is inclined at angle sixty three degrees with the horizontal. The resultant of two vectors is drawn from the tail of the vector A to the head of the vector B.

Figure 3.33 Vector

A

has magnitude

53 . 0 m

and direction

20.0 º

north of the x-axis. Vector

B

has magnitude

34 . 0 m

and direction

63.0 º

north of the x-axis. You can use analytical methods to determine the magnitude and direction of

R

Work it out on paper first. Then open the solution one step at a time, and stop as soon as you can finish on your own.

Strategy
The components of $A$ and $B$ along the x- and y-axes represent walking due east and due north to get to the same ending point. Once found, they are combined to produce the resultant.
Solution
Following the method outlined above, we first find the components of $A$ and $B$ along the x- and y-axes. Note that $A = 53.0 m$ , $θ_{A} = 20.0º$ , $B = 34.0 m$ , and $θ_{B} = 63.0º$ . We find the x-components by using $A_{x} = A cos θ$ , which gives
$\begin{array}{l} A_{x} & = & A cos θ_{A} = (53. 0 m)(cos 20.0º) \\ = & (53. 0 m) (0 .940) = 49. 8 m \end{array}$
3.16

and

$\begin{array}{l} B_{x} & = & B cos θ_{B} = (34 . 0 m)(cos 63.0º) \\ = & (34 . 0 m)(0.454) = 15.4 m. \end{array}$
3.17

Similarly, the y-components are found using $A_{y} = A sin θ_{A}$ :
$\begin{array}{l} A_{y} & = & A sin θ_{A} = (53 . 0 m)(sin 20.0º) \\ = & (53 . 0 m)(0.342) = 18.1 m \end{array}$
3.18

and

$\begin{array}{l} B_{y} & = & B sin θ_{B} = (34 . 0 m)(sin 63.0º) \\ = & (34 . 0 m)(0.891) = 30.3 m. \end{array}$
3.19

The x- and y-components of the resultant are thus

$R_{x} = A_{x} + B_{x} = 49 . 8 m +15 . 4 m =65 . 2 m$
3.20

and

$R_{y} = A_{y} + B_{y} = 18 . 1 m + 30 . 3 m = 48 . 4 m .$
3.21

Now we can find the magnitude of the resultant by using the Pythagorean theorem:

$R = \sqrt{R_{x}^{2} + R_{y}^{2}} = \sqrt{(65.2)^{2} + (48.4)^{2} m}$
3.22

so that

$R = 81.2 m.$
3.23

Finally, we find the direction of the resultant:

$θ = {tan}^{- 1} (R_{y} / R_{x}) =+ {tan}^{- 1} (48 . 4 / 65 . 2) .$
3.24

Thus,

$θ = {tan}^{- 1} (0.742) = 36 . 6 º .$
3.25

Figure 3.34 Using analytical methods, we see that the magnitude of $R$ is $81 . 2 m$ and its direction is $36 . 6º$ north of east.
Discussion
This example illustrates the addition of vectors using perpendicular components. Vector subtraction using perpendicular components is very similar—it is just the addition of a negative vector.

Subtraction of vectors is accomplished by the addition of a negative vector. That is, $A - B \equiv A + (–B)$ . Thus, the method for the subtraction of vectors using perpendicular components is identical to that for addition. The components of $–B$ are the negatives of the components of $B$ . The x- and y-components of the resultant $A - B = R$ are thus
$R_{x} = A_{x} + (- B_{x})$
3.26

and

$R_{y} = A_{y} + (- B_{y})$
3.27

and the rest of the method outlined above is identical to that for addition. (See Figure 3.35.)

0 of 3 steps open

How did it go?

Strategy

Solution

Discussion