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## Q. 1.7

Adding two vectors Now we’ll use the component method to add two vectors. We will use this technique extensively (in Chapter 4) when we begin to consider how forces act upon an object. Vector $\overset{\rightarrow }{\pmb{A}}$ has a magnitude of 50 cm and a direction of 30°, and vector $\overset{\rightarrow}{\pmb{B}}$ has a magnitude of 35 cm and a direction of 110°. Both angles are measured counterclockwise from the positive x axis. Use components to calculate the magnitude and direction of the vector sum (i.e., the resultant) $\overset{\rightarrow }{\pmb{R}}= \overset{\rightarrow }{\pmb{A}} + \overset{\rightarrow }{\pmb{B}}$.

## Verified Solution

SET UP We set up a rectangular (Cartesian) coordinate system and place the tail of each vector at the origin, as shown in Figure 1.20a. We label the known magnitudes and angles on our diagram.

SOLVE We find the x and y components of each vector, using Equations 1.2, and record the results in the table. We add the x components to obtain the x component of the vector sum $\overset{\rightarrow }{\pmb{R}}$ and we add the y components to obtain the y component of $\overset{\rightarrow }{\pmb{R}}$. (But be very careful not to add the x and y components together!) Here is the resulting table:

 Magnitude Angle x component y component A = 50 cm 30° 43.3 cm 25.0 cm B = 35 cm 110° -12.0 cm 32.9 cm $R_x = 31.3 cm$ $R_y = 57.9 cm$

$A_x=A\cos \theta and A_y =A\sin \theta .$          (1.2)

Now we use Equations 1.3 and 1.4 to find the magnitude and direction of the vector sum $\overset{\rightarrow }{\pmb{R}}$. From the Pythagorean theorem,

$A=\sqrt{A^2_x+A_y^2}$       (1.3)

$\tan \theta =\frac{A_y}{A_x} and \theta =\tan^{-1}\frac{A_y}{A_x}$    (direction of vector$\overset{\rightarrow }{\pmb{A}}$)             (1.4)

$R=\sqrt{R^2_x+R_y^2}=\sqrt{(31.3 cm)^2+(59.9 cm)^2}=66 cm.$

From the definition of the inverse tangent, we have

$\theta =\tan^{-1}\frac{R_y}{R_x}=\tan^{-1}\frac{57.9 cm}{31.3 cm}=62° .$

Figure 1.20b shows this resultant vector and its components.

REFLECT We used the components of the vector sum $\overset{\rightarrow }{\pmb{R}}$ to find its magnitude and direction. Note how important it was to recognize that the x component of $\overset{\rightarrow }{\pmb{B}}$ is negative.

Practice Problem: Find the vector sum of these two vectors by using components: 45 m, 55°; 70 m, 135°. Answers: magnitude = 90 m; direction = 105° 