Example from Physics, 5.2 Vector Addition and Subtraction: Analytical Methods

Vector Addition and Subtraction: Analytical Methods: example

Add the vector $A$ to the vector $B$ shown in Figure 5.25, using the steps above. The x-axis is along the east–west direction, and the y-axis is along the north–south directions. A person first walks $53 .0 m$ in a direction $20 .0°$ north of east, represented by vector $A .$ The person then walks $34 .0 m$ in a direction $63 .0 °$ north of east, represented by vector $B .$

Vectors A, B, and R form a triangle with vertex RA at the origin of an x y-axis. The following labels are included: angle (where R and A form an angle) equals thirty-six point six degrees, angle A equals twenty degrees, angle B equals sixty-three degrees, A equals fifty-three meters, B equals thirty-four meters, and A plus B equals R. A compass is shown in the bottom corner for reference.

Figure 5.25 You can use analytical models to add vectors.

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. We will solve for these components and then add them in the x-direction and y-direction to find the resultant.
Solution
First, we find the components of $A$ and $B$ along the x- and y-axes. From the problem, we know that $A = 53.0 m,$ $θ_{A} = {20.0}^{\circ,}$ $B$ = $34 .0 m$ , and $θ_{B} = {63.0}^{\circ}$ . We find the x-components by using $A_{x} = A \cos θ$ , which gives

$\begin{matrix} A_{x} & = & A \cos θ_{A} = (53.0 m) (\cos {20.0}^{\circ}) \\ = & (53.0 m) (0.940) = 49.8 m \end{matrix}$
and

$\begin{matrix} B_{x} & = & B \cos θ_{B} = (34.0 m) (\cos {63.0}^{\circ}) \\ = & (34.0 m) (0.454) = 15.4 m. \end{matrix}$

Similarly, the y-components are found using $A_{y} = A \sin θ_{A}$

$\begin{matrix} A_{y} & = & A \sin θ_{A} = (53.0 m) (\sin {20.0}^{\circ}) \\ = & (53.0 m) (0.342) = 18.1 m \end{matrix}$

and

$\begin{matrix} B_{y} & = & B \sin θ_{B} = (34.0 m) (\sin {63.0}^{\circ}) \\ = & (34.0 m) (0.891) = 30.3 m . \end{matrix}$

The x- and y-components of the resultant are

$R_{x} = A_{x} + B_{x} = 49.8 m + 15.4 m = 65.2 m$

and

$R_{y} = A_{y} + B_{y} = 18.1 m + 30.3 m = 48.4 m .$

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

$R = \sqrt{R_{x}^{2} + R_{y}^{2}} = \sqrt{{(65.2)}^{2} + {(48.4)}^{2}} m$
5.8

so that

$R = \sqrt{6601 m} = 81.2 m .$

Finally, we find the direction of the resultant

$θ = \tan^{- 1} (R_{y} / R_{x}) = + \tan^{- 1} (48.4 / 65.2) .$

This is

$θ = \tan^{- 1} (0.742) = {36.6}^{\circ} .$

0 of 2 steps open

How did it go?

Strategy

Solution

Strategy

Solution