Question

$\overrightarrow{V}$is a vector 24.8 units in magnitude and points at an angle of $23.4^o$above the negative X-axis. (a) Sketch this vector. (b) Calculate$V_x$and$V_y$. Use$V_x$and$V_y$obtain (again) the magnitude and direction of$\overrightarrow{V}$. [Note: Part (c) is a good way to check if you’ve resolved your vector correctly.]

Short answer

(b) $V_{\mathrm{x}}$is –22.76 units and $V_{\mathrm{y}}$is 9.85 units.

(c) The magnitude of $\overrightarrow{V}$is 24.8 units, and $\overrightarrow{V}$is oriented at an angle of $23.4°$with the negative X-axis.

Step-by-step solution

Step 1. Understanding the rectangular components of a vector

Rectangular components of a vector are components in the direction of the sides of a rectangle when the vector itself represents the diagonal to the rectangle.

Step 2. Given data and assumption

The magnitude of vector $\overrightarrow{V}$,$V=24.8\mathrm{units}$

Angle $\overrightarrow{V}$makes with negative direction of X-axis,$\theta =23.4°$

Step 3. Sketching vector V→

Vector $\overrightarrow{V}$can be graphically represented, as shown below (at an angle $\theta =23.4°$, above the negative X-axis and having a length of 24.8 units.)

Step 4. Finding rectangular components

The component of a vector along any direction can be determined by multiplying the magnitude of the vector with the cosine of the angle between vector and direction, along which we need to find the component.

Thus, the component of $\overrightarrow{V}$along the X-direction is

$\begin{array}{c}{V}_{\mathrm{x}}=-V\cos \theta \\ =-24.8\cos 23.4°\\ =-22.76\mathrm{units}.\end{array}$

($V_{\mathrm{x}}$is negative from the graph of $\overrightarrow{V}$)

The component of $\overrightarrow{V}$along the Y-direction is

$\begin{array}{c}{V}_{\mathrm{y}}=V\cos \left(90°-\theta \right)\\ =24.8\cos \left(90°-23.4°\right)\\ =9.85\mathrm{units}.\end{array}$

Thus, the values of $V_{\mathrm{x}}$and $V_{\mathrm{y}}$are $-22.76$units and 9.85 units, respectively.

Step 5. Calculating orientation with respect to negative the X-axis 

The tangent function of angle $\theta$is $\tan \theta =\left|\frac{V_{\mathrm{y}}}{V_{\mathrm{x}}}\right|$.

Substituting the values,

$\begin{array}{c}\theta ={\tan }^{-1}\left|-\frac{9.85}{22.76}\right|\\ =23.4°.\end{array}$

Thus, vector $\overrightarrow{V}$is oriented at an angle of $23.4°$with respect to the negative direction of the X-axis.

The magnitude of $\overrightarrow{V}$is the square root of the sum of squares of its rectangular components. Thus,

$\begin{array}{c}V=\sqrt{V_{\mathrm{x}}^2+V_{\mathrm{y}}^2}\\ =\sqrt{{22.76}^2+{9.85}^2}\\ =24.8\mathrm{units}\end{array}$

Thus, the magnitude of $\overrightarrow{V}$is equal to 24.8 units.

The magnitude concludes that the resolutions are correct as they give the same value as provided in the question.