Question

How do I find the magnitude and direction of the resultant vector of two vectors, 5 Newton and 2 Newton, inclined at an angle of 60 degrees to each other?

Answer 1

To find the magnitude and direction of the resultant vector of two vectors, you can use the following steps:

1. **Magnitude Calculation**: The resultant \( R \) of two vectors \( A \) and \( B \) that are inclined at an angle \( \theta \) can be calculated using the formula:
\[
R = \sqrt{A^2 + B^2 + 2AB \cos(\theta)}
\]
where:
- \( A = 5 \, \text{N} \)
- \( B = 2 \, \text{N} \)
- \( \theta = 60^\circ \) (convert to radians if necessary for calculation, but many calculators can handle degrees)

Plugging in the values:
\[
R = \sqrt{5^2 + 2^2 + 2 \cdot 5 \cdot 2 \cdot \cos(60^\circ)}
\]
We know \( \cos(60^\circ) = 0.5 \), thus:
\[
R = \sqrt{25 + 4 + 2 \cdot 5 \cdot 2 \cdot 0.5}
\]
\[
R = \sqrt{25 + 4 + 10}
\]
\[
R = \sqrt{39}
\]
\[
R \approx 6.24 \, \text{N}
\]

2. **Direction Calculation**: The direction \( \phi \) of the resultant vector can be found using the formula:
\[
\tan(\phi) = \frac{B \sin(\theta)}{A + B \cos(\theta)}
\]
Substituting the values:
\[
\tan(\phi) = \frac{2 \sin(60^\circ)}{5 + 2 \cos(60^\circ)}
\]
\( \sin(60^\circ) = \frac{\sqrt{3}}{2} \) and \( \cos(60^\circ) = 0.5 \), so:
\[
\tan(\phi) = \frac{2 \cdot \frac{\sqrt{3}}{2}}{5 + 2 \cdot 0.5}
\]
\[
= \frac{\sqrt{3}}{5 + 1} = \frac{\sqrt{3}}{6}
\]
Now, to find \( \phi \):
\[
\phi = \tan^{-1}\left(\frac{\sqrt{3}}{6}\right)
\]

You can calculate \( \tan^{-1}\left(\frac{\sqrt{3}}{6}\right) \) using a calculator to find the angle in degrees.

In summary:
- The magnitude of the resultant vector \( R \) is approximately **6.24 Newtons**.
- The direction \( \phi \) can be calculated as \( \tan^{-1}\left(\frac{\sqrt{3}}{6}\right) \).